$f,g \in \mathscr{R}[-1,1] \implies f*g:[-1,1] \to \mathbb{R}$ is continuous on $[-1,1]$

Question

Let $f,g \in \mathscr{R}[-1,1]$. Prove $f*g:[-1,1] \to \mathbb{R}$ is continuous on $[-1,1]$.

Where the convolution is defined as: $\displaystyle (f*g)(x) = > \int_{-1}^1 f(y)g(x-y)dy$.

I haven't been able to figure this out, but I have a few thoughts which are described below.

I think we can do away with $\epsilon-\delta$ proof. So my goal is to show for a given $x_0 \in[-1,1]$, $|(f*g)(x)-(f*g)(x_0)| \lt \epsilon$.

Now, there is a lemma that I can perhaps use:

Lemma:

Let $f \in \mathscr{R}[a,b]$. For every $\epsilon \gt 0$, $\exists$ a continuous function $g:[a,b] \to \mathbb{R}$, s.t. $\displaystyle\int_a^b |f-g| \lt \epsilon$.

$$ \begin{split} \displaystyle |(f*g)(x_)-(f*g)(x_0)| & = \left| \int_{-1}^1 f(y)g(x-y)dy - \int_{-1}^1 f(y)g(x_0-y)dy \right| \\ & = \left| \int_{-1}^1 f(y)\bigl(g(x-y) - g(x_0-y)\bigl) dy \right| \\ & \leq M\left| \int_{-1}^1 \bigl(g(x-y) - g(x_0-y)\bigl) dy \right| \end{split} $$

Where $M \gt 0$, the bound of $f$ since $f \in \mathscr{R}[-1,1]$.

Now my goals is to show $\displaystyle \left| \int_{-1}^1 \bigl(g(x-y) - g(x_0-y)\bigl) dy \right| \lt \frac{\epsilon}{M}$.

Now $g$ is not necessarily continuous so picking a $\delta \gt 0$ s.t. $|x-x_0|\lt\delta$ won't do much good here.

Using the lemma above, I can come up with a continuous function $g_c$ s.t. $\int_{-1}^1 |g-g_c| \lt \epsilon$, or whatever positive quantity we desire. But using the triangle inequality I end up with extra positive terms added.

Can I get some hints, please?

Should I continue in this direction? That is, can I come up with a $g_c$ a continuous function, that will help me show the above quantity to be $\lt \epsilon/M$?

Or should I take a different approach altogether? That is, use sequences (which I tried, but I couldn't get very far with it).

Thank you

Answer 1

Here is my full answer after collecting all the tips and helps I have gotten so far:

Let $\epsilon \gt 0$ be given. $f \in \mathscr[-1,1]$, then $\exists M \gt 0$ s.t. $|f| \leq M$.

By the lemma mentioned in the question, $\exists g_c \in C[-1,1]$ s.t. $\displaystyle \int_{-1}^1 |g-g_c| \lt \frac{\epsilon}{3M}$.

Since $g_c$ is continuous on a compact interval, it is uniformly continuous. So $\exists \delta \gt 0$, s.t. $|x-x_0| \lt \delta \implies |g_c(x) - g_c(x_0)| \lt \frac{\epsilon}{6M}$.

Let $x_0 \in [-1,1]$ be s.t. $|x-x_0| \lt \delta$, then

$$ \displaystyle \begin{split} |(f*g)(x) - (f*g)(x_0)| & = \left|\int_{-1}^1 f(y)g(x-y)dy - \int_{-1}^1 f(y)g(x_0-y)dy \right| \\ & = \left| \int_{-1}^1 f(y) \bigl[ g(x-y)-g(x_0-y) \bigl] dy \right| \\ & \leq M \left| \int_{-1}^1 \bigl[ g(x-y)-g(x_0-y) \bigl] dy \right| \\ & \leq M \int_{-1}^1 \bigl| g(x-y)-g(x_0-y) \bigl| dy \\ & = M \int_{-1}^1 \bigl| g(x-y)-g_c(x-y)-g(x_0-y)+g_c(x_0-y)+g_c(x-y)-g_c(x_0-y) \bigl| dy \\ & \leq M \int_{-1}^1 \Bigl( \bigl|g(x-y)-g_c(x-y)\bigl|+\bigl|g(x_0-y)-g_c(x_0-y)\bigl|+\bigl|g_c(x-y)-g_c(x_0-y)\bigl| \Bigl) dy \\ & = M \left[ \int_{-1}^1 \bigl|g(x-y)-g_c(x-y)\bigl| dy + \int_{-1}^1 \bigl|g(x_0-y)-g_c(x_0-y)\bigl| dy + \int_{-1}^1 \bigl|g_c(x-y)-g_c(x_0-y)\bigl| dy \right] \\ & \lt M \left[ \frac{\epsilon}{3M} + \frac{\epsilon}{3M} + \int_{-1}^1 \frac{\epsilon}{6M} \right] \\ & = M \left[ \frac{\epsilon}{3M} + \frac{\epsilon}{3M} + \frac{\epsilon}{3M} \right] \\ & = \epsilon \end{split} $$

Answer 2

You have extend the functions beyond $[-1,1]$ by periodicity so that the definition of $f*g$ makes sense.

You only need the fact that $g_c$ is uniformly continuous. There exists $\delta >0$ such that $|g_c(x-y)-g_c(x_0-y)| <\frac {\epsilon } {2M}$ whenever $|x-x_0| <\delta$.