1. ## Existence of Convolution

Show that if $\displaystyle f$ is locally integrable and that $\displaystyle g\in C^{\infty}_{c}(\mathbb{R})$, then $\displaystyle f*g(x)<\infty$.

OK, so first thing I'm thinking to do is to use the commutativity of the convolution operator and then say since $\displaystyle g$ has compact support (which in this setting means it vanishes at positive and negative infinity, or that it has support on a bounded interval), we can write the convolution in terms of a finite integral in this way:

$\displaystyle f*g = \int_{-\infty}^{\infty}f(y)g(x-y)dy$

$\displaystyle =\int_{-\infty}^{\infty}f(x-y)g(y)dy$

$\displaystyle =\int_a^bf(x-y)g(y)dy$

For some interval $\displaystyle [a, b]$.

And now, trying to figure out the bounds has me confused, since we know nothing about $\displaystyle f$ other than local integrability (so we can't say that we have less than or equal to $\displaystyle f\cdot\max\{g(x): x\in [a,b]\}$ or anything of that sort), so I'm having issues. :-/

2. Notice $\displaystyle \int_{\mathbb{R}}\left( \int_{\mathbb{R} } |f(y)g(x-y)|dx\right) dy <\infty$ so by Tonelli's theorem $\displaystyle (x,y)\mapsto f(y)g(x-y) \in L^1(\mathbb{R} ^2)$ then a standard theorem tells you that the convolution is almost everywhere finite, but since (it's easy to prove) it is continous it cannot blow up at any number (only at plus/minus infinity).