# Bounding for convolution convergence

by Mika H.   Last Updated June 22, 2018 07:20 AM

Suppose $f\in L^p(\mathbb{R})$ and $K\in L^1(\mathbb{R})$ with $\int_\mathbb{R}K(x)dx=1$. Define $$K_t(x)=\dfrac{1}{t}K\left(\dfrac{x}{t}\right)$$ I'm trying to prove that $\lim_{t\rightarrow 0}\|f\ast K_t-f\|_p=0$.

I choose a compactly supported function $g\in C^\infty$ such that $\|f-g\|_p<\epsilon$ (possible because this class is dense in $L^p(\mathbb{R})$). Then I want to bound $$\|f\ast K_t-f\|_p\leq \|f\ast K_t-g\ast K_t\|_p+\|g\ast K_t-g\|_p+\|g-f\|_p$$

We have of course $\|g-f\|_p<\epsilon$. We have $\|(f-g)\ast K_t\|_p\leq \|f-g\|_p\|K_t\|_1<\epsilon\|K_t\|_1=\epsilon\|K\|_1$.

Now I need to bound $\|g\ast K_t-g\|_p$.

I have $$\left|(g\ast K_t)(x)-g(x)\right|_p=\left|\int_\mathbb{R}(g(x-y)-g(x))K_t(y)dy\right|$$ Since $g$ is continuous and compactly supported, it is uniformly continuous. But what can I do with the $K_t(y)$?

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In the last displayed integral, use the substitution $s:=y/t$. Then $$\lVert g\star K_t-g\rVert_p^p=\int_{\mathbb R}\left|\int_{\mathbb R}(g(x-ts)-g(x))K(s)\mathrm ds\right|^p\mathrm dx.$$ I assume $K\geqslant 0$ (not restrictive). Then by Jensen's inequality, it's enough to prove that $$\lim_{t\to 0}\iint_{\mathbb R^2}|g(x-ts)-g(x)|^pK(s)\mathrm ds\mathrm dx=0.$$ This can be done by an approximation argument: take $\varepsilon\gt 0$: then there is $R$ such that $\int_{\mathbb R\setminus [-R,R]}K(s)\mathrm ds\lt \varepsilon$. Then we have $$\int_{\mathbb R}\int_{\mathbb R\setminus [-R,R]}|g(x-ts)-g(x)|^pK(s)\mathrm ds\mathrm dx\leqslant 2\varepsilon\lVert g\rVert_p.$$ We have $$\int_{\mathbb R}\int_{[-R,R]}|g(x-ts)-g(x)|^pK(s)\mathrm ds\mathrm dx\leqslant 2R\sup|g'|t^p\lVert K\rVert_1.$$

Davide Giraudo
November 17, 2013 19:51 PM

Sorry I'm late to the game, but: there is a nice way to do this when $p = 2$ that does not need a density argument. Assume $K \geq 0$. Recall that the Fourier transform is a unitary operator $\mathcal{F}: L^{2}(\mathbb{R}) \rightarrow L^{2}(\mathbb{R})$, so the question is equivalent to showing $\widehat{f*K_{t}} \rightarrow \hat{f}$ in $L^{2}(\mathbb{R})$. Now,

$$|\widehat{f*K_{t}} - \hat{f}|^{2} = | \hat{f}\hat{K_{t}} - \hat{f} |^{2} = | \hat{f}(\hat{K_{t}} - 1) |^{2}$$

We bound $| \hat{K_{t}}(\zeta) - 1 |$ by:

$$|\hat{K_{t}}(\zeta) - 1 | = | \int_{\mathbb{R}} K_{t}(x)e^{-2\pi i x \zeta} dx- \int_{\mathbb{R}}K_{t}(x)dx | \leq \int_{\mathbb{R}}K_{t}(x)|e^{-2\pi ix\zeta} - 1|dx \leq 2\int_{\mathbb{R}}K_{t}(x)dx = 2.$$

Thus, $|\widehat{f*K_{t}} - \hat{f}|^{2} \leq 4|\hat{f}|^{2}$. Noting $\hat{f} \in L^{2}(\mathbb{R})$, by the dominated convergence theorem, we get

$$\lim_{t \to 0} ||\widehat{f*K_{t}} - \hat{f}||_{L^{2}}^{2} = \int_{\mathbb{R}}|\hat{f}(\zeta)|^{2}\lim_{t \to 0}|\hat{K_{t}}(\zeta) - 1|^{2} d\zeta.$$

Notice that $\hat{K_{t}}(\zeta) - 1 = \int_{\mathbb{R}}K(x)(e^{-2 \pi i t x \zeta} - 1)dx$, by a change of variables. But we have already shown $\int_{\mathbb{R}}K(x)|e^{-2\pi i t x\zeta} - 1|dx \leq 2$, so by the dominated convergence theorem again,

$$\lim_{t \to 0} (\hat{K_{t}}(\zeta)- 1) = \int_{\mathbb{R}}K(x)\lim_{t \to 0}(e^{-2\pi i t x \zeta} - 1)dx = 0.$$

The result follows.

Willie Dong
June 22, 2018 07:07 AM