Caratheodory Criterion

Appendix C. Caratheodory Criterion

We now prove that the Caratheodory criterion in Theorem 2.3.3 is equivalent to the usual definition of differentiability.

Theorem C.0.1. Caratheodory Criterion.

Let \(A \subseteq \R^n\text{,}\) let \(\va \in A\) be an interior point, and let \(f \colon A \to \R^m\text{.}\) Then the following are equivalent.

\(f\) is differentiable at \(\va\text{.}\)
There exists a matrix-valued function \(M(\vx)\text{,}\) defined near \(\va\) and continuous at \(\va\text{,}\) such that

\begin{equation*} f(\vx)-f(\va)=M(\vx)(\vx-\va) \end{equation*}

for all \(\vx\) near \(\va\text{.}\)

When these conditions hold, \(Df(\va)=M(\va)\text{.}\)

Proof.

Suppose first that \(f\) is differentiable at \(\va\text{.}\) Then there exists a linear map \(L \colon \R^n \to \R^m\) such that

\begin{equation*} f(\vx)-f(\va)=L(\vx-\va)+\vr(\vx), \end{equation*}

where

\begin{equation*} \lim_{\vx \to \va}\frac{\norm{\vr(\vx)}}{\norm{\vx-\va}}=0. \end{equation*}

Write \(L=(\ell_{ij})\) in matrix form and \(\vr(\vx)=\langle r_1(\vx),\ldots,r_m(\vx)\rangle\text{.}\) For \(\vx \ne \va\text{,}\) define

\begin{equation*} M_{ij}(\vx) := \ell_{ij} + r_i(\vx)\frac{x_j-a_j}{\norm{\vx-\va}^2}, \end{equation*}

and set \(M_{ij}(\va):=\ell_{ij}\text{.}\) Let \(M(\vx)=(M_{ij}(\vx))\text{.}\) Then for each component \(i\text{,}\)

\begin{align*} \sum_{j=1}^n M_{ij}(\vx)(x_j-a_j) \amp = \sum_{j=1}^n \ell_{ij}(x_j-a_j)\\ \amp\phantom{{}={}} + r_i(\vx)\sum_{j=1}^n \frac{(x_j-a_j)^2}{\norm{\vx-\va}^2}\\ \amp = \sum_{j=1}^n \ell_{ij}(x_j-a_j)+r_i(\vx). \end{align*}

Therefore

\begin{align*} M(\vx)(\vx-\va) \amp = L(\vx-\va)+\vr(\vx)\\ \amp = f(\vx)-f(\va). \end{align*}

It remains to show that \(M\) is continuous at \(\va\text{.}\) For \(\vx \ne \va\text{,}\)

\begin{equation*} |M_{ij}(\vx)-\ell_{ij}| = \left|r_i(\vx)\frac{x_j-a_j}{\norm{\vx-\va}^2}\right| \le \frac{|r_i(\vx)|}{\norm{\vx-\va}} \le \frac{\norm{\vr(\vx)}}{\norm{\vx-\va}} \to 0. \end{equation*}

Hence each entry \(M_{ij}(\vx)\) tends to \(\ell_{ij}\text{,}\) so \(M(\vx) \to L\) as \(\vx \to \va\text{.}\) Thus \(M\) is continuous at \(\va\text{.}\)

Conversely, suppose

\begin{equation*} f(\vx)-f(\va)=M(\vx)(\vx-\va), \end{equation*}

where \(M(\vx)\) is continuous at \(\va\text{.}\) Let \(L:=M(\va)\text{.}\) Then

\begin{equation*} f(\vx)-f(\va)-L(\vx-\va) = (M(\vx)-M(\va))(\vx-\va). \end{equation*}

Therefore

\begin{equation*} \frac{\norm{f(\vx)-f(\va)-L(\vx-\va)}}{\norm{\vx-\va}} \le \norm{M(\vx)-M(\va)}_{\mathrm{op}}. \end{equation*}

Since \(M(\vx)\to M(\va)\text{,}\) the right-hand side tends to \(0\text{.}\) Hence

\begin{equation*} \lim_{\vx \to \va} \frac{\norm{f(\vx)-f(\va)-L(\vx-\va)}}{\norm{\vx-\va}}=0, \end{equation*}

so \(f\) is differentiable at \(\va\text{.}\) By uniqueness of the derivative, \(Df(\va)=L=M(\va)\text{.}\)

Corollary C.0.2. Scalar-Valued Form.

Let \(f \colon A \to \R\text{.}\) Then \(f\) is differentiable at \(\va\) if and only if there exist functions \(A_1,\ldots,A_n\text{,}\) continuous at \(\va\text{,}\) such that

\begin{equation*} f(\vx)-f(\va)=\sum_{i=1}^n A_i(\vx)(x_i-a_i) \end{equation*}

for \(\vx\) near \(\va\text{.}\) In that case,

\begin{equation*} Df(\va)(\vh)=\sum_{i=1}^n A_i(\va)h_i. \end{equation*}

Proof.

This is exactly the case \(m=1\) of the theorem above. The \(1 \times n\) matrix \(M(\vx)\) is simply the row vector \(\langle A_1(\vx),\ldots,A_n(\vx)\rangle\text{.}\)