0
$\begingroup$

Let $f , g : {\mathbb{R}}^n \to \mathbb{R}$ (or $\mathbb{C}$) two measurable functions. We suppose that $f \in L^{\infty}({\mathbb{R}}^n)$, $\mbox{supp} f$ is a compact set in ${\mathbb{R}}^n$ and $g \in L^1(K)$ for all compact set $K$ in ${\mathbb{R}}^n$. Prove that, given $x \in {\mathbb{R}}^n$, $$ (f*g)(x) = \int_{{\mathbb{R}}^n} f(x - y) g(y) dy $$ is well defined, that is, $f$ and $g$ are non-negative or non-positive, or $\varphi \in L^1({\mathbb{R}}^n)$, where $\varphi : y \mapsto f(x - y) g(y)$ for all $y \in {\mathbb{R}}^n$. My idea is the next: we know that exists $M \in [0 , \infty)$ such that $|f(z)| \leq M$ for almost every $z \in {\mathbb{R}}^n$, because $f \in L^{\infty}({\mathbb{R}}^n)$, so, using properties about the integral, $$ |(f*g)(x)| \leq \int_{{\mathbb{R}}^n} |f(x - y)| |g(y)| dy \leq M \int_{{\mathbb{R}}^n} |g(y)| dy = M \int_{\mbox{supp} f} |g(y)| dy + M \int_{{(\mbox{supp} f)}^c} |g(y)| dy\mbox{.} $$ Now, we also know that $g \in L^1(\mbox{supp} f)$ because $\mbox{supp} f$ is a compact set in ${\mathbb{R}}^n$ and $g \in L^1(K)$ for all compact set $K$ in ${\mathbb{R}}^n$. Then $$ M \int_{\mbox{supp} f} |g(y)| dy < \infty\mbox{,} $$ so we only should prove that $$ \int_{{(\mbox{supp} f)}^c} |g(y)| dy < \infty $$ but I have no idea. Thank you very much.

  • 0
    What are your thoughts? – 2017-01-19
  • 0
    I have no idea, sorry – 2017-01-19
  • 0
    Possible duplicate of [Why is a convolution of an $L^1$ and an $L^p$ functions well-defined?](http://math.stackexchange.com/questions/1733782/why-is-a-convolution-of-an-l1-and-an-lp-functions-well-defined) – 2017-01-19
  • 0
    I think this question is different – 2017-01-19

2 Answers 2

3

Let $F$ be the support of $f$ and $x\in\mathbb{R}^n$ fixed. The function $y\mapsto f(x-y)\,g(y)$ is supported in $-F+x$, which is compact. Now $f$ is bounded and $g$ is integrable on compact sets, in particular in $-F+x$; thus, $f(x-y)\,g(y)$ is integrable.

  • 0
    Thanks, so $\varphi \in L^1(- F + x)$, and in ${\mathbb{R}}^n \setminus (- F + x)$? – 2017-01-19
  • 0
    It vanishes outside $-F+x$. – 2017-01-19
  • 0
    You say that $\varphi = 0$ on ${(- F + x)}^c$? – 2017-01-19
  • 0
    Yes, because $f$ is $0$ outside $F$. – 2017-01-19
0

I believe Folland does it this way:

$1).$ Define $\tau_y f(x)=f(x-y)$ and observe that $\left \| \tau_y f \right \|_{\infty }=\left \| f \right \|_{\infty }.$ Then,

$2).\ \left \| f* g \right \|_{1}=\left \| \int f(\cdot -y)\cdot g(y)dy \right \|_1=\left \| \int \tau_y f(\cdot)\cdot g(y)dy \right \|_1\le \left \| f \right \|_{\infty }\int |g(y)|dy=\left \| f \right \|_{\infty }\left \| g \right \|_1$