CAL1_L19

A sufficient condition for differentiability

Theorem 1(Lemma).

A vector valued function $\mathbf{f}=(f_1,\dots ,f_m):{\mathbb{R}}^n\to {\mathbb{R}}^m$ is differentiable at $\mathbf{c}$ iff each $f_k$ is differentiable at $c$.

Proof
Let $\mathbf{f}$ be differentiable at $\mathbf{c}$. Then,

$$\mathbf{f}(\mathbf{c}+\mathbf{v})=\mathbf{f}(\mathbf{c})+{\mathbf{f}}^{\prime }(\mathbf{c})(\mathbf{v})+\Vert \mathbf{v}\Vert {\mathbf{E}}_{\mathbf{c}}(\mathbf{v}),$$

where ${\mathbf{E}}_{\mathbf{c}}(\mathbf{v})\to 0$ as $\mathbf{v}\to 0$. On writing the $m$ component wise equations, we find

$$f_k(\mathbf{c}+\mathbf{v})=f_k(\mathbf{c})+\nabla f_k(\mathbf{c})\cdot \mathbf{v}+\Vert \mathbf{v}\Vert E_{\mathbf{c},k}(\mathbf{v}),$$

where $E_{\mathbf{c},k}(\mathbf{v})\to 0$ as $\mathbf{v}\to 0$. Thus, $f_k$ is differentiable at $\mathbf{c}$, and $f_k^{\prime }(\mathbf{c})=\nabla f_k(\mathbf{c})$.

Conversely, if each $f_k$ is differentiable at $\mathbf{c}$ with $f_k^{\prime }(\mathbf{c})=\nabla f_k(\mathbf{c})$, then the linear transformation given by

$$\begin{array}{r}{\mathbf{T}}_{\mathbf{c}}(\mathbf{v})=\sum _{k=1}^m(\nabla f_k(\mathbf{c})\cdot \mathbf{v}){\mathbf{e}}_k\end{array}$$

satisfies the first order Taylor formula for $\mathbf{f}$, and hence $\mathbf{f}$ is differentiable at $\mathbf{c}$ with derivative ${\mathbf{f}}^{\prime }(\mathbf{c})={\mathbf{T}}_{\mathbf{c}}$.

Theorem 2(Theorem).

Assume that one of the partial derivatives $D_1\mathbf{f},\dots ,D_n\mathbf{f}$ exist at $\mathbf{c}$ and that the remaining $n-1$ partial derivatives exist in some $n$-ball $B(\mathbf{c})$ and are continuous at $\mathbf{c}$. Then $\mathbf{f}$ is differentiable at $\mathbf{c}$.

Note that this is not a necessary condition.

Proof
The existence of $D_j\mathbf{f}$ implies the existence of every $D_j{f}_k$ and the continuity of $D_j\mathbf{f}$ implies the continuity of each $D_j{f}_k$. Thus, thanks to the preceding lemma, it suffices to prove the theorem when $\mathbf{f}$ is real valued.

WLOG, assume that $D_{1}f(\mathbf{c})$ exists at $\mathbf{c}$ and that $D_{2}f,\dots ,D_{n}f$ exist in $B(\mathbf{c})$ and are continuous at $\mathbf{c}$. We know that if $f$ is differentiable at $\mathbf{c}$, then its derivative is given by the Jacobian matrix, which in this case is the gradient $\nabla f(\mathbf{c})$. Thus, we only have to prove that

$$f(\mathbf{c}+\mathbf{v})-f(\mathbf{c})=\nabla f(\mathbf{c})\cdot \mathbf{v}+\Vert \mathbf{v}\Vert E_{\mathbf{c}}(\mathbf{v})$$

where $E_{\mathbf{c}}(\mathbf{v})\to 0$ as $\mathbf{v}\to 0$.

Let $\mathbf{v}=\lambda \mathbf{y}$, where $\Vert \mathbf{y}\Vert =1$ and $\lambda =\Vert \mathbf{v}\Vert$. We keep $\lambda$ small enough so that $\mathbf{c}+\mathbf{v}$ lies in the ball $b(\mathbf{c})$ in which the partial derivatives $D_{2}f$, …, $D_{n}f$ exist. Let $\mathbf{y}=y_1{\mathbf{e}}_1+\cdots +y_n{\mathbf{e}}_n$.

Write $f(\mathbf{c}+\mathbf{v})-f(\mathbf{c})$ as a telescoping sum:

$$f(\mathbf{c}+\lambda \mathbf{y})-f(\mathbf{c})=\sum _{k=1}^n\{f(\mathbf{c}+\lambda {\mathbf{v}}_k)-f(\mathbf{c}+\lambda {\mathbf{v}}_{k-1})\}$$

where ${\mathbf{v}}_0=0$, ${\mathbf{v}}_1=y_1{\mathbf{e}}_1$, ${\mathbf{v}}_2=y_1{\mathbf{e}}_1+y_2{\mathbf{e}}_2$, …, ${\mathbf{v}}_n=\mathbf{y}$.

The first term in the sum is $f(\mathbf{c}+\lambda y_1{\mathbf{e}}_1)-f(\mathbf{c})$. Since the two points $\mathbf{c}$ and $\mathbf{c}+\lambda y_1{\mathbf{e}}_1$ differ only in the first component, and since $D_{1}f(\mathbf{c})$ exists, we can write

$$f(\mathbf{c}+\lambda y_1{\mathbf{e}}_1)-f(\mathbf{c})=\lambda y_1{D}_{1}f(\mathbf{c})+\lambda y_1{E}_1(\lambda ),$$

where $E_1(\lambda )\to 0$ as $\lambda \to 0$.

For $k\ge 2$, the $k$th term in the sum is

$$f(\mathbf{c}+\lambda {\mathbf{v}}_{k-1}+\lambda y_k{\mathbf{e}}_k)-f(\mathbf{c}+\lambda {\mathbf{v}}_{k-1})=f({\mathbf{b}}_k+\lambda y_k{\mathbf{e}}_k)-f({\mathbf{b}}_k)$$

where ${\mathbf{b}}_k=\mathbf{c}+\lambda {\mathbf{v}}_{k-1}$. We could try to proceed as before: notice that ${\mathbf{b}}_k+\lambda y_k{\mathbf{e}}_k$ and ${\mathbf{b}}_k$ differ in only the $k$th component and that $D_{k}f({\mathbf{b}}_k)$ exists, so

$$f({\mathbf{b}}_k+\lambda y_k{\mathbf{e}}_k)-f({\mathbf{b}}_k)=\lambda y_k{D}_{k}f({\mathbf{b}}_k)+\lambda y_1{E}_k^{\prime }(\lambda )$$

where $E_k^{\prime }(\lambda )\to 0$ as $\lambda \to 0$. But this is not helpful; we need $D_{k}f(\mathbf{c})$, not $D_{k}f({\mathbf{b}}_k)$. This is where we use the continuity hypothesis. From the vanilla mean value theorem, we have

$$f({\mathbf{b}}_k+\lambda y_k{\mathbf{e}}_k)-f({\mathbf{b}}_k)=\lambda y_k{D}_{k}f({\mathbf{a}}_k),$$

where ${\mathbf{a}}_k\in L({\mathbf{b}}_k,{\mathbf{b}}_k+\lambda y_k{\mathbf{e}}_k)$. Note that since ${\mathbf{b}}_k\to \mathbf{c}$ as $\lambda \to 0$, ${\mathbf{a}}_k\to \mathbf{c}$ as $\lambda \to 0$. Now, since $D_{k}f$ is continuous at $\mathbf{c}$,

$$D_{k}f({\mathbf{a}}_k)=D_{k}f(\mathbf{c})+E_k(\lambda ),$$

where $E_k(\lambda )\to 0$ as $\lambda \to 0$.

We finally have

$$\begin{array}{rl}f(\mathbf{c}+\mathbf{v})-f(\mathbf{c})& =\lambda \sum _{k=1}^n{y}_k{D}_{k}f(\mathbf{c})+\lambda \sum _{k=1}^n{y}_k{E}_k(\lambda ).\\ & =\nabla f(\mathbf{c})\cdot \mathbf{v}+\Vert \mathbf{v}\Vert E(\mathbf{v}),\end{array}$$

where

$$E(\mathbf{v})=\sum _{k=1}^n{y}_k{E}_k(\Vert \mathbf{v}\Vert )\to 0\text{ as }\mathbf{v}\to 0.$$

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A sufficient condition for equality of mixed partial derivatives

Definition 3(Definition).

$$D_{r,k}\mathbf{f}\equiv D_r(D_k\mathbf{f}).$$

Theorem 4(Theorem).

If both partial derivatives $D_r\mathbf{f}$ and $D_k\mathbf{f}$ exist in an $n$-ball $B_{\delta }(\mathbf{c})$ and if both are differentiable at $\mathbf{c}$, then

$$D_{r,k}\mathbf{f}(\mathbf{c})=D_{k,r}\mathbf{f}(\mathbf{c}).$$

Proof
In light of the previous lemma, it is sufficient to prove the theorem for real valued $\mathbf{f}$. Also, since only two components are involved, it suffices to consider the case $n=2$. Assume $\mathbf{c}=(0,0)$; the same argument can be repeated for arbitrary $\mathbf{c}$.

Choose $h\ne 0$ so that the square with vertices $(0,0),(h,0),(h,h)$, and $(0,h)$ lies in the $2$-ball $B_{\delta }(0)$. Consider the quantity

$$\Delta h=f(h,h)+f(0,0)-f(h,0)-f(0,h).$$

We will show that ${\lim }_{h\to 0}\Delta h/h^2$ is equal to both $D_{2,1}f(0)$ and $D_{1,2}f(0)$.

Define $G(x)=f(x,h)-f(x,0)$ and note that

$$\Delta h=G(h)-G(0).$$

The existence of $D_{1}f$ in $L((0,h),(h,h))$ implies the differentiability of the map $x\mapsto f(x,h)$ on $[0,h]$. Similarly, the map $x\mapsto (x,0)$ is differentiable on $[0,h]$. It follows that $G(x)$ is differentiable on $[0,h]$. From the one-dimensional mean value theorem, we have

$$G(h)-G(0)=h{G}^{\prime }(x_1)=h\{D_{1}f(x_1,h)-D_{1}f(x_1,0)\},$$

where $x_1\in (0,h)$. Since $D_{1}f$ is differentiable at $(0,0)$, we have the first order Taylor formulas

$$\begin{array}{rl}{D}_{1}f(x_1,h)& =D_{1}f(0,0)+\nabla D_{1}f(0,0)\cdot (x_1,h)+\sqrt{x_1^2+h^2}{E}_1(h)\\ & =D_{1}f(0,0)+D_{1,1}f(0,0)x_1+D_{2,1}f(0,0)h+\sqrt{x_1^2+h^2}{E}_1(h),\end{array}$$

and

$$\begin{array}{rl}{D}_{1}f(x_1,0)& =D_{1}f(0,0)+\nabla D_{1}f(0,0)\cdot (x_1,0)+|x_1|E_2(h)\\ & =D_{1}f(0,0)+D_{1,1}f(0,0)x_1+|x_1|E_2(h),\end{array}$$

where $E_1(h)\to 0$ and $E_2(h)\to 0$ as $h\to 0$. So,

$$\begin{array}{r}\Delta h=h^2{D}_{2,1}f(0,0)+E(h),\end{array}$$

where $E(h)\equiv h\sqrt{x_1^2+h^2}{E}_1(h)-h|x_1|E_2(h)$. Since $|x_1|\le h$,

$$0\le |E(h)|\le h^2\sqrt{2}|E_1(h)|+h^2|E_2(h)|,$$

so

$$\underset{h\to 0}{\lim }\frac{\Delta h}{h^2}=D_{2,1}f(0,0).$$

Next, define $H(x)=f(h,x)-f(0,x)$ and note that $\Delta h=H(h)-H(0)$. The same procedure yields ${\lim }_{h\to 0}\Delta h/h^2=D_{1,2}f(0,0)$.

Theorem 5(Corollary).

If both partial derivatives $D_{r,k}\mathbf{f}$ and $D_{k,r}\mathbf{f}$ exist in an $n$-ball $B(\mathbf{c})$ and if both $D_{r,k}\mathbf{f}$ and $D_{k,r}\mathbf{f}$ are continuous at $\mathbf{c}$, then

$$D_{r,k}\mathbf{f}(\mathbf{c})=D_{k,r}\mathbf{f}(\mathbf{c}).$$

As in the previous theorem, it suffices to prove this when $n=2$ and $\mathbf{f}$ is real valued. The proof is very similar to the previous one; use a second application of MVT.