Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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...

share|cite|improve this question
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

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.