How to show that $H(x)=\int_0^1 f(y-x)g(y)~dy$ is bounded and continuous on $\mathbf{R}$ 
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.
 A: As you saw, the boundedness is a consequence of Cauchy-Schwarz inequality. For the continuity, fix $\varepsilon >0$. We can find $f_0$ continuous with compact support such that $\lVert f-f_0\rVert_{L^2(\mathbb R)}\leq \varepsilon$. We have for $x,h\in\mathbb R$
\begin{align*}
|H(x+h)-H(x)|&=\left|\int_{\left[0,1\right]}f(y-(x+h))g(y)dy-\int_{\left[0,1\right]}f(y-x)g(y)dy\right|\\
&=\left|\int_{\left[0,1\right]}\left[f(y-(x+h))-f(y-x)\right]g(y)dy\right|
\\
&\leq\lVert g\rVert_{L^2}\left(\int_{\left[0,1\right]}\left[f(y-x-h)-f(y-x)\right]^2dy\right)^{\frac 12}\\
&=\lVert g\rVert_{L^2}\left(\int_{\left[-x,1-x\right]}\left[f(t-h)-f(t)\right]^2dy\right)^{\frac 12}\\
&\leq \lVert g\rVert_{L^2}\left(2\lVert f-f_0\rVert_{L^2}+\left(\int_{\operatorname{supp}f_0}\left[f_0(t-h)-f_0(t)\right]^2dy\right)^{\frac 12}\right),
\end{align*}
and you can conclude applying the uniform continuity of $f_0$ on the compact $\operatorname{supp}f_0$, since we can choose $h$ smalll enough to get $\left(\int_{\operatorname{supp}f_0}\left[f_0(t-h)-f_0(t)\right]^2dy\right)^{\frac 12}\leq \varepsilon$, so $|H(x+h)-H(x)|\leq 3\lVert g\rVert_{L^2}\varepsilon$ for $h$ small enough.
