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Let $\phi\in C^\infty_c(\mathbb R)$ be a smooth function with compact support.

For $h>0$ define the difference quotient $\phi_h\in C^\infty_c(\mathbb R)$ by $\phi_h(t)=\dfrac{\phi(t+h)-\phi(t)}{h}$.

By definition, for fixed $t\in\mathbb R$, we have $\phi_h(t)\to\phi'(t)$ as $h\to 0$.

Question: Can we conclude, that $\phi_h\to\phi'$ uniformly on $\mathbb R$ (as $h\to 0)$?

Motivation: This is used in a proof of Stone's Theorem on the existence of generators of operator groups I'm trying to understand.

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    $\begingroup$ I think the answer is yes: apply fundamental thm of calculus to write $\phi_h$ as an integral involving $\phi'$ and then use the fact that $\phi'$ is uniformly continuous (being continuous with compact support). Does that work? $\endgroup$
    – Yemon Choi
    Commented Jul 5, 2010 at 20:19
  • $\begingroup$ @Yemon Choi: Yes, it works. Thank you! $\endgroup$
    – Rasmus
    Commented Jul 5, 2010 at 20:33

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By Taylor's theorem $$\phi(t+h)=\phi(t)+h\phi'(t)+h^2\phi''(t+u(h,t)h)/2$$ where $0\le u(h,t)\le1$. So $$\phi_h(t)=\phi'(t)+h\phi''(t+u(h,t)h)/2.$$ As $\phi''$ is in $C^\infty_c$ it's pretty clear that $\phi_h\to\phi'$ uniformly.

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  • $\begingroup$ This works as well. Thank you! This is an impressive demonstration of the power of those big theorems that one is taught in a first course on analysis. $\endgroup$
    – Rasmus
    Commented Jul 5, 2010 at 20:45
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More generally, any function with uniformly continuous first derivative satisfies $$\|\phi_h-\phi'\|_\infty\leq \omega(|h|),$$ where $\omega$ is any modulus of continuity of $\phi'$ (just recall that $\phi_h(x)$ equals a value of $\phi'$ in a point within a distance $|h|$ from $x$).

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