Show the uniform convergence of the partial derivatives

Question

Let $f\in C_1\left(\mathbb{R}^n,\mathbb{R}\right)$, is it true that if $[a,b]$ is a closed interval and $\left(x_2,...,x_n\right)$ is fix, then for all $\varepsilon>0$ there exists $\delta$ such that: $$ \left|\frac{f\left(x_1+\delta,x_2,...,x_n\right)-f\left(x_1,x_2,...,x_n\right)}{\delta}-\frac{\partial}{\partial x_1}f\left(x_1,x_2,...,x_n\right)\right|<\varepsilon $$ for all $x_1\in[a,b]$. If yes, how to prove it? If no, are there similar but weaker statements which are true? Or are there even stronger ones?

In a way this expresses uniform convergence of the partial derivatives. But I wasn't able to remove the dependency on $x_1$ which is exactly the purpose of this assertion.

Answer 1

It can be proofed in a closed interval though this problem doesn't have to be a partial derivative because all other variables are fixed.

problem: let $f\in C_1\left(\mathbb{R},\mathbb{R}\right)$ and $[a,b]$ is a closed interval. For each $\epsilon \gt 0$, there exists a $\delta$ s.t if $|x-x_1| \lt \delta$, $|\frac {f(x)-f(x_1)}{x-x_1}-\frac{df(x_1)}{dx}| \lt \epsilon$ for all $x_1 \in [a,b]$

We understand that, by definition, there exists such a $\delta$ for each $x_1$ or $\delta(x)$ is defined in closed interval $[a,b]$. Our question is not changed to if there exists a $\delta$ s.t. $\delta(x) \ge \delta$ in interval $[a,b]$ This is true because the function is a $C_1$ function.

Answer 2

This result is true. Let $g:\mathbb{R}\rightarrow\mathbb{R}$ be defined by $g(x)=\frac{\partial}{\partial x_1}f(x_1,x_2,...,x_n)$. $g$ is continuous and hence $g$ restricted to the compact set $[a,b+1]$ is uniformly continuous. Therefore, given $\epsilon >0$, $\exists 1>\delta >0$ such that $\forall y,z\in[a,b+1]$, $|y-z|<\delta \implies |g(y)-g(z)|< \epsilon$. Now, $\forall x_1\in[a,b]$, by the legrange mean value theorem, we have $$\left|\frac{f\left(x_1+\delta,x_2,...,x_n\right)-f\left(x_1,x_2,...,x_n\right)}{\delta}-\frac{\partial}{\partial x_1}f\left(x_1,x_2,...,x_n\right)\right|$$ $$=\left|\frac{\partial}{\partial x_1}f\left(x_1+\gamma,x_2,...,x_n\right)-\frac{\partial}{\partial x_1}f(x_1,x_2,...,x_n)\right| (where\gamma\in(0,\delta))$$ $$=|g(x_1+\gamma)-g(x_1)|$$ $$<\epsilon$$