# How to prove and what are the necessary hypothesis to prove that $\frac{f(x+te_i)-f(x)}{t}\to\frac{\partial f}{\partial x_i}(x)$ uniformly?

Let $U\subset\mathbb{R}^n$ be a open set and $f:U\to\mathbb{R}$ a function in $C^\infty_c(U)$. Evans PDE book uses the following result

$$\frac{f(x+te_i)-f(x)}{t}\to\frac{\partial f}{\partial x_i}(x)\text{ uniformly as }h\to 0\;\;\;[\#]$$

The definition of uniformly convergence that I know uses sequences, but in this case no sequence is mentioned (and the book doesn’t presents a definition). I think it’s means that for each $\varepsilon>0$, there exists $\delta>0$ (which depends on $\varepsilon$ and doesn’t depend on $x$) such that

$$x,x+te_i\in U, \; 0<|t|<\delta \Rightarrow \left|\frac{f(x+te_i)-f(x)}{t}-\frac{\partial f}{\partial x_i}(x)\right|<\varepsilon$$

My first question is: is the above definition correct?

My second question is: is it enough $f\in C^1_c(U)$ to conclude $[\#]$? I think so, but some posts suggests that $f\in C^2_c(U)$ it’s necessary (see here, here and here). Then:

My third question is: is the proof below correct?

Proof of $[\#]$ (based on this post): Suppose $f\in C^1_c(U)$. Let $g:\mathbb{R}^n\to\mathbb{R}$ be an extension of $f$ defined by

$$g(x)=\left\{\begin{matrix} f(x), &\text{if } x\in U\\ 0, & \text{if }x\in\mathbb{R}^n\backslash U \end{matrix}\right.\;\;\;(1)$$

So, $g\in C^1_c(\mathbb{R}^n)$. Consequently $\frac{\partial g}{\partial x_i}\in C^1_c(\mathbb{R}^n)$. Therefore $\frac{\partial g}{\partial x_i}$ is uniformly continuous. Follows that given $\varepsilon>0$, there exists $\delta>0$ (which depends on $\varepsilon$ and doesn’t depend on $x$) such that

$$y,x\in\mathbb{R}^n,\;\|y-x\|<\delta\Rightarrow \left|\frac{\partial g}{\partial x_i}(y)-\frac{\partial g}{\partial x_i}(x)\right|<\varepsilon\;\;\;(2)$$

By Mean Value Theorem, given $x\in\mathbb{R}^n$ and $t\in\mathbb{R}$, there exists $\theta_{x,t}\in (0,1)$ such that

$$\frac{\partial g}{\partial x_i}(x+\theta_{x,t}te_i)=\frac{g(x+te_i)-g(x)}{t}\;\;\;(3)$$

Thus we can conclude that

\begin{align*} x\in\mathbb{R}^n,\;0<|t|<\delta &\Rightarrow\|(x+\theta_{x,t}te_i)-x\|<|t|<\delta\\ &\overset{(2)}\Rightarrow \left|\frac{\partial g}{\partial x_i}(x+\theta_{x,t}te_i)-\frac{\partial g}{\partial x_i}(x)\right|<\varepsilon\\ &\overset{(3)}\Rightarrow \left|\frac{g(x+te_i)-g(x)}{t}-\frac{\partial g}{\partial x_i}(x)\right|<\varepsilon \end{align*}

Particularly,

\begin{align*} x,x+te_i\in U,\;0<|t|<\delta &\Rightarrow \left|\frac{g(x+te_i)-g(x)}{t}-\frac{\partial g}{\partial x_i}(x)\right|<\varepsilon \\ &\overset{(1)}\Rightarrow \left|\frac{f(x+te_i)-f(x)}{t}-\frac{\partial f}{\partial x_i}(x)\right|<\varepsilon \end{align*}

Thanks.

#### Solutions Collecting From Web of "How to prove and what are the necessary hypothesis to prove that $\frac{f(x+te_i)-f(x)}{t}\to\frac{\partial f}{\partial x_i}(x)$ uniformly?"

Here there is the answer: the key fact is that $\frac{\partial f}{\partial x_i}$ must be uniformly continuous, which it is if $f\in C^1_c(U)$.

Actually, none of the three posts that you link in your question suggests that $f\in C^2_c(U)$ is necessary, instead they all say what I have reported here.