# A proof of a sufficient condition to have that $f : A\subset\mathbb{R}^n\to\mathbb{R}$ is differentiable

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 !

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

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 $$0 Thus $$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.