Taylor remainders, Cauchy mean values and convexity

Continuous real functions on a closed interval are uniformly continuous

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.

Let $a\le b$ be finite real numbers and let $f:[a,b]\to\mathbb R$ be continuous relative to this interval. Uniform continuity means that for every $\varepsilon>0$ there is $\delta>0$ such that all $x,y\in[a,b]$ with $|x-y|<\delta$ satisfy $|f(x)-f(y)|<\varepsilon$. Fix $\varepsilon>0$. For each $c\in[a,b]$, relative continuity gives $r_c>0$ such that $|f(x)-f(c)|<\varepsilon/2$ for interval points with $|x-c|<r_c$. The intervals $(c-r_c/2,c+r_c/2)$ cover the compact interval. Choose a finite subcover with centres $c_1,\ldots,c_k$, and set $\delta=\min_i r_{c_i}/2>0$. The cover is nonempty since $[a,b]$ is nonempty. Given $x,y$ as above, choose $i$ with $|x-c_i|<r_{c_i}/2$. Then $|y-c_i|<\delta+r_{c_i}/2\le r_{c_i}$, so the triangle inequality gives $|f(x)-f(y)|<\varepsilon/2+\varepsilon/2=\varepsilon$. The same $\delta$ works for all points. This includes a singleton interval and uses the actual finite-cover compactness theorem, rather than citing an unproved scalar Heine–Cantor assertion.

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.

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.

Fundamental theorem of calculus

Integration by parts under the actual continuous-derivative assumptions

Let $u,v$ be real $C^1$ functions on the closed interval with endpoints $a,b$. The functions $u'v$ and $uv'$ are continuous and therefore properly Riemann integrable. For nonzero real $h$ the product difference quotient is

Continuity of $u$ and existence of both derivatives justify this limit, proving $(uv)'=u'v+uv'$. The already proved continuous-function fundamental theorem, applied to $uv$, and linearity of the proper integral give

and hence

For reversed endpoints use oriented integrals, and for equal endpoints all terms are zero. This is sufficient for the Taylor induction in U15: for fixed real $x$, take $u(t)=(x-t)^n/n!$ and $v(t)=f^{(n)}(t)$ when $f\in C^{n+1}$ on the segment. It does not claim integration by parts for merely differentiable functions with unproved integrability of their derivatives.

For a sequence $u_n\ge0$, define

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 $L$.

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

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.

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.

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.

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

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 $x+h$:

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.

Define

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.

Full flat-function differentiation at zero

Use the previously proved real exponential series, positivity, reciprocal law and derivative identity $\exp'=\exp$. Define $F(0)=0$ and $F(x)=\exp(-1/x^2)$ for $x\ne0$. For every real $r\ge0$, set $u=1/x^2$. Choose an integer $k>r/2$. The nonnegative exponential series gives $\exp u\ge u^k/k!$ for $u>0$, hence

At nonzero inputs, induction gives $F^{(n)}(x)=P_n(1/x)\exp(-1/x^2)$ for a polynomial $P_n$, starting with $P_0=1$ and using $P_{n+1}(z)=-z^2P_n'(z)+2z^3P_n(z)$. Suppose the order-$n$ derivative has this formula off zero and value zero at zero. Every polynomial has finite degree, so the estimate above first shows $F^{(n)}(x)\to0$ and then $F^{(n)}(h)/h\to0$. Thus $F^{(n)}$ is continuous at zero, its derivative at zero exists and equals zero, and the recurrence gives the next off-zero formula. Induction proves $F$ is smooth with every derivative zero at zero. Its Taylor series there is identically zero, whereas $F(x)>0$ for each $x\ne0$. This completes the claimed counterexample without treating smoothness as analyticity.