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.


2 Answers 2


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 difference quotients to the derivative for $C_c^1(\mathbb{R}^n)$ functions. The step for the second derivative follows from this case:

The key idea is to use that for $f\in C_c^1(\mathbb{R}^n)$ we find that $\nabla f$ (and also $f$) is 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 $|\partial_i f(x) -\partial_i f(y)|< \varepsilon$. Choosing $h\leq\delta$ proves the statement.

  • 2
    $\begingroup$ Basically correct, but here are some corrections and suggested improvements: (1) The wording "uniform convergence of the derivative" suggests that the derivative is converging to something. More correct would be "uniform convergence of the difference quotient [to the derivative]". (2) The claim in Evans isn't just proved similarly, it's a particular case of your result. (3) The uniform continuity of $f$ is irrelevant. The mean value theorem uses the differentiability of $f$. (4) One of the $\partial_i f(x)$ in the last paragraph should be a $\partial_i f(y)$ instead. $\endgroup$
    – epimorphic
    Commented Sep 8, 2015 at 23:10
  • $\begingroup$ Thanks for the comment. Corrected the wording. Fixed the uniform continuity $\endgroup$ Commented Sep 24, 2015 at 15:17
  • $\begingroup$ Why $f\in C^{1}_c(\mathbb{R}^n)$ implies that $\nabla f$ is uniformly continuous? $\endgroup$
    – Plantation
    Commented Mar 23 at 13:41
  • $\begingroup$ The condition $f\in C^{1}_c(\mathbb{R}^n)$ is sufficient? Isn't condition $f\in C^{2}_c(\mathbb{R}^n)$ required? $\endgroup$
    – Plantation
    Commented Mar 23 at 13:45
  • $\begingroup$ Since continuity with compact support implies uniform continuity ? $\endgroup$
    – Plantation
    Commented Mar 23 at 13:55

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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