Given that $f$ and $g$ belong to $L^2(\mathbf{R})$, how can I show that $$ H(x)=\int_0^1 f(y-x)g(y)~dy$$ is a bounded and continuous function on $\mathbf{R}$.
My attempt for the boundedness part:
$$\begin{align*}
|H(x)| = \left|\int_0^1 f(y-x)g(y)~dy\right| &\leqslant \int_0^1|f(y-x)g(y)|~dy \\
& \leqslant\left(\int_0^1|f(y-x)|^2~dy\right)^{1/2}\left(\int_0^1 |g(y)|^2~dy\right)^{1/2}\\
& = \|f\|_2 ~ \|g\|_2.
\end{align*}$$
Hence $G(x)$ is bounded.
Is what I've done for the boundedness part okay? I'll also need help in the continuous portion. Thanks
Added after the comments below:
$$\begin{align*}
|H(x)-H(t)|
&= \left| \int_0^1 f(y-x)(y)~dy)-\int_0^1 f(y-t)g(y)~dy\right| \\
&= \left| \int_0^1\left[f(y-x)-f(y-t)\right] g(y)~dy \right|\\
& \ldots
\end{align*}$$
I guess this is where I have to use translation, but I'm unaware of it. Probably, because , my class haven't gotten there yet. Maybe, someone would be kind enough to 'spoon-feed' a little...
Thanks.