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Consider $f : O\subset\mathbb{R}^{n}\to\mathbb{R}$ a function continuously differentiable on $O$ an open set of $\mathbb{R}^n$ that is all its partial derivatives exist on $O$ and they are continuous. We want to show that it is a sufficient condition to have that $f$ is differentiable on $O$.

To do this, we consider the open set $N(X_0, r)$ where $X_0\in O$. The idea is to use the theorem discussed here : Kind of Taylor expansion for functions of several variables?

This theorem says the following :

Consider $f:A\subset\mathbb{R}^n\to\mathbb{R}$ where all its partial derivatives exists on the open ball $N(X_o, r)$ with $X_0\in A$. Consider $Z$ a vector of $\mathbb{R}^n$ with $\lVert Z\rVert\leq r$. Then we have

$f(X_0 + Z) = f(X_0) + \sum_{i}^{n}f_{x_i}^{'}(X_0 + V_i)z_i$ with $V_i=(z_1, ..., z_{i-1}, \theta z_{i}, 0, ..., 0),\quad 0<\theta<1$

Now, consider $Z\in\mathbb{R}^n$ such that $\lVert Z\rVert < r$ . Then, we can use the theorem above at the point $X_0$, it follows that $f(X_0 + Z) = f(X_0) + \sum_{i=1}^{n}f_{x_i}^{'}(X_0 + V_i)z_i$. If we consider the approximation $f_{x_i}^{'}(X_0)$ instead of $f_{x_i}^{'}(X_0 + V_i)$ we can write the following : $\epsilon(X_0, Z) = f(X_0 + Z) - f(X_0) - \sum_{i=1}^{n}f_{x_i}^{'}(X_0)z_i $

And then we make appear the equality of interest for the differentiability that is : $\epsilon_1(X_0, Z) =\frac{1}{\lVert Z\rVert}\left[ f(X_0 + Z) - f(X_0) - \sum_{i=1}^{n}f_{x_i}^{'}(X_0)z_i\right] $

Using the fact that $f(X_0 + Z) - f(X_0) = \sum_{i}^{n}f_{x_i}^{'}(X_0 + V_i)z_i$ we have :

$\lvert\epsilon_1(X_0, Z)\rvert =\left\lvert\frac{1}{\lVert Z\rVert}\left[ \sum_{i}^{n}f_{x_i}^{'}(X_0 + V_i)z_i - \sum_{i=1}^{n}f_{x_i}^{'}(X_0)z_i\right]\right\rvert$

$\quad\quad\quad\quad\quad = \left\lvert\sum_{i=1}^{n}\frac{z_i}{\lVert Z\rVert}\left[f_{x_i}^{'}(X_0 + V_i) - f_{x_i}^{'}(X_0)\right]\right\rvert $

$\quad\quad\quad\quad\;\;\;\leq\sum_{i=1}^{n}\left\lvert\frac{z_i}{\lVert Z\rVert}\right\rvert\left\lvert\left[f_{x_i}^{'}(X_0 + V_i) - f_{x_i}^{'}(X_0)\right]\right\rvert\leq\sum_{i=1}^{n}\left\lvert\left[f_{x_i}^{'}(X_0 + V_i) - f_{x_i}^{'}(X_0)\right]\right\rvert $

But $\forall 1\leq i\leq n : \lVert V_i\rVert\leq\lVert Z\rVert\implies\lim_{Z\to 0_{\mathbb{R}^n}} V_i =0$ Thus using the continuity of the absolute value and of $f_{x_i}^{'}$ at $X_0$ we get $\lim_{Z\to 0_{\mathbb{R}^n}}\lvert\epsilon_1(X_0, Z)\rvert = 0 $

And this holds for all $X\in O$, which concludes the proof.

Is this seems correct or do you see some improvement possible for this proof ?

Thank you a lot !

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    $\begingroup$ I read through the proof and it seems correct to me. $\endgroup$ Jan 2, 2023 at 23:55
  • $\begingroup$ Thank you a lot ! $\endgroup$
    – coboy
    Jan 3, 2023 at 0:32

1 Answer 1

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Take two points in your region, say $A(a_1, ... ,a_n),B=(b_1, ... ,b_n)$. Let $\mathbf v=[b_1-a_1, ... ,b_n-a_n]$, Then define $$g: \mathbb R \to \mathbb R \text{ by } g(t)=f(A+t \mathbf v)$$ Then $g(0)=f(A),g(1)=f(B)$. The chain rule is valid under your assumption of the existence of partial drivatives and gives $$g^{\prime}(t)=\mathbf v \bullet \nabla f(A+t \mathbf v).$$ The mean-value theorem for functions of one variable gives $g(1)-g(0)=g^{\prime}(t_*)$ for some $0<t_*<1.$ Thus $$f(B)-f(A)=\mathbf v \bullet \nabla f(A+t_*\mathbf v).$$ Note that the point $A+t_*\mathbf v $ is on the line joining $A$ and $B$ and is closer to $A$ then $B$ is. This preliminary geometry should give you what you need to finish the proof.

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  • $\begingroup$ Thank you a lot for your answer really. The geometry is clear now ! $\endgroup$
    – coboy
    Jan 3, 2023 at 10:46

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