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What are the TVM hypotheses that you are using? I think, $E$ should be a convex set and $f$ differentiable or Am I confused? $\textbf{My question is: Can I use TVM here? Is it version of TVM?:}$

I know that: Suppose $f$ maps a convex open set $E \subset \mathbb{R}^{n}$ into $\mathbb{R}^{m}$, $f$ differentiable in $E$, and there is a real number $M$ such that $$||f'(x)||\leq M$$ for every $x \in E$. Then $$||f(b)-f(a)||\leq M ||b-a||$$ for all $a,b \in E$.

Partial derivatives bounded implies continuity

Now,

$\textbf{An attempt:}$ Let $\vec{x},\vec{y}\in E$

Let's see that $$||f(\vec{x})-f(\vec{y})||\leq ||f(x_{1},x_{2},...,x_{n})-f(y_{1},x_{2},...,x_{n})||+\cdots+||f(y_{1},...,y_{n-1},x_{n})-f(y_{1},...,y_{n})||$$

Now eventually bound $\| f(\vec{x}) - f(\vec{y})\|$ by $n$ many such quantities. Notice that each of them are of the form $\|f(\vec{w_i}) - f(\vec{z_i})\|$ where $\vec{w_i}, \vec{z_i}$ differ only at one coordinate, say the $i$-th one. Now the $i$-th partial derivative is bounded, say by $M_i.$ So by MVT ($\textbf{I don't sure about it}$) we get $$\| f(\vec{w_i}) - f(\vec{z_i})\| \le M_i \|\vec{w_i} - \vec{z_i}\| \leq M_i \|\vec{x} - \vec{y}\|.$$ Hence $$\|f(\vec{x}) - f(\vec{y})\| \leq \sum_{i=1}^n \|f(\vec{w_i})-f(\vec{z_i})\| \leq \sum_{i=1}^n M_i \|\vec{x} - \vec{y}\| = M \|\vec{x} - \vec{y}\|.$$So, $f$ is Lipschitz $\implies$ $f$ is continuous function?

$\textbf{I don't understand if I can use the TVM in this problem how I used.}$ The problem you can find it: math.stackexchange.com/q/1510457/798113 In my attempt I used TVM but I am not sure if I have all the hypothesis to used it (TVM). For example I need that $E$ be convex set, and $f$ differentiable function but I don't have (I think that) it hypothesis.or maybe need I to use other version of TVM?

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    What are you trying to prove? The mean-value theorem? THe fact that Lipschitz implies continuity? You need to clarify what your question is. – peek-a-boo Aug 09 '20 at 11:48
  • I don't understand if I can use the TVM in this problem how I used. –  Aug 09 '20 at 11:49
  • The problem you can find it: https://math.stackexchange.com/q/1510457/798113 –  Aug 09 '20 at 11:50
  • In my attempt I used TVM but I am not sure if I have all the hypothesis to used it (TVM). For example I need that $E$ be convex set, but I don't have it hypothesis. –  Aug 09 '20 at 11:53
  • or maybe need I to use other version of TVM? –  Aug 09 '20 at 11:55
  • Yes, the idea is right (though all these comments should really be part of your question). About convexity, note that continuity is a local property, which means you can prove it pointwise. So, pick a point $x_0\in E$, then since $E$ is open, there is an open ball $B$ such that $x_0\in B\subset E$. Since $B$ is convex, you can apply the mean-value theorem there. By your reasoning, $f$ is Lipschitz on $B$ (hence continuous at $x_0$). Since $x_0\in E$ was arbitrary, the proof is done. – peek-a-boo Aug 09 '20 at 11:56
  • and the hypothesis of differentiation in $f$ in the TVM? –  Aug 09 '20 at 12:00
  • I understand now, about the convex set and the hypothesis. –  Aug 09 '20 at 12:00
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    Also, regarding differentiability, note that while $f$ may not be differentiable, all its partial derivatives exist on $E$ (and bounded), so for example, consider the curve $\gamma(t)=(x_1+t(y_1-x_1), x_2, \dots, x_n)$, with $t\in[0,1]$, and the composite function $f\circ \gamma$. Then, $f\circ\gamma$ is differentiable because of the hypothesis that all partial derivatives of $f$ exist. Now, apply the MVT to $f\circ \gamma$. Apply similar reasoning for all the other terms. – peek-a-boo Aug 09 '20 at 12:01

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