1
$\begingroup$

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 $\mathbb R$ ?

Thanks,

$\endgroup$
  • 2
    $\begingroup$ 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
2
$\begingroup$

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 $(f * \phi)(x) = \frac{1}{n^3}>0$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.