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) 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 $h(z):=f(t)$.

A: For each $x\in\mathbb{R}$ we define $h_{x}:\mathbb{R}\to\mathbb{R}$ by $h_{x}(t)=f(x-t)g(t)$. Observe that $h_{x}(t)=0$ except possibly on $[a,b]$ and $h_{x}$ is continuous being the product of continuous functions. Let $C_{i}=[-i,i]$. Then $C_{i}\subset{}\text{Int}\left(C_{i+1}\right)$ and so we have
\begin{align*}
\left(f*g\right)(x)=\int_{\mathbb{R}}h_{x}(t)dt=\lim_{i\to\infty}\int_{-i}^{i}h_{x}(t)dt
\end{align*}
exists and is finite provided we can show that $\int_{-i}^{i}\left\lvert h_{x}(t)\right\rvert dt$ is uniformly bounded. To see that this is the case observe that for all sufficiently large $i$ we must have $[a,b]\subset[-i,i]$ by compactness. We conclude that for all sufficiently large $i$ that
\begin{align*}
\int_{-i}^{i}\left\lvert h_{x}(t)\right\rvert dt&=\int_{a}^{b}\left\lvert h_{x}(t)\right\rvert dt=\int_{a}^{b}\left\lvert f(x-t)\right\rvert\left\lvert g(t)\right\rvert{}dt\\
&=\int_{a}^{b}\left\lvert f(x-t)\right\rvert\left\lvert{}g(t)\right\rvert{}dt\\
&\le\left\lvert\left\lvert f\right\rvert\right\rvert_{L^{\infty}\left([x-b,x-a]\right)}\left\lvert\left\lvert g\right\rvert\right\rvert_{L^{\infty}\left([a,b]\right)}(b-a)
\end{align*}
which means that $\int_{-i}^{i}\left\lvert h_{x}(t)\right\rvert{}dt$ is uniformly bounded. Note also that the integral agrees with integration over $[a,b]$ since for all sufficiently large $i$ in the limit this is true. In particular we observe that
\begin{align*}
\left\lvert\left(f*g\right)(x)\right\rvert&\le\int_{a}^{b}\left\lvert f(x-t)\right\rvert\left\lvert{}g(t)\right\rvert{}dt=\int_{x-b}^{x-a}\left\lvert f(t)\right\rvert\left\lvert{}g(x-t)\right\rvert{}dt\\
&\le\left\lvert\left\lvert g\right\rvert\right\rvert_{L^{\infty}\left([a,b]\right)}\int_{x-b}^{x-a}\left\lvert f(t)\right\rvert dt
\end{align*}
which means that if $f\in L^{1}\left(\mathbb{R}\right)$ then the convolution tends to $0$ as $x\to\infty$.
