Showing a convolution is well defined Let $f: \Bbb{R}\to \Bbb{R}$ and $g: \Bbb{R}\to \Bbb{R}$ be continuous functions, where $g(x) = 0$ for all $x \notin [a,b]$ for some interval $[a,b]$.
a) Show that the convolution $$(g*f)(x):=\int_{-\infty}^{\infty} f(t)g(x-t) \,dt$$ is well defined.
I understand how to use change of variables to simplify the problem to a non-infinite integral, but I am not sure how to show that it is well-defined. I am struggling to understand how the definition of well-defined and how to show it.
b) Suppose $$\int_{-\infty}^{\infty} \lvert f(t) \rvert \,dt < \infty$$ Prove that $\lim_{x\to-\infty} (g*f)(x)=0$ and $\lim_{x\to\infty} (g*f)(x)=0$.
I understand that there is some relation to a sequence/series which should make this straightforward, however, I cannot figure out how to get there.
Any help would be greatly appreciated!

a) A function $f:A\to B$ is well-defined if for each $x\in A$ then $f(x)\in B$ and $f(x)$ is unique.
If $f$ and $g$ are continuous and the improper integral can be written as a proper integral of Riemann in a compact set of the domain of $f$ and $g$ then the integral exists, so the convolution is well-defined.
b) Use the Cauchy-Schwarz inequality (rewrite the convolution as an inner product in a suitable space) together with the observation that if $\int_{-\infty}^\infty |f(x)|\,\mathrm dx<\infty$ then $\lim_{x\to\pm\infty}f(x)=0$.

 Observe that $$\langle h,g\rangle:=\int_a^b h(z)g(z)\,\mathrm dz$$ define an inner product for bounded real-valued functions defined in $[a,b]$. Then it remains to set $z:=x-t$ and $h(z):=f(t)$.

