Pointwise A.E. Convergence of Convolution

Question

The following is a prelim question, which I can't seem to show under the hypotheses given.

Problem. Let $f$ and $g$ be bounded measurable functions on $\mathbb{R}^{n}$. Assume that $g$ is integrable and satisfies $\int g=0$. Define $g_{k}(x):=k^{n}g(kx)$ for $k\in\mathbb{N}$. Show that $f\ast g_{k}\rightarrow 0$ pointwise almost everywhere, as $k\rightarrow\infty$.

It's easy to show that $(f \ast g_{k})(x)\rightarrow 0$ for every point of continuity of $f$. Moreover, if $g$ satisfies some decay at $\infty$ condition like $O(|x|^{-n-\delta})$, for some $\delta>0$. Then we can show that $(f\ast g_{k})(x)\rightarrow 0$ for a.e. $x$ (in fact, every point in the Lebesgue set of $f$). Indeed, sSince $f$ is locally integrable, we have by the Lebesgue differentiation theorem that for a.e. $x\in \mathbb{R}^{n}$, $$\lim_{r\rightarrow 0}r^{n}\int_{|y|\leq r}|f(x-y)-f(x)|dy=0$$ Using the hypothesis $\int g=0$, we have \begin{align*} \left|(f\ast g_{k})(x)\right|&=\left|k^{n}\int_{\mathbb{R}^{n}}[f(x-y)-f(x)]g(ky)dy\right|\\ &\leq k^{n}\int_{|y|\leq 1/k}|f(x-y)-f(x)||g(ky)|dy+k^{n}\int_{|y|>1/k}|f(x-y)-f(x)||g(ky)|dy\\ &\leq\|g\|_{L^{\infty}}k^{n}\int_{|y|\leq 1/k}|f(x-y)-f(x)|dy+k^{n}\int_{|y|>1/k}|f(x-y)-f(x)||g(ky)|dy\\ \end{align*} We estimate the second term by \begin{align*} &\lesssim\sum_{j=0}^{\infty}k^{n}\int_{2^{j}k^{-1}<|y|\leq 2^{j+1}k^{-1}}|f(x-y)-f(x)||ky|^{-n-\delta}dy\\ &\leq \sum_{j=0}^{\infty}2^{-j(n+\delta)}k^{n}\int_{|y|\leq 2^{j+1}k^{-1}}|f(x-y)-f(x)|dy\\ &=\sum_{j=0}^{\infty}2^{-j(n+\delta)}2^{(j+1)n}(2^{-j-1}k)^{n}\int_{|y|\leq 2^{j+1}k^{-1}}|f(x-y)-f(x)|dy\\ &\lesssim\sum_{j=0}^{\infty}2^{-j\delta}(2^{-j-1}k)^{n}\int_{|y|\leq 2^{j+1}k^{-1}}|f(x-y)-f(x)|dy \end{align*} The rest of the proof follows by using the $L^{\infty}$ bound for $f$ and the summability of the resulting series to reduce to a finite sum, then appealing to Lebesgue differentiation.

Answer 1

Let $\epsilon>0$ be given. Using the hypothesis $\int g=0$, we have the estimate

$$\left|k^{n}\int_{\mathbb{R}^{n}}f(x-y)g(ky)dy\right|\leq k^{n}\int_{\mathbb{R}^{n}}|f(x-y)-f(x)||g(ky)|dy$$

Let $R\gg 1$ be a parameter, the value of which will be determined later. Then by dilation invariance,

\begin{align*} k^{n}\int_{\mathbb{R}^{n}}|f(x-y)-f(x)||g(ky)|dy&=\int_{\mathbb{R}^{n}}|f(x-y/k)-f(x)||g(y)|dy\\ &\leq2\|f\|_{L^{\infty}}\int_{|y|> R}|g(y)|dy+\|g\|_{L^{\infty}}\int_{|y|\leq R}|f(x-y/k)-f(x)|dy\\ &=2\|f\|_{L^{\infty}}\int_{|y|> R}|g(y)|dy+\|g\|_{L^{\infty}}k^{n}\int_{|y|\leq R/k}|f(x-y)-f(x)|dy, \end{align*} where the ultimate line follows from dilation invariance. Choose $R>0$ so that the first term is $<\epsilon/2$. For the second term, $$k^{n}\int_{|y|\leq R/k}|f(x-y)-f(x)|dy=R^{n}\dfrac{1}{(R/k)^{n}}\int_{|y|\leq R/k}|f(x-y)-f(x)|dy$$ By the Lebesgue differentiation theorem, the RHS $\rightarrow 0$ as $k\rightarrow\infty$ for a.e. $x\in \mathbb{R}^{n}$.

Answer 2

What I was going to say is the following; I think it's really the same as the argument you came up with, expressed somewhat differently.

Let $A_N=\{|x|\le N\}$ and set $$g^N=g\chi_{A_N}.$$Let $\alpha_N=\int g^N$, and note that $\alpha_N\to0$.

Now since $g_N$ is bounded and has compact support, that argument you know very well shows that for every $N$ we have $$\lim_{k\to\infty}g^N_k*f=\alpha_Nf$$almost everywhere. But $$||g_k*f-g_k^N*f||_\infty\le||f||_\infty||g-g^N||_1,$$so $g^N_k*f(x)\to g_k*f(x)$ uniformly in $x$ and $k$ as $N\to\infty$.

QED: First choose $N$ so $|g^N_k*f-g_k*f|<\epsilon/2$...