It is well-known that, if $f$ and $g$ are compactly supported continuous functions, then their convolution exists, and is also compactly supported and continuous (Hörmander 1983, Chapter 1).

Next, Suppose $f\in L^{1}(\mathbb R)$ is given.

My Question is:

Can we expect to choose $\phi \in C_{c}^{\infty}(\mathbb R)$ with $\int_{\mathbb R}\phi(t)dt=1$ and the support of $f\ast \phi$ is contained in a compact set, that is, $\operatorname{supp} f\ast \phi \subset K;$ where $K$ is some compact set in $\mathbb R$ ?


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    $\begingroup$ Do you have an example of a non-negative, continuous, integrable $f$ and a compactly-supported, non-negative, continuous function $\phi$ with unit integral such that the support of their convolution is compact? $\endgroup$ – Andrew D. Hwang Sep 4 '14 at 12:33

Not in general. Let Consider the following function, where $n = 1,2,...$:

$$f(x) = \begin{cases} 0 & \text{ if } x < 0 \\ \frac{1}{n^3}& \text{ if }\frac{n^2-n}{2} \leq x < \frac{n^2+n}{2}\end{cases}$$

This function is $L^1$ with total integral $\frac{\pi^2}{6}$. However, it is constant and non-zero on arbitrarily long intervals. Thus for any $\phi$ as specified, there is sufficiently large $N$, such that for all $n >N$, there is some $x$ in the interval $\frac{n^2-n}{2} < x<\frac{n^2+n}{n}$ with $(f * \phi)(x) = \frac{1}{n^3}>0$.

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  • $\begingroup$ Could you justify the existence of this $x$ that makes $(f*\phi)(x)$ be $1/n^3$? $\endgroup$ – Rodolfo Ferreira de Oliveira Sep 24 '19 at 7:25

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