Multivariable differentiability and mean value estimates

Several directional rates do not automatically assemble into a good linear approximation. The central analytical question is whether one linear map controls the error for every sufficiently small displacement, not just along a chosen line. This distinction is the foundation for chain rules, Taylor estimates and multivariable optimization.

1 Norms limits and the derivative

Use the Euclidean norm on finite-dimensional spaces. For an open set $U\subset\mathbb R^n$ and $f:U\to\mathbb R^m$, continuity at $a$ means $\|f(x)-f(a)\|\to0$ as $\|x-a\|\to0$. A map is differentiable at $a$ if there is a linear map $L$ with

$$\lim_{h\to0}\frac{\|f(a+h)-f(a)-Lh\|}{\|h\|}=0.$$

The map $L=Df(a)$ is the total or Fréchet derivative. It is unique: if $L$ and $M$ both work, substitute $h=tv$ to obtain $\|(L-M)v\|=0$ for every $v$. For $h=\sum_{j=1}^n h_je_j$, linearity and the triangle inequality give $\|Lh\|\le\sum_j|h_j|\|Le_j\|\le(\sum_j\|Le_j\|)\|h\|$. This proves boundedness directly in finite dimension. The derivative definition then implies continuity.

The Jacobian matrix has entries $\partial f_i/\partial x_j$. Differentiability implies these partial derivatives exist and are the entries of $L$, by testing $h=te_j$. If all partial derivatives are continuous in a neighborhood, then $f$ is differentiable there. For scalar $f$, split the increment into $n$ coordinate steps and apply the one-variable mean value theorem to each. The resulting coefficients are partial derivatives at points tending to $a$; continuity makes the error at most $\sum o(1)|h_j|=o(\|h\|)$. Apply this componentwise for vector outputs.

2 Why line tests can fail

Define $f(0,0)=0$ and

$$f(x,y)=\frac{x^2y}{x^2+y^2}\quad\text{for }(x,y)\ne(0,0).$$

Because $|f(x,y)|\le|y|$, it is continuous at zero. Both coordinate partials at zero are zero. If it were differentiable, its derivative would therefore be zero. But along $h=(t,t)$, $f(t,t)=t/2$, and

$$\frac{|f(t,t)|}{\|(t,t)\|}=\frac1{2\sqrt2},$$

which does not tend to zero. It is not differentiable. In fact every directional derivative exists, but their dependence on direction is not linear, another obstruction to a total derivative.

3 The chain rule with error control

Theorem. If $f$ is differentiable at $a$ and $g$ is differentiable at $f(a)$, with the composition defined near $a$, then

$$D(g\circ f)(a)=Dg(f(a))Df(a).$$

Proof. Write $f(a+h)-f(a)=Lh+r(h)$, with $r(h)=o(\|h\|)$. This increment $k$ is $O(\|h\|)$. The second approximation gives $g(f(a)+k)-g(f(a))=Mk+s(k)$, with $s(k)=o(\|k\|)$. Hence the remainder after subtracting $MLh$ is $Mr(h)+s(k)=o(\|h\|)$. When $k=0$, the second remainder is zero by definition, so no division-by-zero exception is needed. $\square$

For $f(x,y)=(xy,x+y)$ and $g(u,v)=u^2+v$, the derivative of the composition is $(2xy^2+1,2x^2y+1)$, either by differentiating $x^2y^2+x+y$ or by multiplying Jacobians.

4 Mean value along a segment

Scalar mean value theorem. If the segment $[a,b]$ lies in an open domain and $f:U\to\mathbb R$ is differentiable there, then some $t\in(0,1)$ satisfies

$$f(b)-f(a)=Df(a+t(b-a))(b-a).$$

Apply the one-variable theorem to $\phi(t)=f(a+t(b-a))$. The chain rule gives its derivative. For vector outputs there need not be a single point giving an analogous vector equality: $f(t)=(\cos t,\sin t)$ on $[0,2\pi]$ has zero total increment but never zero derivative.

A valid vector statement is the estimate: if $f$ is $C^1$ along the segment and $\|Df(x)\|_{\rm op}\le M$, then

$$\|f(b)-f(a)\|\le M\|b-a\|.$$

Indeed, integrate the derivative of the path componentwise and use the norm inequality. If the same constant $M$ bounds $\|Df(x)\|_{\rm op}$ for every $x$ in an open convex domain, the estimate applies to every joining segment and gives a global Lipschitz constant $M$ on that domain. A bound chosen separately on one segment is not a global bound. A disconnected or nonconvex domain may not contain the needed segment.

Exercises with complete solutions

E1 Foundation. Prove $f(x,y)=xy$ is differentiable at $(a,b)$ directly from the definition.

S1. The increment is $bh+ak+hk$ at displacement $(h,k)$. The linear part is $bh+ak$ and $|hk|\le(h^2+k^2)/2$, so the remainder divided by $\sqrt{h^2+k^2}$ tends to zero.

E2 Counterexample. For the function in Section 2, calculate the directional derivative along a unit vector $v=(p,q)$ and show it is not linear in $v$.

S2. For $v\ne0$, $f(tp,tq)/t=p^2q/(p^2+q^2)$; for a unit vector this is $p^2q$. It vanishes at both coordinate directions but not at $(1,1)/\sqrt2$, so it cannot be a linear functional. The formula extends to a homogeneous but nonadditive function of arbitrary nonzero directions.

E3 Estimate. For $R\ge0$, prove $f(x,y)=(x^2,y^2)$ is Lipschitz on the square $[-R,R]^2$ with constant $2R$.

S3. With $R\ge0$, the Jacobian is $\operatorname{diag}(2x,2y)$ with operator norm at most $2R$. Every segment between two square points lies in the square, so the segment estimate applies. One can also bound each difference of squares directly and square-sum the bounds.

E4 Proof. If $Df=0$ on a convex open set, prove $f$ is constant there, including vector-valued $f$.

S4. For any $a,b$ in the set, restrict each component to their segment. Its one-variable derivative is zero, so the component has equal endpoint values by the mean value theorem. Hence $f(a)=f(b)$.

Summary

Differentiability is a uniform first-order approximation by one linear map. Partial or directional derivatives alone do not guarantee it. Segment arguments transfer one-variable theorems carefully; scalar equalities and vector norm estimates must not be confused.