Proof that continuous partial derivatives implies differentiability

To simplify life, use the $\|\cdot\|_1$ norm. To reduce clutter, let $\phi_k(h) = (h_1,...,h_k,0,...0)$. Let $\phi_0(h) = 0$, and note that $\|\phi_k(h)-\phi_j(h)\|_1 \le \|h\|_1$.

First suppose $m=1$.

Let $\epsilon>0$.

By continuity, we can choose $\delta>0$ such that $|D_kf(a+h)-D_kf(a)| < \epsilon$ for all $k$ and $\|h\| < \delta$.

Let $A=(D_1f(a),...,D_n f(a))$ and suppose $\|h\| < \delta$, then we have \begin{eqnarray} |f(a+h)-f(a)-Ah| &=& |\sum_{k=1}^n (f(a+\phi_k(h))-f(a+\phi_{k-1}(h))-D_kf(a)h_k)| \\ &\le& \sum_{k=1}^n |f(a+\phi_k(h))-f(a+\phi_{k-1}(h))-D_kf(a)h_k| \end{eqnarray} By the mean value theorem, there are $c_k \in [a+\phi_{k-1}(h), a+\phi_k(h)]$ (that is, each $c_k$ lies on the line segment) such that $f(a+\phi_k(h))-f(a+\phi_{k-1}(h)) = D_k f(c_k) h_k$. Note that $\|c_k -a\|_1 \le \|h\|_1$. Continuing: \begin{eqnarray} |f(a+h)-f(a)-Ah| &\le& \sum_{k=1}^n |D_k f(c_k) h_k-D_kf(a)h_k| \\ &=& \sum_{k=1}^n |D_k f(c_k) -D_kf(a)||h_k| \\ &<& \epsilon \sum_{k=1}^n |h_k| \\ &=& \epsilon \|h\|_1 \end{eqnarray} Since $\epsilon>0$ was arbitrary, this shows that $f$ is differentiable at $a$ and $Df(a)h = Ah$.

It is straightforward to show that if $f_1,...f_m$ are differentiable, then so is $f(x) = (f_1(x),...,f_n(x))$, and $Df(x)h = (Df_1(x)h, ..., D f_m(x)h)$.