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.