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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.

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  • $\begingroup$ Please tell me this proof as an answer to my question. You are free to do so. By the way "Do you see any error in my proof"? Please tell me this also. $\endgroup$ – Arnab Chatterjee. Dec 24 '17 at 8:52
  • $\begingroup$ sorry I mean a proof for $f:\mathbb{R^2}\to\mathbb{R}$ to extend at $f:\mathbb{R^n}\to\mathbb{R}$ and then $f:\mathbb{R^n}\to\mathbb{R^m}$ $\endgroup$ – gimusi Dec 24 '17 at 8:54
  • $\begingroup$ Can it be extended to the case of $\mathbb R^n$ to $\mathbb R^m$? $\endgroup$ – Arnab Chatterjee. Dec 24 '17 at 8:56
  • $\begingroup$ In fact I think you can prove just for $f:\mathbb{R^n}\to\mathbb{R}$ since for $f:\mathbb{R^n}\to\mathbb{R^m}$ it's a simply extension to each component $\endgroup$ – gimusi Dec 24 '17 at 8:56
  • $\begingroup$ Ok! that's correct. $\endgroup$ – Arnab Chatterjee. Dec 24 '17 at 8:57
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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.

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  • $\begingroup$ Nice proof! But you haven't commented yet about my proof about whether you consider it as correct or not. $\endgroup$ – Arnab Chatterjee. Dec 24 '17 at 16:04
  • $\begingroup$ @ArnabChatterjee. I'm sorry but I can't fully understand the step of your proof to comment it. $\endgroup$ – gimusi Dec 24 '17 at 19:58
  • $\begingroup$ What do you not understand? $\endgroup$ – Arnab Chatterjee. Dec 25 '17 at 7:34
  • $\begingroup$ @ArnabChatterjee. It seems that you are only showing that $||Df(x)\cdot z-Df(y)\cdot z|| \to 0$ which is trivially true and do not guarantees differentiability. You need to show that $f(x+h)-f(x)-Df(x) \cdot h=r(h)$ and $\frac{|r(h)|}{|h|}\to 0$. $\endgroup$ – gimusi Dec 25 '17 at 8:38
  • $\begingroup$ Not that @gimusi. I will explain it after some time. Do you aware of Frechet differentiability? $\endgroup$ – Arnab Chatterjee. Dec 25 '17 at 9:11

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