# Convolution converges in infinity norm?

Assume $\phi$ to be a nonnegative continuous function on the real line with compact support. Also assume that integral of $\phi$ over $\mathbb{R}$ is normalized to $1$. Let $\phi_e(x) = \frac{1}{e}\phi\left(\frac{x}{e}\right)$.

I succeed to prove that $\phi_e$ is an approximation to the identity & the convoluton of $\phi_e$ with $L^p$ function $g$ converges to $g$ in $L^p$ norm.

What I really need help with is the following. Prove or disprove when $p=\infty$. I thought my proof didn't work for $p = \infty$, so I tried all night to find counterexamples. But I failed. Can anyone give an answer or idea for my question?

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Hint: Uniform limits of continuous functions are continuous. – Jose27 Oct 16 '12 at 5:35

Take $g(x)=\chi_{(0,+\infty)}(x)$. Then $h_e(x):= g\star\phi_e(x)=\int_{-\infty}^{x/e}\phi(t)dt$. We have, if $x\in (0,\delta/k)$, that $$|h(x)-h_{k^{-1}}(x)|=\int_{xk}^{+\infty}\phi(t)dt\geq \int_{\delta}^{+\infty}\phi(t)dt.$$ We choose $\delta$ such that $\int_{\delta}^{+\infty}\phi(t)dt\geq 1/2$ (if it's not possible, consider $\psi(t)=\phi(-t)$.
As the measure of $(0,\delta/k)$ is not $0$, $\lVert h-h_{k^{-1}}\rVert\geq 1/2$.
Davide, how do you define $h(x)$？ – Topoguy Feb 13 '15 at 7:22