Let $f: E \subset_{open} \mathbb R^n \longrightarrow \mathbb R^m$ be a differentiable function on $E$. Then $f \in \mathcal C'(E)$ if and only if each $D_j f_i$ exists and is continuous on $E$ for $1 \leq i \leq m$, $1 \leq j \leq n$, where $f_i$'s are the components of $f$.
I have proved "$\implies$" part as it was in Rudin's "Principles of Mathematical Analysis". But I have tried to prove "$\impliedby$" part in some other way. Here's my attempt $:$
Since we know that any two norms in $\mathbb R^m$ are equivalent so if we can prove the result for the usual or Euclidean norm on $\mathbb R^m$ then we are done. Since each $D_jf_i$ exists and is continuous on $E$. We fix $x \in E$. Then for a given $\epsilon>0$ we can find a $\delta>0$ such that $|D_jf_i(y) – D_jf_i(x)| < \frac {\epsilon} {n \sqrt m}$ whenever $\|y-x\| < \delta$ for any $1 \leq i \leq m$, $1 \leq j \leq n$. Take any $z \in \mathbb R^n$ with $\|z\| < 1$. Let $z = \sum_{j=1}^{n}z_je_j$ where $e_j$ is the $j$-th coordinate vector of $\mathbb R^n$ for$1 \leq j \leq n$. Then clearly $|z_j| < 1$ for $1 \leq j \leq n$. If $u_i$ is the $i$-th coordinate vector of $\mathbb R^m$ for $1 \leq i \leq m$ then we have $$\left \|[f'(y)-f'(x)](z) \right \| = \left \|\sum_{i=1}^m \left \{\sum_{j=1}^{n} [D_jf_i(y) – D_jf_i(x)]z_j \right \} u_i \right \|$$ $$\implies \|[f'(y)-f'(x)](z) \| = \left [\sum_{i=1}^{m} \left \{\sum_{j=1}^{n} [D_jf_i(y)-D_jf_i(x)]z_j \right \}^{2} \right ]^{\frac {1} {2}} < \epsilon$$ whenever $\|y-x\| < \delta$. This proves that $\|f'(y)-f'(x)\|_{op} < \epsilon$ whenever $\|y-x\| < \delta$. This proves that $f'$ is continuous at $x$. Since we fixed $x \in E$ arbitrarily so $f'$ is continuous on $E$ and therefore $f$ is continuously differentiable on $E$ i.e. $f \in \mathcal C'(E)$.
This completes the proof.
Is my above reasoning correct at all? Please check it.
Thank you in advance.
Best Answer
Consider $f:\mathbb{R^2}\to\mathbb{R}$
$$f(x_0+h,y_0+k)-f(x_0,y_0)=\\=f(x_0+h,y_0+k)-f(x_0+h,y_0)+f(x_0+h,y_0)-f(x_0,y_0)=$$
$$\text{by MVT} \quad\exists c,d\in\mathbb{R} \quad \text{such that}$$
$$=kf_y(x_0+h,y_0+c)+hf_x(x_0+d,y_0)=$$
$$=kf_y(x_0,y_0)+k[f_y(x_0+h,y_0+c)-f_y(x_0,y_0)]+\\+hf_x(x_0,y_0)+h[f_x(x_0+d,y_0)-f_x(x_0,y_0)]=$$
$$=hf_x(x_0,y_0)+kf_y(x_0,y_0)+r(h,k)$$
with:
$$r(h,k)=h[f_x(x_0+d,y_0)-f_x(x_0,y_0)]+k[f_y(x_0+h,y_0+c)-f_y(x_0,y_0)]$$
and
$$\lim_{(h,k)\to(0,0)} \frac{r(h,k)}{\sqrt {h^2+k^2}}=0 \quad \square$$
The same proof can easily extend to $f:\mathbb{R^n}\to\mathbb{R}$ by adding and subtracting n-1 terms in order to apply MVT $n$ times.
The extension to $f:\mathbb{R^n}\to\mathbb{R^m}$ is immediate by the same proof for each component.