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Let $g: \mathbb{R} \rightarrow \mathbb{R}$ be integrable with$\int_{\mathbb{R}}g(x)dx=1$ and $|g(x)| \leq \frac{C}{(1+|x|)^{1+h}}$ for $x \in \mathbb{R} $, where $C, h>0$ are constants.

Let $g_t(x)=\frac{1}{t} g(\frac{x}{t})$ for $x \in \mathbb{R}$, $t>0$.

I want to show that:

If $f\in L^p$, where $1\leq p\leq \infty$, then $f*g_t(x) \rightarrow f(x)$ a.e.

I have tried in this way:

Let $x\in \mathbb{R}$ be the Lebesgue point of $f$, that is $lim_{r\rightarrow 0} \frac{1}{r} \int_{B(x,r)} |f(y)-f(x)|dx=0$, then

$$ |f*g_t(x)-f(x)|\leq \int_{\mathbb{R}} g_t(x-y)|f(y)-f(x)|dy =I_1+I_2, $$

where

$I_1=\int_{B(x,t)} g_t(x-y)|f(y)-f(x)|dy \leq\frac{1}{t} \int_{B(x,t)} \frac{C}{(1+\|\frac{x-y}{t}\|)^{1+h}} |f(y)-f(x)|dy $

$ \leq C\frac{1}{t}\int_{B(x,t)} |f(y)-f(x)|dy \rightarrow 0 \ as \ t \rightarrow 0;$

$I_2=\int_{\mathbb{R}\setminus B(x,t)} g_t(x-y)|f(y)-f(x)|dy .$

I don't know how to estimate the integral $I_2$.

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Is there a reason why you used $B(x,t)$ to split the integral (with the same $t$ as in $g_t$? –  Dirk Jul 13 '12 at 20:37
1  
I think this paper closely related to your question. See theorem 3.4. Using your denotions we can say that author proves this result for arbitrary function $g\in L^1(\mathbb{R})$, without any restrictions on decrease rate. But he consider only positive functions $g$. –  no identity Jul 24 '12 at 10:34

1 Answer 1

up vote 6 down vote accepted
+50

Since it is a local result, we may assume without loss of generality that $p=1$.

The Hardy-Littlewood centered maximal function of $f$ is defined as $$ f^*(x)=\sup_{r>0}\frac1{2\,r}\int_{x-r}^{x+r}|f(y)|\,dy=\sup_{r>0}\frac1{2\,r}\int_{|y|\le r}|f(x-y)|\,dy. $$ It satisfies the weak one-one inequality $$ |\{x:f^*(x)>\delta\}|\le\frac{M}{\delta}\|f\|_1\quad\forall\delta>0, $$ where $M>0$ is a constant independent of $f$ and $\delta$. We first bound $g_t\ast f(x)$ in terms of $f^*(x)$. $$\begin{align*} |g_t\ast f(x)|&\le \frac{C}{t}\int_{|y|\le t}\frac{|f(x-y)|}{(1+y/t)^{1+h}}\,dy+\frac{C}{t}\sum_{k=0}^\infty\int_{2^kt<|y|\le 2^{k+1}t}\frac{|f(x-y)|}{(1+y/t)^{1+h}}\,dy\\ &\le\frac{C}{t}\int_{|y|\le t}|f(x-y)|\,dy+\frac{C}{t}\sum_{k=0}^\infty\int_{|y|\le 2^{k+1}t}\frac{|f(x-y)|}{(1+2^k)^{1+h}}\,dy\\ &\le 2\,C\,f^*(x)+C\Bigl(\sum_{k=0}^\infty\frac{2^{k+2}}{(1+2^k)^{1+h}}\Bigr)f^*(x)\\ &\le K\,f^*(x), \end{align*}$$ where $$ K=2\,C+C\sum_{k=0}^\infty\frac{2^{k+2}}{(1+2^k)^{1+h}}<\infty. $$

Now we mimic the proof of the Lebesgue differentiation theorem. Given $\epsilon>0$ choose a continuous function $\phi$ with compact support such that $\|f-\phi\|_1<\epsilon$. Then $$ g_t\ast f(x)-f(x)=g_t\ast(f-\phi)(x)+\bigl(g_t\ast\phi(x)-\phi(x)\bigr)+(\phi(x)-f(x)), $$ from where $$ |g_t\ast f(x)-f(x)|\le K(f-\phi)^*(x)+|g_t\ast\phi(x)-\phi(x)|+|\phi(x)-f(x)| $$ Since $\phi$ is continuous, the middle term converges to $0$ as $t\to0$ for all $x$. Let $\delta>0$. Then $$ \{x:\limsup_{t\to0}|g_t\ast f(x)-f(x)|>2\,\delta\}\subset\{x:(f-\phi)^*(x)>\frac\delta{K}\}\cup\{x:|\phi(x)-f(x)|>\delta\}. $$ By Chebychev's inequality $$ |\{x:|\phi(x)-f(x)|>\delta\}|\le\frac1\delta\|\phi-f\|_1\le\frac\epsilon\delta. $$ By the weak $1$-$1$ inequality for the maximal function $$ |\{x:(f-\phi)^*(x)>\frac\delta{K}\}|\le\frac{K\,M}\delta\|\phi-f\|_1\le\frac{K\,M\,\epsilon}\delta. $$ Finally $$ |\{x:\limsup_{t\to0}|g_t\ast f(x)-f(x)|>2\,\delta\}|\le\frac{K\,M+1}{\delta}\,\epsilon. $$ Since $\epsilon$ was arbitrary, it follows that $$ |\{x:\limsup_{t\to0}|g_t\ast f(x)-f(x)|>2\,\delta\}|=0\quad\forall\delta>0. $$

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