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I'm currently reading Evans' PDE book. In it he claims that for $f \in C^2_c(\mathbb{R}^n)$ $$\frac{f(x + he_i) - f(x)}{h} \to \frac{\partial}{\partial x_i}f(x)$$ and $$\frac{\frac{\partial}{\partial x_i}f(x + he_j) - \frac{\partial}{\partial x_i}f(x)}{h} \to \frac{\partial^2}{\partial x_jx_i}f(x)$$ uniformly as $h \to 0$.

My question is why must the convergence be uniform?

Thanks in advance.

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2 Answers

up vote 1 down vote accepted

Using Taylor's formula, you get for the first claim :

$$|\frac{f(x+h e_i)-f(x)}{h} - \frac{\partial f}{\partial x_i}(x)| \leq |h| \sup |D^2f|$$

the supremum being finite and well defined since $f$ is $C^2$ with compact support. That gives uniform convergence.

If you add another order of regularity to your hypothesis, the same argument works for the second part. However, as it is, it doesn't work.

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I know that this, is an old question but one can improve the answer to the case mentioned in Evans book. In particular, one can show the uniform convergence of the derivative for $C_c^1(\mathbb{R}^n)$ functions. The step for the second derivative is then similar:

The key idea is to use that for $f\in C_c^1(\mathbb{R}^n)$ we find that $f$ and $\nabla f$ are uniformly continuous. Hence, we receive with the mean value theoren:

$$| \frac{f(x+he_i)-f(x)}{h} -\partial_i f(x)| = |\partial_i f(y) -\partial_i f(x)| $$ for some $y$ in the line from $x$ to $x+he_i$ and consequently for some $y\in B_h(x)$.

Now let $\varepsilon >0$ be given and $x\in \mathbb{R}^n$ be arbitrary. Then, by the uniform continuity there is a $\delta >0$ (independent of $x$) such that for all $y\in B_\delta(x)$ we find $|\partial_i f(x) -\partial_i f(x)|< \varepsilon$. Choosing $h\leq\delta$ proves the statement.

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