Learning goals and preparation

We assume derivatives, the chain and product rules, elementary antiderivatives, and definite integrals. This chapter treats a first-order ordinary differential equation as a rule for change. You will solve separable and linear equations, retain equilibrium solutions that division can hide, and calculate Euler approximations while distinguishing numerical stability from exact dynamics.

A differential equation connects an unknown function to its derivative. An initial-value problem additionally specifies the function at a starting time. A solution must satisfy the equation and initial condition on a stated interval; merely obtaining a formula after integration is not enough.

Why the real numbers need more than arithmetic

The rational numbers support the familiar ordered arithmetic, yet the set $\{q\in\mathbb Q:q>0,\ q^2<2\}$ has no rational least upper bound. The real number system adds a completeness property: bounded nonempty sets have least upper bounds. This is the source of many existence results later in analysis, not simply another rule for manipulating inequalities.

Definition

Upper bound and supremum

For a nonempty $A\subset\mathbb R$, $u$ is an upper bound if $a\le u$ for every $a\in A$. A number $s$ is $\sup A$ if it is an upper bound and $s\le u$ for every upper bound $u$. A maximum is an element of $A$ that is also an upper bound. The maximum, when it exists, is the supremum; a supremum need not belong to $A$. Lower bounds, infimum and minimum are defined by reversing inequalities.

Theorem

Completeness axiom

Every nonempty subset of $\mathbb R$ that is bounded above has a real supremum. Equivalently, every nonempty subset bounded below has a real infimum, via $\inf A=-\sup(-A)$.

For example, $\sup(0,1)=1$ although $1$ is not a maximum of that set. Conversely, $\sup[0,1]=1$ and it is a maximum. The empty set and an unbounded-above set do not have a real supremum under this definition.

Theorem

Supremum approximation criterion

An upper bound $s$ of a nonempty set $A$ is $\sup A$ if and only if, for every $\varepsilon>0$, there exists $a\in A$ such that $s-\varepsilon<a\le s$.

Proof

Proving both directions

If $s=\sup A$ and no element exceeds $s-\varepsilon$, then $s-\varepsilon$ is a smaller upper bound, contradiction. Conversely, suppose $s$ is an upper bound and the approximation property holds. If another upper bound $u$ were smaller than $s$, choose $\varepsilon=s-u>0$. The property gives $a>u$, contradicting that $u$ is an upper bound. Thus $s$ is the least upper bound.

Worked example

Use both halves of a supremum proof

Let $A=\{n/(n+1):n\in\mathbb N, n\ge1\}$. Every element is less than $1$, so $1$ is an upper bound. Given $\varepsilon>0$, choose $n>1/\varepsilon$ by the Archimedean property below. Then $1-n/(n+1)=1/(n+1)<\varepsilon$. Thus $\sup A=1$. No element equals $1$, so $A$ has no maximum. An upper-bound calculation alone would not prove leastness: $2$ is also an upper bound.

Archimedean property and density

Theorem

Archimedean property

For every real $x$ there exists a positive integer $n>x$. Consequently, for every $\varepsilon>0$ there is $n\in\mathbb N$ such that $1/n<\varepsilon$.

Proof

Completeness forces integers to be unbounded

If the positive integers were bounded above, completeness would give $s=\sup\mathbb N$. Since $s-1$ is not an upper bound, some $n\in\mathbb N$ satisfies $n>s-1$. But $n+1>s$ and $n+1\in\mathbb N$, contradicting that $s$ bounds the set. Apply this to $1/\varepsilon$ to obtain the second statement.

Between any real numbers $x<y$ there is a rational number. Indeed choose $n$ with $n(y-x)>1$; the Archimedean property and the integer-part property yield an integer $m$ with $nx<m<ny$, so $x<m/n<y$. The integer-part property follows from the unboundedness of integers in both directions and the well-ordering of the natural numbers. An irrational also lies between $x$ and $y$: choose a rational $q$ with $x-\sqrt2<q<y-\sqrt2$; then $q+\sqrt2$ is irrational and lies in the interval. Thus density does not say rationals and irrationals are equal sets.

Worked example

An infimum via reflection

For $B=\{2+1/n:n\ge1\}$, every element exceeds $2$. Given $\varepsilon>0$, choose $n>1/\varepsilon$ to obtain $2<2+1/n<2+\varepsilon$. Therefore $2$ is the infimum, but not a minimum. The maximum is $3$, attained at $n=1$.

Quantifiers in convergence

Definition

Convergence of a real sequence

A sequence $(x_n)$ converges to $L\in\mathbb R$ if for every $\varepsilon>0$ there exists $N\in\mathbb N$ such that for every $n\ge N$, $|x_n-L|<\varepsilon$. Write $x_n\to L$. The index $N$ may depend on $\varepsilon$ but must work for all later indices.

To negate convergence to $L$, there must be some $\varepsilon_0>0$ such that for every $N$ there is an $n\ge N$ with $|x_n-L|\ge\varepsilon_0$. A few early terms do not affect convergence; a recurring failure arbitrarily late does.

Worked example

Construct an epsilon–N proof

For $x_n=(3n+1)/(n+2)$, guess $L=3$. Since

$$\left|\frac{3n+1}{n+2}-3\right|=\frac5{n+2}<\frac5n,$$

choose $N>5/\varepsilon$. Then every $n\ge N$ gives $|x_n-3|<\varepsilon$. The guess is motivation; the inequality and quantified choice establish the proof.

Theorem

Uniqueness, boundedness and limit laws

A convergent real sequence has a unique limit and is bounded. If $x_n\to x$ and $y_n\to y$, then $x_n+y_n\to x+y$, $x_ny_n\to xy$, and $x_n/y_n\to x/y$ when $y\ne0$ and the quotient is defined eventually. If $x_n\le z_n\le y_n$ eventually and both outer sequences converge to $L$, then $z_n\to L$.

Proof

Uniqueness and boundedness

If $x_n\to L$ and $x_n\to M$, then for any $\varepsilon>0$, sufficiently late terms satisfy both $|x_n-L|<\varepsilon/2$ and $|x_n-M|<\varepsilon/2$. The triangle inequality yields $|L-M|<\varepsilon$. Since this holds for every positive $\varepsilon$, $L=M$. For boundedness choose $N$ so that $|x_n-L|<1$ for $n\ge N$. The finite initial terms and the bound $|x_n|\le|L|+1$ for the tail together give a finite bound for the whole sequence.

The converse of boundedness is false: $(-1)^n$ is bounded but does not converge. If it converged to $L$, its even subsequence would converge to $1$ and its odd subsequence to $-1$, violating uniqueness.

Proof

Product law mechanism

A convergent sequence $(y_n)$ is bounded, say $|y_n|\le M$. Write

$$|x_ny_n-xy|\le|y_n||x_n-x|+|x||y_n-y|.$$

Both terms tend to zero. For the quotient, $y_n\to y\ne0$ implies $|y_n|\ge|y|/2$ eventually; this keeps the denominator away from zero and gives $1/y_n\to1/y$.

Subsequences

Definition

Subsequence

A subsequence of $(x_n)$ is $(x_{n_k})$ where $n_1<n_2<\cdots$ are positive integers. If $x_n\to L$, then every subsequence converges to $L$ because $n_k\ge k$ and therefore eventually lies beyond any prescribed index.

Theorem

Bolzano–Weierstrass

Every bounded real sequence has a convergent subsequence.

Proof

Nested-interval proof

Place every term in a closed bounded interval $I_1$. Bisect it. At least one half contains infinitely many terms; choose that half as $I_2$. Continue, obtaining nested closed intervals $I_k$ of lengths tending to zero, each containing infinitely many terms. Choose indices $n_k$ strictly increasing with $x_{n_k}\in I_k$; this is possible because each interval contains infinitely many terms. Completeness gives a common point $L$ in all intervals: the supremum of their left endpoints lies in each interval because every left endpoint is at most every right endpoint. Since $x_{n_k},L\in I_k$, their distance is at most the interval length and tends to zero.

The conclusion is about a subsequence, not necessarily the whole sequence. For example, $(-1)^n$ has constant subsequences converging to $1$ and $-1$, but does not converge.

Cauchy criterion

Definition

Cauchy sequence

A sequence $(x_n)$ is Cauchy if for every $\varepsilon>0$ there exists $N$ such that $|x_m-x_n|<\varepsilon$ whenever $m,n\ge N$. Unlike the limit definition, no candidate limit is named.

Theorem

Cauchy completeness of $\mathbb R$

A real sequence converges if and only if it is Cauchy.

Proof

Both directions

If $x_n\to L$, choose a tail within $\varepsilon/2$ of $L$ and apply the triangle inequality to $x_m-L$ and $L-x_n$. Conversely, a Cauchy sequence is bounded: choose a tail within $1$ of one fixed tail term, then include the finite initial terms. By Bolzano–Weierstrass it has a subsequence $x_{n_k}\to L$. Given $\varepsilon>0$, choose $N$ so all pairs of terms beyond $N$ differ by less than $\varepsilon/2$, and choose $k$ with $n_k\ge N$ and $|x_{n_k}-L|<\varepsilon/2$. For every $n\ge N$, $|x_n-L|\le|x_n-x_{n_k}|+|x_{n_k}-L|<\varepsilon$.

This equivalence uses completeness of $\mathbb R$. It fails as a statement that every rational Cauchy sequence converges to a rational number. For instance, decimal truncations of $\sqrt2$ are rational and Cauchy but have no rational limit.

Worked example

Convergence without knowing the sum

For $s_n=\sum_{k=1}^n2^{-k}$ and $m>n$,

$$|s_m-s_n|\le\sum_{k=n+1}^\infty2^{-k}=2^{-n}.$$

Given $\varepsilon>0$, choose $N$ with $2^{-N}<\varepsilon$. If $m,n\ge N$, the tail between them is at most $2^{-N}$, so $(s_n)$ is Cauchy and converges. The exact sum is $1$, but the Cauchy argument establishes existence without guessing it. The geometric-tail identity can also be verified by a finite geometric sum and taking the elementary limit $2^{-j}\to0$.

Real closed sets: the sequence criterion and the open complement

Work on the real line in the usual classical set-theoretic setting, with countable witness selections available. A set $U$ is open when every $y$ in $U$ has some $r>0$ with $(y-r,y+r)$ contained in $U$. We use the Archimedean consequence that $1/n$ tends to zero. For every $K$ subset of the real line, the following conditions are equivalent:

1. Every sequence of points of $K$ that converges in the real line has its limit in $K$.
2. The complement of $K$ is open.

Proof of 2 implies 1. Suppose $x_n$ belongs to $K$ and $x_n$ tends to $x$. If $x$ were outside $K$, openness would give $r>0$ with $(x-r,x+r)$ disjoint from $K$. Convergence gives a threshold after which $|x_n-x|<r$, contradicting $x_n$ in $K$. Therefore $x$ belongs to $K$.

Proof of 1 implies 2. Fix $y$ outside $K$. If no positive-radius interval around $y$ avoided $K$, then for every positive integer $n$ there would be a point $x_n$ in $K$ with $|x_n-y|<1/n$. Choose these witnesses as a sequence. Given $\varepsilon>0$, choose $N$ with $1/N<\varepsilon$. For $n\ge N$, $|x_n-y|<1/n\le1/N<\varepsilon$, so $x_n$ tends to $y$. Condition 1 would force $y$ into $K$, a contradiction. Hence some interval around $y$ avoids $K$. This holds for every point of the complement, proving it open. The witness-sequence step uses the stated classical setting; this proof is not offered as a theorem about choice-free foundations. If $K$ is empty there is no sequence in $K$, and its complement is the whole open line. If $K$ is the whole line, its empty complement is open vacuously. Both edge cases also satisfy the equivalence.

Learning goals and definitions

A set $K\subseteq\mathbb R$ is compact if every family of open sets whose union contains $K$ has a finite subfamily still covering $K$. A set is closed if it contains all limits of convergent sequences of its points, equivalently if its complement is open. Sequential compactness means every sequence in $K$ has a subsequence converging to a member of $K$.

The empty set is compact, closed and bounded, and sequentially compact vacuously. In proofs using a selected point or an infinite sequence, treat that case separately. We may assume the real completeness property and its nested-interval and Bolzano–Weierstrass consequences from the earlier sequence units.

Theorem: Heine–Borel in the real line

For $K\subseteq\mathbb R$, the following are equivalent: compactness by open covers; closedness and boundedness; sequential compactness.

Compact implies closed and bounded. The intervals $(-n,n)$, $n\ge1$, cover $\mathbb R$, hence $K$. A finite subcover gives a bound. To show closedness, take $y\notin K$. For each $x\in K$, the interval $B(x,|x-y|/2)$ avoids $y$. A finite subfamily covers $K$. Let $\delta$ be the minimum of their positive half-distances to $y$. Every point in the chosen interval centered at $x$ has distance from $y$ strictly greater than $|x-y|/2\ge\delta$. Thus $B(y,\delta)$ misses $K$, showing its complement open. If $K$ is empty this is immediate.

Closed and bounded implies compact. Enclose nonempty $K$ in a closed bounded interval $I_0$. Suppose an open cover of $K$ has no finite subcover. Bisect $I_0$. At least one closed half meets $K$ in a set not finitely coverable; otherwise two finite collections would cover $K$. Choose such a half and repeat. We obtain nested closed intervals $I_n$ with lengths tending to zero, and $K\cap I_n$ not finitely coverable. Each intersection is nonempty; choose $x_n\in K\cap I_n$. The nested-interval property gives a unique common point $p$, and $x_n\to p$ because both lie in $I_n$. Closedness gives $p\in K$. Some member $O$ of the cover contains $p$, and openness gives a ball around $p$ inside $O$. For large $n$, the whole interval $I_n$ lies in that ball, so $O$ alone covers $K\cap I_n$, a contradiction.

Closed and bounded implies sequentially compact. Every sequence in $K$ is bounded, so Bolzano–Weierstrass supplies a convergent subsequence. Closedness retains its limit in $K$.

Sequentially compact implies closed and bounded. If unbounded, choose $x_n\in K$ with $|x_n|>n$; no subsequence can converge to a finite real number, a contradiction. If $p$ is a limit of points in $K$, choose $x_n\in K$ with $|x_n-p|<1/n$. A subsequence converges to a point of $K$ by sequential compactness, but it also converges to $p$; uniqueness of real limits gives $p\in K$. Thus $K$ is closed. This completes all implications.

Compact subsets of the real line

A set $K\subset\mathbb R$ is compact if every open cover has a finite subcover. In $\mathbb R$, Heine–Borel says this is equivalent to being closed and bounded; it is also equivalent to every sequence in $K$ having a subsequence converging to a point of $K$. The closedness requirement keeps the subsequential limit in the set; boundedness permits extraction. We use the sequential form in the proofs below.

Theorem

Extreme value theorem

If $K\subset\mathbb R$ is nonempty and compact and $f:K\to\mathbb R$ is continuous, then $f$ is bounded and attains both a maximum and a minimum on $K$.

Proof

Extract a maximizing subsequence

If $f$ were unbounded above, choose $x_n\in K$ with $f(x_n)>n$. A subsequence $x_{n_k}\to x\in K$ exists; continuity would give $f(x_{n_k})\to f(x)$, contradicting divergence to infinity. The same reasoning applied to $-f$ gives boundedness below. Let $s=\sup f(K)$. Choose $x_n\in K$ with $s-1/n<f(x_n)\le s$. A convergent subsequence tends to some $x\in K$, and continuity implies $f(x)=s$. Apply the argument to $-f$ for a minimum.

Worked example

Which hypothesis fails?

The continuous function $f(x)=x$ on $(0,1)$ is bounded but has no maximum or minimum: $0$ and $1$ are not in the domain. The set is bounded but not closed, hence not compact. On $[0,1]$, both endpoints are present and the extrema are attained. Compactness is a sufficient global hypothesis, not a condition required for every individual function to attain extrema.

Fix an integer $n\ge1$ and the Euclidean norm on $\mathbb R^n$. A set is bounded if it lies in some finite-radius ball, closed if its complement is open, and compact if every cover by open subsets of $\mathbb R^n$ has a finite subcover. We use real completeness, including the nested closed interval property.

Heine–Borel in finite-dimensional Euclidean space

Theorem. A set $K\subseteq\mathbb R^n$ is compact if and only if it is closed and bounded. The empty set satisfies all three properties.

Compact implies bounded. The open balls $B(0,m)$, $m\ge1$ an integer, cover $K$. A finite subcover puts $K$ inside the largest of these balls.

Compact implies closed. Suppose $K\ne\varnothing$ and $y\notin K$. The balls $B(x,\|x-y\|/2)$, $x\in K$, cover $K$. Choose a finite subcover with centers $x_1,\ldots,x_r$ and set $\delta=\min_i\|x_i-y\|/2>0$. If $z$ lies in the ball centered at $x_i$, the triangle inequality gives $\|z-y\|>\|x_i-y\|/2\ge\delta$. Hence $B(y,\delta)$ misses $K$, so the complement is open.

Closed and bounded implies compact. Let nonempty $K$ be closed and bounded. Enclose it in a closed cube $Q_0=[-M,M]^n$ with $M>0$. Suppose an open cover of $K$ has no finite subcover. Divide each coordinate interval into two closed halves. The resulting $2^n$ closed subcubes cover $Q_0$. At least one subcube has intersection with $K$ that is not finitely coverable by the given family; otherwise the finite union of the finite subcovers would cover $K$. Select such a subcube and repeat. This produces nested closed cubes $Q_j$ with side length $2M/2^j$ and $K\cap Q_j$ not finitely coverable, hence nonempty.

For each coordinate, the nested interval property gives a common coordinate value. Together these give $p\in\bigcap_jQ_j$. Every pair of points in $Q_j$ has distance at most $2M\sqrt n/2^j$, so the intersection contains only $p$. Choose $x_j\in K\cap Q_j$. Then $\|x_j-p\|\le2M\sqrt n/2^j\to0$. A closed set contains such a limit: if $p$ were in its open complement, a ball around $p$ would eventually contain $x_j$ outside $K$, a contradiction. Thus $p\in K$.

One member $O$ of the cover contains $p$, and openness gives $B(p,\varepsilon)\subseteq O$ for some $\varepsilon>0$. For sufficiently large $j$, every point of $Q_j$ is within $\varepsilon$ of $p$, so $O$ alone covers $K\cap Q_j$. This contradicts the selection of $Q_j$. Therefore every open cover has a finite subcover. The number $2^n$ is finite; replacing it by infinitely many subpieces would invalidate this argument.

Continuous images and extreme values

Theorem. If $K\subseteq\mathbb R^n$ is compact and $F:K\to\mathbb R^m$ is continuous relative to $K$, then $F(K)$ is compact. In particular, a continuous $f:K\to\mathbb R$ on nonempty compact $K$ is bounded and attains its maximum and minimum.

Proof. An open cover $(O_i)$ of $F(K)$ pulls back to relatively open sets $F^{-1}(O_i)=K\cap U_i$, with $U_i$ open in $\mathbb R^n$. These $U_i$ cover $K$, so finitely many suffice. Their corresponding $O_i$ cover $F(K)$, proving compactness. For scalar $f$, the image is nonempty, closed and bounded by the theorem already proved in dimension one. Let $a=\sup f(K)$. For each $j\ge1$, the definition of supremum supplies $y_j\in f(K)$ with $a-1/j<y_j\le a$. Then $y_j\to a$, and closedness places $a$ in the image. Thus the maximum is attained. Applying the same argument to $-f$ gives the minimum. This proves existence, not a formula for the maximizing input.

Fermat's necessary condition. At a differentiable interior local maximum or minimum, $f'(a)=0$.

Proof. At a local maximum, sufficiently small positive increments give nonpositive difference quotients and negative increments give nonnegative ones. A two-sided derivative can satisfy both sign limits only by being zero. Negating the function treats a minimum. $\square$

The interior hypothesis matters: $f(x)=x$ on $[0,1]$ has extrema at endpoints despite derivative 1. The converse fails: $x^3$ has zero derivative at zero but no extremum there.

Rolle's theorem. If $a<b$, $f$ is continuous on $[a,b]$, differentiable on $(a,b)$ and $f(a)=f(b)$, then $f'(c)=0$ for some $c\in(a,b)$.

Proof. The extreme-value theorem supplies maximum and minimum. If the function is constant, every interior point works. Otherwise a value differs from the equal endpoint values, so either a maximum or a minimum occurs in the interior. Fermat's condition applies there. $\square$

Mean value theorem. Under the same continuity and differentiability hypotheses but without equal endpoint values, some $c\in(a,b)$ satisfies $f(b)-f(a)=f'(c)(b-a)$.

Subtract the secant line $f(a)+[f(b)-f(a)](x-a)/(b-a)$ from $f$ and apply Rolle. This proves the theorem without assuming $f'$ is continuous. Consequences include: derivative zero on an interval makes the function constant; nonnegative derivative makes it nondecreasing; and $|f'|\le M$ on an interval gives $|f(y)-f(x)|\le M|y-x|$ for points in it. An interval-wide derivative bound cannot be inferred from one derivative value.

Theorem

Intermediate value theorem

If $f$ is continuous on $[a,b]$ and $y$ lies between $f(a)$ and $f(b)$, there is $c\in[a,b]$ with $f(c)=y$.

Proof

Completeness locates a crossing

If $y=f(a)$ or $y=f(b)$, take the corresponding endpoint. Otherwise, assume $f(a)<y<f(b)$; the reverse order follows by replacing both $f$ by $-f$ and $y$ by $-y$. The set $S=\{x\in[a,b]:f(x)\le y\}$ is nonempty and bounded above. Let $c=\sup S$. If $f(c)<y$, continuity gives nearby points to the right with value below $y$; since $c<b$, this contradicts the upper-bound property. If $f(c)>y$, continuity gives a neighbourhood of $c$ with value above $y$; the supremum approximation property supplies an element of $S$ in that neighbourhood, a contradiction. Thus $f(c)=y$.

Uniform continuity

Definition

One delta for the entire domain

A function $f:A\to\mathbb R$ is uniformly continuous on $A$ if for every $\varepsilon>0$ there exists $\delta>0$ such that for all $x,y\in A$, $|x-y|<\delta$ implies $|f(x)-f(y)|<\varepsilon$. The chosen $\delta$ can depend on $\varepsilon$ but not on $x$ or $y$.

Theorem

Heine–Cantor theorem

Every continuous real-valued function on a compact set is uniformly continuous.

Proof

Contradiction using compactness

If uniform continuity failed, some $\varepsilon_0>0$ would permit, for every $n$, points $x_n,y_n\in K$ with $|x_n-y_n|<1/n$ but $|f(x_n)-f(y_n)|\ge\varepsilon_0$. Compactness supplies a subsequence $x_{n_k}\to x\in K$. Since the paired distances tend to zero, also $y_{n_k}\to x$. Continuity gives both $f(x_{n_k})\to f(x)$ and $f(y_{n_k})\to f(x)$, contradicting the output separation.

Worked example

Pointwise but not uniform on an unbounded domain

The function $f(x)=x^2$ is continuous on $\mathbb R$ but not uniformly continuous there. Take $x_n=n$ and $y_n=n+1/n$. Then $|x_n-y_n|=1/n\to0$, but $|f(y_n)-f(x_n)|=2+1/n^2\not\to0$. On any closed bounded interval, Heine–Cantor does give uniform continuity. It is also direct: if $|x|,|y|\le M$, then $|x^2-y^2|\le2M|x-y|$.

Learning goals and definitions

Develop the proof behind integrability statements and distinguish boundedness from integrability. Let $a<b$ and $f$ be bounded on $[a,b]$. For a partition $P:a=x_0<\cdots<x_n=b$, put $m_i=\inf_{[x_{i-1},x_i]}f$, $M_i=\sup_{[x_{i-1},x_i]}f$, and define $L(f,P)=\sum m_i\Delta x_i$, $U(f,P)=\sum M_i\Delta x_i$. The lower integral $\underline I$ is the supremum of all lower sums; the upper integral $\overline I$ is the infimum of all upper sums. Refining a partition makes lower sums nondecreasing and upper sums nonincreasing.

For any two partitions $P,Q$, their common refinement gives $L(f,P)\le U(f,Q)$, so $\underline I\le\overline I$. Boundedness ensures these quantities are finite. The function is Riemann integrable when they agree.

Theorem: Darboux criterion

The bounded function is integrable if and only if for every $\varepsilon>0$ there is a partition with $U(f,P)-L(f,P)<\varepsilon$.

Proof. If the two integrals equal $I$, choose a lower sum above $I-\varepsilon/2$ and an upper sum below $I+\varepsilon/2$, using supremum and infimum approximation. Their common refinement has gap less than $\varepsilon$. Conversely the inequalities $0\le\overline I-\underline I\le U(f,P)-L(f,P)$ force the integral gap to be zero when such partitions exist for every positive $\varepsilon$.

Theorem: finitely many discontinuities are permitted

If $f$ is bounded and continuous except possibly at finitely many points, it is integrable.

Proof. Let $|f|\le M$, taking $M>0$. Surround the exceptional points by finitely many small intervals, with endpoints included in the partition, having total length less than $\varepsilon/(4M)$. At endpoints of $[a,b]$ use one-sided intervals. Their contribution to the upper-lower gap is at most $2M$ times their total length, hence less than $\varepsilon/2$. On the remaining finitely many closed intervals, $f$ is continuous and therefore uniformly continuous. Subdivide each sufficiently finely that its oscillation on every resulting subinterval is less than $\varepsilon/[2(b-a)]$. Their total gap is less than $\varepsilon/2$. Combining all partition points gives gap below $\varepsilon$. If an exceptional interval endpoint would itself be exceptional, move it slightly so all actual exceptional points lie inside the selected intervals or at the domain endpoints. This keeps the remaining closed intervals continuous.

Continuous functions and all tagged meshes

Let $f:[a,b]\to\mathbb R$ be continuous, where $a<b$. The extreme value theorem makes $f$ bounded, and Heine–Cantor makes it uniformly continuous. For a partition $P$, write $L(f,P)$ and $U(f,P)$ for its lower and upper Darboux sums. The proved Darboux integrability theorem for continuous functions supplies the common integral $I$.

Fix $\varepsilon>0$. Uniform continuity supplies $\delta>0$ such that $|f(x)-f(y)|<\varepsilon/[2(b-a)]$ whenever $|x-y|<\delta$. If the mesh of $P$ is less than $\delta$, every subinterval has oscillation at most $\varepsilon/[2(b-a)]$. Therefore

$$U(f,P)-L(f,P)\le\frac{\varepsilon}{2}<\varepsilon.$$

For every selection of tags, its tagged sum $S(P)$ lies between $L(f,P)$ and $U(f,P)$; so does $I$. Thus $|S(P)-I|\le U(f,P)-L(f,P)<\varepsilon$. The same $\delta$ works for every partition of sufficiently small mesh and every choice of tags. This proves the tagged-mesh definition for continuous functions on the stated closed interval.

For $a<b$, take a partition $a=x_0<x_1<\cdots<x_n=b$, with $\Delta x_i=x_i-x_{i-1}$, tags $\xi_i\in[x_{i-1},x_i]$, and mesh $\|P\|=\max_i\Delta x_i$. A finite number $I$ is the definite integral $\int_a^bf(x)dx$ if, for every $\varepsilon>0$, there is $\delta>0$ such that every partition with $\|P\|<\delta$ and every choice of tags satisfy $|\sum_i f(\xi_i)\Delta x_i-I|<\varepsilon$. In that case $I$ is the common limit. Continuous functions on a closed interval are integrable.

Let $f$ be continuous on $[a,b]$ and $A(x)=\int_a^xf(t)dt$. For $x\in(a,b)$ and nonzero $h$ with $x+h\in[a,b]$, the integral estimate gives

$$\left|\frac{A(x+h)-A(x)}h-f(x)\right|\le\sup_{t\in[\min(x,x+h),\max(x,x+h)]}|f(t)-f(x)|\longrightarrow0\quad(h\to0).$$

Thus $A'(x)=f(x)$ in the interior; at the endpoints the same argument gives one-sided derivatives. If $F$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $F'=f$ there, the mean value theorem makes $F-A$ constant on $[a,b]$. Therefore $\int_a^bf=F(b)-F(a)$.

For real $r$ with $|r|<1$, $r^n\to0$. For $0<q=|r|<1$, put $h=q^{-1}-1>0$. The elementary inequality $(1+h)^n\ge1+nh$ gives

$$|r^n|=q^n\le\frac1{1+nh}\to0.$$

The inequality follows by multiplying $1+nh$ by $1+h$ and retaining the extra nonnegative $nh^2$ at each step, starting at $n=1$. For $r=0$, all positive-index powers are zero. At $r=1$ the sequence is one; at $r=-1$ it oscillates; for $|r|>1$ its magnitude grows without bound. These boundary cases must not be hidden in the formula for $|r|<1$.

6.4 An infinite series is a limit of finite totals

Definition

For terms $a_1,a_2,\ldots$, define the $N$th partial sum

$$S_N=\sum_{n=1}^{N}a_n.$$

The infinite series $\sum_{n=1}^{\infty}a_n$ converges to $S$ when the sequence $(S_N)$ converges to $S$. Its sum is that limit. If the partial sums do not have a finite limit, the series diverges. The notation is a question about a limit, not permission to finish an infinite number of operations.

A necessary condition for convergence is $a_n\to0$. If $S_N\to S$, then for $n\ge2$,

$$a_n=S_n-S_{n-1}\longrightarrow S-S=0.$$

Common mistake

Thus terms that fail to approach zero prove divergence. Terms that approach zero do not prove convergence.

Worked example

Worked example 4: distinguish terms from totals. The sequence $a_n=1/n$ converges to zero. Nevertheless its series diverges. Group a partial sum ending at $2^m$ as

$$1+\frac12+\left(\frac13+\frac14\right) +\left(\frac15+\cdots+\frac18\right)+\cdots.$$

The block from $2^{j-1}+1$ through $2^j$ has $2^{j-1}$ terms, each at least $1/2^j$, so each block contributes at least $1/2$. Consequently $S_{2^m}\ge1+m/2$, which is unbounded. These are finite groupings of selected partial sums, not an unexplained rearrangement of an infinite series. This proves divergence of the harmonic series.

Consider $a+ar+ar^2+\cdots$, with partial sum $S_N=\sum_{k=0}^{N-1}ar^k$. If $a=0$, every term is zero and the series converges to zero for every real $r$. Suppose now that $a\ne0$.

A geometric progression has $a_1=a$ and $a_{k+1}=ra_k$. For $k\ge2$, $a_k=ar^{k-1}$. Describing $a_1$ separately also avoids an unnecessary $0^0$ expression when $r=0$. Its finite sum is derived by shifting and subtracting:

$$S_n=a+ar+\cdots+ar^{n-1},\quad rS_n=ar+ar^2+\cdots+ar^n.$$

Consequently $(1-r)S_n=a(1-r^n)$, giving

$$S_n=\frac{a(1-r^n)}{1-r}\quad(r\ne1),\qquad S_n=na\quad(r=1).$$

This is a finite identity for any real $a,r$ and positive integer $n$; no convergence assumption is required. Negative $r$ causes alternating signs. The exceptional ratio $1$ is not handled by division by zero.

When $|r|<1$, $r^N\to0$, and therefore

$$\sum_{k=0}^{\infty}ar^k=\frac a{1-r}.$$

When $|r|\ge1$, the terms $ar^k$ do not approach zero, so the series diverges. This includes $r=1$ and $r=-1$. Writing $a/(1-r)$ at $r=2$ gives an ordinary number but does not make it the sum of the divergent series.

For $|r|<1$, the exact remainder after the first $N$ terms is

$$R_N=S-S_N=\frac{ar^N}{1-r},\qquad |R_N|=\frac{|a|\,|r|^N}{|1-r|}.$$

The simpler estimate $|R_N|\le |a||r|^N/(1-|r|)$ also holds. The exact expression retains the sign and accounts correctly for negative ratios.

Monotone sequences

Theorem

Monotone convergence theorem

Every increasing real sequence bounded above converges to the supremum of its set of terms. Every decreasing sequence bounded below converges to the infimum.

Proof

From supremum to eventual closeness

Let $s=\sup\{x_n:n\in\mathbb N\}$. Given $\varepsilon>0$, the supremum criterion gives an index $N$ with $s-\varepsilon<x_N\le s$. Monotonicity gives $s-\varepsilon<x_N\le x_n\le s$ for all $n\ge N$, proving $x_n\to s$.

Worked example

A recursive sequence

Let $x_1=1$ and $x_{n+1}=(x_n+2)/2$. Induction gives $x_n<2$, and $x_{n+1}-x_n=(2-x_n)/2>0$, so the sequence increases and is bounded above. It converges to some $L$. Taking limits in the recurrence yields $L=(L+2)/2$, hence $L=2$. Solving the fixed-point equation without first proving convergence would not by itself show the sequence has a limit.

Partial sums and tails

Definition

Series convergence

The series $\sum_{n=1}^\infty a_n$ converges to $s$ if its partial sums $s_N=\sum_{n=1}^N a_n$ converge to $s$. It converges absolutely if $\sum|a_n|$ converges, and conditionally if it converges but not absolutely.

The Cauchy criterion for partial sums says convergence is equivalent to: for every $\varepsilon>0$ there exists $N$ such that $|\sum_{k=m}^n a_k|<\varepsilon$ for all $n\ge m\ge N$. In particular, convergence forces $a_n\to0$, but this necessary condition is not sufficient: the harmonic series diverges.

Theorem

Comparison and absolute convergence

For $0\le a_n\le b_n$ eventually, convergence of $\sum b_n$ implies convergence of $\sum a_n$; divergence of $\sum a_n$ implies divergence of $\sum b_n$. If $\sum|a_n|$ converges, then $\sum a_n$ converges.

Proof

Tail control proves absolute convergence

If $\sum|a_n|$ converges, its nonnegative tails are arbitrarily small. By the triangle inequality, $|\sum_{k=m}^n a_k|\le\sum_{k=m}^n|a_k|$, so the signed partial sums are Cauchy and converge in $\mathbb R$. The comparison statements follow from monotone bounded partial sums and, for the divergence conclusion, contraposition.

Worked example

Geometric, harmonic and p-series

For $|r|<1$, $\sum_{n=0}^N r^n=(1-r^{N+1})/(1-r)$ tends to $1/(1-r)$. For the harmonic series, grouping terms from $1/2$ onward as $1/2$, then $1/3+1/4$, then $1/5+\cdots+1/8$, and so on gives each group at least $1/2$. Partial sums are unbounded, so it diverges although its terms tend to zero. The standard $p$-series $\sum n^{-p}$ converges for $p>1$ and diverges for $p\le1$; the integral or dyadic comparison test establishes this threshold.

4. The limit superior is a tail bound, not a guessed limit

For a sequence $u_n\ge0$, define

$$s_N=\sup\{u_n:n\ge N\},\qquad L=\inf_N s_N=\limsup_{n\to\infty}u_n,$$

allowing the value $+\infty$. The tail suprema are nonincreasing, so their infimum is their extended limit. Two consequences are needed.

• If $L<q$, then $u_n<q$ eventually. Indeed some tail supremum is below $q$; otherwise their infimum would be at least $q$.
• If $q<L$, then $u_n>q$ for infinitely many indices. Otherwise a tail would have every term at most $q$, forcing its supremum and hence $L$ to be at most $q$.

These statements remain useful when the original sequence has no limit. A finite number of unusually large terms affects some early tail suprema, but not $L$.

Ratio and alternating tests

If $a_n\ne0$ eventually and $\limsup|a_{n+1}/a_n|<1$, the series converges absolutely by comparison with a geometric series. If $|a_{n+1}/a_n|\ge1$ eventually, the terms cannot tend to zero, so the series diverges. A limiting ratio of $1$ is inconclusive: both $\sum1/n$ and $\sum1/n^2$ have ratio tending to $1$.

Proof of the convergent ratio case. Suppose $a_n\ne0$ for all sufficiently large $n$ and $\limsup |a_{n+1}/a_n|<1$. Choose $0<q<1$ strictly larger than this limsup. For some $N$, every $n\ge N$ satisfies $|a_{n+1}/a_n|\le q$. Iterating gives

$$|a_{N+k}|\le |a_N|q^k\qquad(k\ge0).$$

The geometric series $\sum_{k\ge0}|a_N|q^k$ converges, so comparison proves convergence of the absolute tail. A finite prefix does not affect convergence; hence $\sum a_n$ converges absolutely. A uniform conclusion for a family of series requires a bound valid uniformly in the input.

If $b_n\ge0$ is nonincreasing and tends to zero, then $\sum(-1)^{n-1}b_n$ converges, and the error after $N$ terms has magnitude at most $b_{N+1}$. The even partial sums are nondecreasing, the odd partial sums are nonincreasing, and $S_{2n+1}-S_{2n}=b_{2n+1}\to0$. Thus the alternating harmonic series converges but not absolutely.

Worked example

Choose the right test

For $\sum_{n=1}^\infty n/3^n$, the ratio of successive positive terms is $(n+1)/(3n)\to1/3<1$, so the series converges. For $\sum(-1)^{n-1}/n$, the alternating test gives convergence while harmonic divergence rules out absolute convergence. A ratio tending to $1$ in the latter case cannot decide it.

For real functions on a nonempty set $A$, uniform convergence means $\sup_{x\in A}|f_n(x)-f(x)|\to0$, with the errors eventually finite. Uniform Cauchy means every $\varepsilon>0$ has $N$ such that $|f_n(x)-f_m(x)|<\varepsilon$ for all $x\in A$ and all $m,n\ge N$. If $A=\varnothing$, the epsilon–N conditions are vacuously satisfied; no real supremum of the empty set is used.

Integration, differentiation and series

If Riemann-integrable $f_n$ converge uniformly to $f$ on $[a,b]$, then $f$ is integrable and $\int_a^b f_n\to\int_a^b f$. Indeed the integral difference is bounded by $(b-a)\sup|f_n-f|$; integrability of the uniform limit follows by combining a sufficiently small uniform error with a partition for a fixed $f_N$.

Uniform convergence of $f_n$ alone does not guarantee convergence of derivatives. For example $f_n(x)=\sin(nx)/n$ converges uniformly to zero on $\mathbb R$, but $f_n'(0)=1$ for every $n$, whereas the limit's derivative is zero. A useful sufficient theorem is: if $a<b$, $f_n$ are $C^1$ on $[a,b]$, $f_n(x_0)$ converges for one $x_0\in[a,b]$, and $f_n'$ converges uniformly to $g$, then $f_n$ converges uniformly to $f$, differentiable on $(a,b)$ with $f'=g$ there. The fundamental theorem and the uniform-integration result give $f(x)=\lim f_n(x_0)+\int_{x_0}^xg(t)dt$; continuity of $g$ follows from uniform convergence of continuous derivatives.

For a series of functions, uniform convergence means uniform convergence of its partial sums. The Weierstrass M-test says that if $|u_n(x)|\le M_n$ for all $x\in A$ and $\sum M_n$ converges, then $\sum u_n$ converges uniformly and absolutely at each point. Its proof uses the uniform Cauchy tail bound $|\sum_{k=m}^n u_k(x)|\le\sum_{k=m}^nM_k$.

Worked example

A uniform series bound

For $u_n(x)=x^n/2^n$ on $[0,1]$, $|u_n(x)|\le2^{-n}$. The geometric majorant converges, so the M-test gives uniform convergence. Each partial sum is continuous, so the limit is continuous. In fact its sum is $x/(2-x)$ by the geometric formula.

Derivative interchange without $C^1$ hypotheses

Theorem

Let $a<b$. Suppose every $f_n$ is continuous on $[a,b]$ and differentiable on $(a,b)$. If $f_n(x_0)$ converges at one fixed $x_0\in[a,b]$ and $f_n'$ converges uniformly on $(a,b)$ to $g$, then $f_n$ converges uniformly on $[a,b]$ to a continuous $f$, differentiable on $(a,b)$, and $f'=g$. Continuity of the individual derivatives is not required.

Proof of existence. Uniform convergence of derivatives makes them uniformly Cauchy. For $m,n$ sufficiently large, the mean value theorem applied to $f_n-f_m$ between $x_0$ and any $x$ gives

$$|f_n(x)-f_m(x)|\le|f_n(x_0)-f_m(x_0)|+(b-a)\sup_{(a,b)}|f_n'-f_m'|.$$

The anchor sequence is Cauchy, so the right side is uniformly small. Uniform completeness gives a uniform limit $f$, which is continuous as a uniform limit of continuous functions.

Proof of the derivative identity. Fix an interior point $x$ and $\varepsilon>0$. Choose $N$ with $|f_n'(t)-f_N'(t)|\le\varepsilon$ for all interior $t$ and all $n\ge N$, and also $|g(x)-f_N'(x)|\le\varepsilon$. For sufficiently small nonzero $h$, apply the mean value theorem to $f_n-f_N$ on the segment from $x$ to $x+h$:

$$\left|\frac{f_n(x+h)-f_n(x)}h-\frac{f_N(x+h)-f_N(x)}h\right|\le\varepsilon.$$

Pass $n\to\infty$ at these two fixed inputs. The same inequality holds with $f$ replacing $f_n$. Since the fixed $f_N$ is differentiable at $x$, its difference quotient is within $\varepsilon$ of $f_N'(x)$ for small enough $h$. The triangle inequality bounds the difference quotient of $f$ from $g(x)$ by $3\varepsilon$. Arbitrary $\varepsilon$ proves differentiability and equality. This argument does not integrate $g$, so it does not silently assume $g$ continuous or Riemann integrable.

Learning goals and allowed earlier results

Construct rather than assume an elementary function, prove its algebraic laws, and justify the logarithm as an inverse. We use the previously proved ratio test, uniform convergence of power and derivative series on compact intervals, the mean value theorem, and the intermediate value theorem.

Define

$$E(x)=\sum_{n=0}^{\infty}\frac{x^n}{n!},\qquad x\in\mathbb R.$$

For fixed $x\ne0$, the ratio of successive absolute terms is $|x|/(n+1)\to0$, so the series converges absolutely; at $x=0$ it equals one directly. On any fixed interval $|x|\le M$, both its series and derivative series have summable factorial majorants. The uniform derivative theorem therefore applies to polynomial partial sums and gives $E'(x)=E(x)$ everywhere, with $E(0)=1$.

Theorem: positivity and exponential law

For every real $x,y$, $E(x)>0$, $E(-x)=1/E(x)$, and $E(x+y)=E(x)E(y)$.

Proof. Differentiate $H(x)=E(x)E(-x)$ to obtain $H'(x)=E(x)E(-x)-E(x)E(-x)=0$. The mean value theorem makes $H$ constant; at zero it equals one. Thus $E$ never vanishes. Since it is continuous and $E(0)>0$, an intermediate-value argument prevents it from becoming negative anywhere, proving positivity and the reciprocal identity. For fixed $y$, differentiate $K(x)=E(x+y)E(-x)$; the same cancellation gives zero derivative. At zero $K(0)=E(y)$, so $E(x+y)E(-x)=E(y)$. Multiplication by $E(x)$ yields the addition law. No rearrangement of a double infinite series was needed.

Since $E'=E>0$, it is strictly increasing. For $x\ge0$, all terms are nonnegative and $E(x)\ge1+x$, so $E(x)\to\infty$ at positive infinity. The reciprocal identity gives $E(x)\to0$ at negative infinity. Continuity and monotonicity therefore make $E:\mathbb R\to(0,\infty)$ a bijection.

Define $\exp=E$, $e=E(1)$, and $\ln:(0,\infty)\to\mathbb R$ as its inverse. The inverse is continuous: if $x_0=E(t_0)$, then for any $\varepsilon>0$, $E(t_0-\varepsilon)<x_0<E(t_0+\varepsilon)$; taking $x$ within the smaller output gap confines $\ln x$ between those inputs.

Theorem: logarithmic derivative and laws

For $x>0$, $\ln'(x)=1/x$. For positive $x,y$ and real $r$, $\ln(xy)=\ln x+\ln y$ and $\ln(x^r)=r\ln x$, where $x^r$ is defined as $E(r\ln x)$.

Proof. Write $x_0=E(t_0)$, $x=E(t)$ and use inverse continuity to ensure $t\to t_0$ as $x\to x_0$. Then

$$\frac{\ln x-\ln x_0}{x-x_0}=\frac{t-t_0}{E(t)-E(t_0)}\longrightarrow\frac1{E'(t_0)}=\frac1{x_0}.$$

The exponential law gives $E(\ln x+\ln y)=xy$, and uniqueness of inverse values proves the product law. The real-power definition and inverse identity give the other formula. Differentiating that definition yields $(x^r)'=rx^{r-1}$ for $x>0$. This derivation did not first assume the real-power derivative to define the logarithm.

3. Dyadic condensation: the monotonicity is doing work

Theorem 2. If $a_n\ge0$ and $a_n$ is nonincreasing, then

$$\sum_{n=1}^{\infty}a_n\quad\hbox{converges if and only if}\quad\sum_{k=0}^{\infty}2^k a_{2^k}\quad\hbox{converges}.$$

Proof. The block $B_k=\sum_{n=2^k}^{2^{k+1}-1}a_n$ contains exactly $2^k$ terms. Every term is at most $a_{2^k}$ and at least $a_{2^{k+1}}$, so

$$\frac12\bigl(2^{k+1}a_{2^{k+1}}\bigr)\le B_k\le2^k a_{2^k}.$$

If the condensed series converges, the upper bound makes the partial sums of the block series bounded. If the block series converges, the lower bound makes the condensed series from $k=1$ onward bounded. Adding its one missing term does not change convergence. Finally, the block endpoints are a subsequence of the original nonnegative partial sums. Bounded block-endpoint sums bound every intermediate partial sum, because it lies between neighboring endpoints. Hence the block series converges precisely when the original series does. $\square$

The full p-series threshold by dyadic grouping

For real $p\le0$, $n^{-p}\ge1$, so the terms do not tend to zero and the series diverges. For $p>0$, $a_n=n^{-p}$ is nonnegative and nonincreasing. The proved dyadic test applies, and its condensed terms are $2^k a_{2^k}=2^{k(1-p)}$. Their geometric series converges exactly when $2^{1-p}<1$, equivalently $p>1$. When $0<p\le1$, the condensed terms are at least one and their series diverges, so the original series diverges too. This proves the threshold for every real $p$ without invoking an unproved improper-integral test.

3.1 Rates, initial values, and verification

A first-order equation has the form $y'=f(t,y)$. Its order is one because the highest derivative present is the first. A differentiable function $y:I\to\mathbb R$ is a solution when $y'(t)=f(t,y(t))$ for every $t$ in its interval. The interval must avoid undefined expressions in the equation or the proposed solution.

For $y'=ky$, the functions $y=Ce^{kt}$ are solutions. Differentiating verifies the equation. With $y(0)=y_0$, the initial value fixes $C=y_0$. The exponent requires compatible units: $k$ has units inverse time.

Worked example 1: decay and its time scale. A hypothetical quantity follows $y'=-0.2y$, $y(0)=50$, with time in days. The solution is $y(t)=50e^{-0.2t}$. To reach $10$, solve $e^{-0.2t}=1/5$, giving $t=5\ln5$ days. This solution remains positive for finite time; it approaches zero but never equals zero at a finite time.

1. The local theorem and its assumptions

A solution of $y'=f(t,y)$ is a differentiable real function on an interval whose graph lies in the domain of $f$ and satisfies the equation there. The initial condition is $y(t_0)=y_0$. Local uniqueness means that any two such solutions with this initial value agree on some common neighborhood of $t_0$.

Theorem. Let $U\subset\mathbb R^2$ be open and contain $(t_0,y_0)$. Suppose $f:U\to\mathbb R$ is continuous, its partial derivative $f_y$ exists on $U$, and $f_y$ is continuous there. Then the initial-value problem has a solution on an open interval about $t_0$, unique in the local sense just defined. No derivative with respect to $t$ is required.

2. Two uniform-limit facts

For a nonempty compact interval $I$, write $C(I)$ for the continuous real functions on $I$ and $\|u-v\|_\infty=\sup_{t\in I}|u(t)-v(t)|$. Uniform convergence is convergence in this distance. A Cauchy sequence in this distance is uniformly Cauchy.

Theorem: uniform completeness

A sequence of real-valued functions is uniformly convergent if and only if it is uniformly Cauchy.

Proof. Uniform convergence gives the Cauchy estimate by two errors of size $\varepsilon/2$. Conversely, uniform Cauchy implies that for each fixed $x$, $(f_n(x))$ is a real Cauchy sequence, so define $f(x)=\lim_n f_n(x)$. Fix $\varepsilon>0$ and use uniform Cauchy with tolerance $\varepsilon/2$. For every $x$ and $n\ge N$, pass $m\to\infty$ in the estimate to get $|f_n(x)-f(x)|\le\varepsilon/2<\varepsilon$. The same $N$ works everywhere, proving uniform convergence. Completeness of the target $\mathbb R$ supplied the pointwise limits.

Theorem

Uniform limit preserves continuity

If each $f_n$ is continuous at $a\in A$ relative to $A$ and $f_n\to f$ uniformly on $A$, then $f$ is continuous at $a$.

Proof

The three-term estimate

Given $\varepsilon>0$, choose $N$ so $|f_N(x)-f(x)|<\varepsilon/3$ for all $x\in A$. By continuity of the fixed function $f_N$ at $a$, choose $\delta$ so $|f_N(x)-f_N(a)|<\varepsilon/3$ whenever $|x-a|<\delta$. Then

$$|f(x)-f(a)|\le|f(x)-f_N(x)|+|f_N(x)-f_N(a)|+|f_N(a)-f(a)|<\varepsilon.$$

These facts use completeness of the real numbers. They will also preserve the closed value constraint imposed below.

3. Proof by an integral iteration

Rectangle and quantitative constants. Choose $a,b>0$ so the closed rectangle $Q=[t_0-a,t_0+a]\times[y_0-b,y_0+b]$ lies in $U$. The extreme-value theorem supplies finite constants $M,L\ge0$ with $|f|\le M$ and $|f_y|\le L$ on $Q$. The one-variable mean value theorem in the $y$ coordinate gives

$$|f(t,y)-f(t,z)|\le L|y-z|\qquad(t\in[t_0-a,t_0+a],\ y,z\in[y_0-b,y_0+b]).$$

All intervals in this use lie in $Q$; no differentiability in $t$ has been assumed. Put

$$h=\min\left\{a,\frac b{M+1},\frac1{2(L+1)}\right\}>0,\quad I=[t_0-h,t_0+h],\quad q=hL<\frac12.$$

Then $hM<b$. The added ones avoid division by zero when $M=0$ or $L=0$.

A complete space of permitted functions. Let $X=\{u\in C(I):\|u-y_0\|_\infty\le b\}$, where $y_0$ denotes the constant function. If $(u_n)$ is Cauchy in this distance, uniform completeness gives a uniform limit $u$. The uniform-limit continuity theorem makes $u$ continuous, and passage to the limit in $|u_n(t)-y_0|\le b$ gives $|u(t)-y_0|\le b$ for every $t$. Thus $u\in X$ and $X$ is complete.

The integral map. Define

$$(Tu)(t)=y_0+\int_{t_0}^{t}f(s,u(s))\,ds.$$

The integrand is continuous, since the graph of $u$ stays in $Q$. Thus $Tu$ is continuous, and

$$\|Tu-y_0\|_\infty\le hM<b,\qquad\|Tu-Tv\|_\infty\le hL\|u-v\|_\infty=q\|u-v\|_\infty.$$

For $t<t_0$ these bounds follow by reversing the integral limits before taking absolute values. Therefore $T$ maps $X$ into itself and contracts distances.

Convergence and a fixed point. Start with $u_0(t)=y_0$ and set $u_{n+1}=Tu_n$. If $q=0$, the distance estimate says $Tu=Tv$ for all $u,v\in X$; hence $Tu_1=Tu_0=u_1$, already a fixed point. Suppose $0<q<1$ and put $d=\|u_1-u_0\|_\infty$. Repeated application of the contraction estimate gives $\|u_{j+1}-u_j\|_\infty\le q^jd$ for every integer $j\ge0$. Consequently, for $p\ge1$,

$$\|u_{n+p}-u_n\|_\infty\le\sum_{j=n}^{n+p-1}q^jd\le\frac{q^n d}{1-q}\longrightarrow0\quad(n\to\infty).$$

Completeness gives $u_n\to u\in X$. Since

$$\|Tu-u\|_\infty\le q\|u-u_n\|_\infty+\|u_{n+1}-u\|_\infty\longrightarrow0,$$

we have $Tu=u$. Passing $p\to\infty$ above also proves $\|u-u_n\|_\infty\le q^nd/(1-q)$. For $q=0$, $u_1$ is exact instead.

The differential equation and uniqueness. The fixed-point equation and the fundamental theorem of calculus give $u'(t)=f(t,u(t))$ for interior $t\in I$, and $u(t_0)=y_0$. Restricting to the interior of $I$ supplies the claimed local solution. If $u,v\in X$ are both fixed points, $\|u-v\|_\infty\le q\|u-v\|_\infty$ and $q<1$ force equality of the functions.

Finally, any other local solution is continuous at $t_0$ and has value $y_0$, so its graph stays in $Q$ on a sufficiently small common closed interval $J=[t_0-\eta,t_0+\eta]\subset I$, with $\eta>0$. The differential equation integrated from $t_0$ gives the same integral equation there. On $J$ the contraction estimate has factor $\eta L\le q<1$, so it agrees with $u$ on $J$. This establishes local uniqueness without assuming an alternative solution remains in $Q$ for all future time. $\square$

Local existence does not assert global continuation. A later domain boundary or unbounded growth requires a separate continuation argument. The derivative condition is sufficient, not a necessary condition for every uniquely solvable equation.

3.2 Separation and the solutions it can lose

For $y'=g(t)h(y)$, on an interval where $h(y)\ne0$, separation gives

$$\frac{y'}{h(y)}=g(t).$$

For $y'=g(t)h(y)$, where $h(y)\ne0$ along the solution interval,

$$\int\frac{dy}{h(y)}=\int g(t)dt.$$

This notation abbreviates an application of the chain rule to an antiderivative of $1/h$. Before dividing, check all zeros of $h$ for constant equilibrium solutions. They may be lost by separation.

Worked example 2: finite-time blow-up. Solve $y'=y^2$, $y(0)=1$. Since the required solution is positive near the initial point, divide by $y^2$ and integrate:

$$-1/y=t+C.$$

The initial value gives $C=-1$, hence $y=1/(1-t)$. Direct substitution verifies it for $t<1$, and this is the maximal connected interval containing zero on which the formula remains finite. It does not define a solution through $t=1$. The zero function is also a solution of the differential equation, but not of this initial-value problem.

Worked example 3: a bounded growth model. Suppose $y'=ry(1-y/K)$ with $r,K>0$. Both $y=0$ and $y=K$ are equilibria. For $0<y<K$, separation and partial fractions give

$$\frac1{y(1-y/K)}=\frac1y+\frac1{K-y},\qquad \ln\frac{y}{K-y}=rt+C.$$

With initial value $y(0)=y_0>0$, the resulting formula is

$$y(t)=\frac{K}{1+(K/y_0-1)e^{-rt}}.$$

Substitution verifies it for every $y_0>0$, including $y_0=K$, and the denominator is positive for $t\ge0$. If $0<y_0<K$, the solution increases toward $K$; if $y_0>K$, it decreases toward $K$. The separately checked zero equilibrium supplies the case $y_0=0$. The separation derivation must not be used to claim these equilibria were included without checking them.

3.3 Linear equations and integrating factors

For a linear equation $y'+p(t)y=q(t)$ with continuous coefficients, choose

$$\mu(t)=\exp\left(\int p(t)dt\right).$$

Then $\mu'=p\mu$ and the product rule turns the equation into $(\mu y)'=\mu q$. With initial value at $t_0$,

$$y(t)=\frac{\mu(t_0)y_0+\int_{t_0}^t\mu(s)q(s)ds}{\mu(t)}.$$

The integrating factor is nonzero; multiplying it by a constant does not change the solution formula. The formula's interval must retain continuity of coefficients.

Here the initial condition is $y(t_0)=y_0$.

Worked example 4: an inventory balance. Let a hypothetical inventory receive $12$ units/day and lose stock at rate $0.3I$ units/day. Then

$$I'+0.3I=12,\qquad I(0)=10.$$

Using $\mu=e^{0.3t}$ yields $I(t)=40-30e^{-0.3t}$. The equilibrium is $40$ units. Differentiating gives $I'=9e^{-0.3t}=12-0.3I$, and the initial value is $10$. To reach $30$ units, $e^{-0.3t}=1/3$, so $t=(\ln3)/0.3$ days. The formula is a consequence of the stated constant inflow and proportional loss assumptions.

More generally, for $y'=a-by$, $b>0$, and $y(0)=y_0$, the equilibrium is $a/b$ and the solution is

$$y(t)=a/b+(y_0-a/b)e^{-bt}.$$

The deviation from equilibrium decays exponentially. This exact identity establishes stability for the model without relying on a sketch of a slope field.

3.4 Euler's method

If an exact formula is inconvenient, approximate the derivative over a short step by the slope at the beginning. With step size $h>0$,

$$t_{n+1}=t_n+h,\qquad y_{n+1}=y_n+h f(t_n,y_n).$$

This is the forward Euler method. It replaces a varying slope over the step with one fixed slope. For a sufficiently smooth exact solution, Taylor's theorem gives

$$y(t+h)=y(t)+hy'(t)+O(h^2),$$

so the one-step defect is of second order. Over a fixed interval, accumulated error is typically first order under suitable smoothness and Lipschitz bounds. This is a conditional theorem, not an accuracy guarantee for every equation and step size.

The local defect of forward Euler

Let $h>0$, let $N$ be a positive integer, put $t_n=t_0+nh$ and $T=t_0+Nh$, and suppose a real scalar exact solution $y$ is $C^2$ on an open neighborhood of $[t_0,T]$. Assume $|y''(t)|\le M$ on that interval, with $M\ge0$. Define its one-step defect by

$$d_n=y(t_{n+1})-y(t_n)-h f(t_n,y(t_n)),\qquad 0\le n<N.$$

Since $y'=f(t,y)$, the fundamental theorem gives

$$d_n=\int_{t_n}^{t_{n+1}}[y'(s)-y'(t_n)]\,ds,\qquad |d_n|\le\int_{t_n}^{t_{n+1}}M(s-t_n)\,ds=\frac{Mh^2}{2}.$$

The derivative difference bound follows from the mean value theorem applied to $y'$. This is a statement about the exact solution over one step, before numerical errors from earlier steps have been propagated.

A precise global error estimate

Suppose $f$ is continuous on $[t_0,T]\times\mathbb R$ and has one constant $L\ge0$ satisfying

$$|f(t,u)-f(t,v)|\le L|u-v|\quad(t\in[t_0,T],\ u,v\in\mathbb R).$$

Assume the exact solution and its bound $M$ stated above exist on the entire interval. Start Euler with the exact initial value, $Y_0=y(t_0)$, and set $Y_{n+1}=Y_n+h f(t_n,Y_n)$. The global-in-$y$ domain and Lipschitz hypothesis apply to these numerical values as well as the exact ones; a bound known only along the exact solution would not justify the following comparison.

Writing $e_n=|Y_n-y(t_n)|$, subtract the exact increment and use the local defect bound:

$$e_{n+1}\le(1+Lh)e_n+\frac{Mh^2}{2},\qquad e_0=0.$$

Induction, adding one geometric term at each step, gives

$$e_n\le\frac{Mh^2}{2}\sum_{j=0}^{n-1}(1+Lh)^j.$$

The empty sum at $n=0$ is zero. If $L>0$, summing the finite geometric series and using $1+Lh\le e^{Lh}$ yield

$$e_n\le\frac{Mh}{2L}[(1+Lh)^n-1]\le\frac{Mh}{2L}[e^{L(T-t_0)}-1].$$

If $L=0$, use the finite sum directly: $e_n\le Mnh^2/2\le M(T-t_0)h/2$. In both cases the bound is a fixed constant times $h$ on the fixed time interval. This proves the stated first-order global accuracy under explicit hypotheses. It does not assert a global existence theorem for every equation, preservation of every physical constraint, or a floating-point roundoff bound. On a restricted state domain, an independent argument must first keep both exact and numerical values inside a region with one common Lipschitz bound.

Worked example 5: exact decay, approximate steps. For $y'=-y$, $y(0)=1$, the exact solution at $t=1$ is $e^{-1}$. Euler gives $y_{n+1}=(1-h)y_n$, so $y_n=(1-h)^n$ at $t_n=nh$. With $h=1/2$, two steps give $1/4$. With $h=1/4$, four steps give $(3/4)^4=81/256\approx0.31641$, closer to $e^{-1}\approx0.36788$ but still not exact.

For fixed $h$, Euler's values decay to zero only when $|1-h|<1$, or $0<h<2$. They remain nonnegative for nonnegative initial values when $0<h\le1$. A step $h=3$ gives alternating values of growing magnitude even though the exact solution always decays positively. Numerical stability and positivity therefore require analysis, not just a working iteration formula.

A useful practical check is to reduce $h$ and compare results at the same final time, while keeping any known physical bounds and analytic checks in view. Agreement of two meshes is evidence, not an unconditional proof of accuracy.

3.5 Graded exercises

Foundation

1. Verify that $y(t)=3e^{2t}$ solves $y'=2y$ and state its initial value at zero.
2. Solve $y'=-ky$, $y(0)=y_0>0$, with $k>0$. Find its half-life.
3. List all equilibrium solutions of $y'=y(2-y)$ before separating variables.

Core

4. Solve $y'=y^2$, $y(0)=2$, and state the maximal interval containing zero.
5. Solve $y'+2y=6$, $y(0)=1$, and verify both conditions.
6. Solve $y'+y=t$, $y(0)=0$, using an integrating factor.
7. For $y'=y(1-y)$ and $y(0)=1/4$, find the solution and its limit as $t\to\infty$.

Synthesis

8. Use Euler with $h=1/2$ for $y'=2-y$, $y(0)=0$, through $t=1$. Compare with the exact solution there.
9. For $y'=-3y$, derive the Euler recurrence. Find step sizes giving asymptotic decay, and the smaller range preserving nonnegative values from positive initial data.
10. A student divides $y'=y(1-y)$ by $y(1-y)$ and concludes every solution is strictly between zero and one. Identify the lost cases, give a solution starting above one, and explain why its denominator is positive for future time.

3.6 Full solutions

1. Differentiating gives $y'=6e^{2t}=2y$, and $y(0)=3$.
2. The solution is $y=y_0e^{-kt}$. Half the initial value occurs when $e^{-kt}=1/2$, so $t=\ln2/k$.
3. A constant equilibrium satisfies $y(2-y)=0$, giving $y=0$ and $y=2$.
4. Integration yields $-1/y=t-1/2$, hence $y=2/(1-2t)$. It has a pole at $t=1/2$. The maximal connected interval containing zero is $(-\infty,1/2)$.
5. The equilibrium is $3$, so $y=3-2e^{-2t}$. Then $y'=4e^{-2t}$ and $y'+2y=4e^{-2t}+6-4e^{-2t}=6$. The initial value is $1$.
6. Multiply by $e^t$: $(e^ty)'=te^t$. Integrating from zero gives $e^ty=(t-1)e^t+1$, so $y=t-1+e^{-t}$. At zero it is zero, and direct differentiation yields $y'+y=t$.
7. The formula is $y=1/(1+3e^{-t})$. Its denominator tends to one and remains positive, so the limit is $1$; it starts at $1/4$ and increases because the differential equation is positive between zero and one.
8. Euler gives $y_1=0+(1/2)(2)=1$ at $t=1/2$, then $y_2=1+(1/2)(1)=3/2$ at $t=1$. The exact solution is $2(1-e^{-t})$, giving $2(1-e^{-1})\approx1.26424$ at one. The Euler approximation is larger by $2/e-1/2\approx0.23576$.
9. Euler is $y_{n+1}=(1-3h)y_n$. Asymptotic decay requires $|1-3h|<1$, so $0<h<2/3$. Nonnegative values from positive initial data require $1-3h\ge0$, giving $0<h\le1/3$. At $h=1/3$, the numerical sequence becomes zero after one step; this is not the positive exact solution at that time.
10. The division loses the equilibrium solutions $y=0$ and $y=1$ and restricts the derivation's interval. Starting at $y_0=2$, a verified solution is $y=1/(1-(1/2)e^{-t})$. For $t\ge0$, $e^{-t}\le1$, so the denominator is at least $1/2>0$. This solution remains above one and decreases toward it. A derivation made under $0<y<1$ cannot exclude a separately verified solution outside that range.

Summary

Always verify the differential equation, initial value, and interval. Separation is valid only away from the factors divided out; inspect equilibria first. Integrating factors turn linear equations into product derivatives. Euler's method is a local-slope approximation whose step size affects accuracy, stability, and preservation of model constraints.