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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...

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Yes your proof of boundedness is correct. For continuity the method is quite similar to the boundedness proof, except instead of making the estimate on plain $ |H(x)|$ you make it on $ |H(x) - H(t)| $ and show that this tends to $0$ as $t\to x. $ You will need to use the theorem on continuity of translation for $L^p(\mathbb{R})$ functions. –  Ragib Zaman Nov 30 '11 at 15:18
You can use the density of the continuous functions with compact support in $L^2(\mathbb R)$. –  Davide Giraudo Nov 30 '11 at 15:20
@RagibZaman: Could you please help out? I unaware of the translation you talk about.Probably because, we haven't discussed it yet.Could you show me it's done, if you don't mind? Thanks. –  Jack Nov 30 '11 at 15:34
Boundedness is almost correct, except that $||f||_2$ is not the $\int_0^1$ but $\int_{-\infty}^{\infty}$ so the last step is also an inequality, rather than equality. –  Thomas Andrews Nov 30 '11 at 15:52
@ThomasAndrews: Thanks for the correction. Any help with the second part? –  Jack Nov 30 '11 at 18:26

1 Answer 1

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.

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