# convolution a continuous function?

define $$h(x)=\int_0^{2\pi}f(x-y)g(y)dy=f*g(x)$$ if $f,g \in L^2$ are $2\pi$ periodic, show that h is continuous on $[0,2\pi)$

so let $x_n \to x$, then

$$|h(x)-h(x_n)|=|\int f(x-y)g(y)-\int f(x_n-y)g(y)|$$

i tried to manipulate this but couldnt get anywhere. I was also thinking of using Holder's inequality somehow but im not sure...

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Note first that $|(f \ast g)(x)| \leq \Vert f \Vert_2 \, \Vert g\Vert_2$ by Hölder's inequality. Therefore convolution is a continuous bilinear map from $L^2([0,2\pi])$ to the space of bounded functions.

Since the space of continuous functions is dense in $L^2$, we may find a sequence of continuous $2\pi$-periodic functions such that $f_{n} \to f$ in $L^2$. Then the above inequality gives $f_{n} \ast g \to f \ast g$ uniformly on $[0,2\pi]$. Since a uniform limit of continuous functions is continuous, it suffices to prove that $f \ast g$ is continuous if $f$ is continuous.

If $f$ is continuous then $f$ is uniformly continuous (because $[0,2\pi]$ is compact), that is: for each $\varepsilon \gt 0$ there is $\delta \gt 0$ such that for all $|z - z'| \lt \delta$ we have $|f(z) - f(z')| \lt \varepsilon$. In other words, $\|f(\cdot - x) - f(\cdot - x_{n})\|_{\infty} \lt \varepsilon$.

Let $\varepsilon \gt 0$. For $n$ so large that $|x_{n} - x| \lt \delta$ we then have $|(f \ast g)(x_{n}) - (f \ast g)(x)| \leq \Vert f(\cdot - x) - f(\cdot - x_{n})\Vert_{\infty} \Vert g \Vert_{1} \lt \varepsilon \cdot \Vert g \Vert_1$ again by Hölder (note that $L^2([0,2\pi)] \subset L^1([0,2\pi])$ so that $\|g\|_{1}$ is finite). This shows that $f \ast g$ is continuous if $f$ is continuous and as argued in the second paragraph this shows that $f \ast g$ is continuous for all $f,g \in L^2([0,2\pi])$.

Added: To avoid confusion, let me mention that this argument is not circular. In order to show that the space of continuous functions is dense in $L^2$ one often uses convolution with a mollifying sequence. Of course, we should avoid that and this can be done without too much pain (the crux of the proof lies in showing that the step functions are dense in $L^p$ for $1 \leq p \lt \infty$), see e.g. Royden, Real Analysis, Proposition 8 of Chapter 4, on p.128 of the third edition.

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I think the second part of the proof can be slightly simplified if we approximate continuous functions by smooth functions, and use the fact that a convolution with a smooth function is smooth, therefore continuous a fortiori. – timur Mar 23 '14 at 17:46

I believe I've found an alternative proof for this fact (using some arguably much-less-elementary facts).

Given $f,g\in L^2(S^1)$, we can write their Fourier series $(\hat{f_n}),(\hat{g_n})$, defined by $$f(x)=\sum_{n\in \mathbb{Z}} \hat{f_n} e^{inx}$$ and same for $g$.

Then by the convolution theorem, $h\equiv f\star g$ has Fourier coefficients $\hat{h_n}=2\pi\hat{f_n}\hat{g_n}$. It follows that $\{\hat{h_n}\}$ is absolutely summable, since

$$\sum_{n\in\mathbb{Z}} |\hat{h_n}|=\sum_{n\in\mathbb{Z}} 2 \pi |\hat{f_n}\hat{g_n}|\le 2\pi\langle f,g\rangle <\infty$$ by Parseval's identity.

Finally, the Lebesgue dominated convergence theorem gives

\begin{align} \lim_{x\to y} h(x) &= \lim_{x\to y} \sum_{n\in\mathbb{Z}} \hat{h_n}e^{inx} \\ &= \sum_{n\in\mathbb{Z}} \lim_{x\to y} \hat{h_n}e^{inx} \\ &= \sum_{n\in\mathbb{Z}} \hat{h_n}e^{iny} = h(y) \end{align}

since for all $x$, the sequence $\{\hat{h_n}e^{inx}\}$ is dominated in absolute value by the summable sequence $\{|\hat{h_n}|\}$. Hence, $h=f\star g$ is continuous.

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