For each $p\in[1,+\infty)$ , we have $\lVert k_{\varepsilon}*f\rVert_{L^{p}({\bf R}^{n}~)}\longrightarrow0$ as $\varepsilon\longrightarrow0$

Question

Let $k_{\varepsilon}$ be a family on $\mathbb{\bf R}^{n}$ that satisfies the following properties :

(i) There exists a constant $c>0$ such that $\lVert k_{\varepsilon} \rVert_{L^{1}({\bf{R}}^{n}~)}\le c$ for all $\varepsilon>0.$

(ii)For each $\varepsilon>0$ , we have$$\int_{{\bf R}^{n}}k_{\varepsilon}(x)~dx=0.$$

(iii) For any neighborhood $V$of the identity element $0$ of ${\bf R}^{n},$ one has $$\int_{V^{c}}|k_{\varepsilon}(x)|~dx\longrightarrow0~~as~~\varepsilon\longrightarrow0,$$ where $V^{c}={\bf R}^{n}-V.$

If $f\in L^{p}({\bf R}^{n})$ for $p\in[1,+\infty)$ , then we have $\lVert k_{\varepsilon}*f\rVert_{L^{p}({\bf R}^{n}~)}\longrightarrow0$ as $\varepsilon\longrightarrow0$

"Here's my attempt" :(I edited)

Let $\delta>0$ be arbitrary given.

Firstly , we recall that $C_{00}({\bf R}^{n})$ is dense in $L^{p}({\bf R}^{n})$ ,where $C_{00}({\bf R}^{n})$ is the space of all continuous functions on ${\bf R}^{n}$ with compact support .

So, for $g\in C_{00}({\bf R}^{n})$ with compact support $L$ we have $$|g(x-y)-g(x)|^{p}\le (2\lVert g\rVert_{L^{\infty}})^{p}\chi_{LW^{-1}}$$ for $y$ in a relatively compact neighborhood $W$ of the identity element $0\in {\bf R}^{n}.$

Then by Lebesgue's Dominated Convergence Theorem to obtain $$\lim_{y\rightarrow0}\int_{{\bf R}^{n}}|g(x-y)-g(x)|^{p}~dx=0$$ Now, approximate a given $f\in L^{p}({\bf R}^{n})$ by a continuous function with compact support $g$ to deduce that $$\lim_{y\rightarrow0}\int_{{\bf R}^{n}}|f(x-y)-f(x)|^{p}~dx=0.$$

Choose a sufficiently small $\alpha>0$ such that $\lVert f(x-y)-f(x)\rVert_{L^{p}({\bf R}^{n~})}^{p}<\bigg(\frac{\delta}{2c}\bigg)^{p}~$ when $|y|\le\alpha$

Since $k_{\varepsilon}$ has integral zero for all $\varepsilon>0$ , we have \begin{align} |k_{\varepsilon}*f|(x)&=\bigg|\int_{{\bf R}^{n}}k_{\varepsilon}(y)f(x-y)~dy-0~\bigg|\\ &=\bigg|\int_{{\bf R}^{n}}k_{\varepsilon}(y)f(x-y)~dy-\int_{{\bf R}^{n}}k_{\varepsilon}(y)f(x)~dy\bigg|\\ &=\bigg|\int_{{\bf R}^{n}}k_{\varepsilon}(y)\bigg(f(x-y)-f(x)\bigg)~dy\bigg|\\ &\le\bigg|\int_{|y|\le\alpha}k_{\varepsilon}(y)\bigg(f(x-y)-f(x)\bigg)~dy\bigg|~+\bigg|\int_{|y|>\alpha}k_{\varepsilon}(y)\bigg(f(x-y)-f(x)\bigg)~dy\bigg| \end{align}

Therefore, \begin{align} \bigg\lVert \int_{|y|\le\alpha}k_{\varepsilon}(y)\bigg(f(x-y)-f(x)\bigg)~dy\bigg\rVert_{L^{p}({\bf R}^{n},~dx)}&\le\int_{|y|\le\alpha}|k_{\varepsilon}(y)|\lVert f(x-y)-f(x)\rVert_{L^{p}({\bf R}^{n},~dx)}~dy\\ &< \int_{|y|\le\alpha}\frac{\delta}{2c}|k_{\varepsilon}(y)|~dy\le\frac{\delta}{2} \end{align}

and

\begin{align} \bigg\lVert \int_{|y|>\alpha}k_{\varepsilon}(y)\bigg(f(x-y)-f(x)\bigg)~dy\bigg\rVert_{L^{p}({\bf R}^{n},~dx)}&\le\int_{|y|>\alpha}|k_{\varepsilon}(y)|\lVert f(x-y)-f(x)\rVert_{L^{p}({\bf R}^{n},~dx)}~dy\\ &\le \int_{|y|>\alpha}2|k_{\varepsilon}(y)|\lVert f\rVert_{L^{p}({\bf R}^{n})}~dy<\frac{\delta}{2} \end{align}

provided we have that $\displaystyle\int_{|y|>\alpha} |k_{\varepsilon}(x)|~dy<\frac{1}{4}\cdot\frac{\delta}{1+\|f\|_{L^{p}({\bf R}^{n})}}~~$, for each $\varepsilon >0~~~~~-(1).$

$(1)$ actually holds since we use the nice property of the mollifier $k_{\varepsilon}$ , which is given in (iii) .

Now choose $\varepsilon_{0}>0$ such that $(1)$ is true for each $\varepsilon>0$ with $\varepsilon<\varepsilon_{0}~.$

Then for all $\delta>0$ and all $\varepsilon>0$ with $\varepsilon<\varepsilon_{0}$ , we have $\lVert k_{\varepsilon}*f\rVert_{L^{p}({\bf R}^{n}~)}<\delta,$

so,$$\lim_{\varepsilon\rightarrow0}\lVert k_{\varepsilon}*f\rVert_{L^{p}({\bf R}^{n}~)}\le\delta.$$

Whence, our conclusion follows as $\delta\longrightarrow0.$

If you have the time , please , check my proof for validity. Any valuable suggestion or advice will be appreciated. Thanks a lot.

Answer 1

Property (1) does not follow from the stated assumptions. In fact if (1) and (2) both hold, then $\|k_\varepsilon\|_{L^1}\to 0$ as $\varepsilon\to 0$ which of course implies the result, without any splitting of integrals in two parts.

The attempt to justify (1) by saying $k_\varepsilon \in L^1$ doesn't work because for each $\varepsilon$ you have a different function $k_\epsilon$, and it will need a different $\delta_1$ for the conclusion to be true.

Generally: when we split the region of integration in two parts, it is because different estimates will be made on each part. By looking at the proof and seeing you do exactly the same thing for $|x|\le \alpha$ as for $|x|>\alpha$ one can tell this isn't likely to work. This is without even getting into how $|x|<\alpha$ should have been $|y|<\alpha$, etc.

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Notice you haven't used the assumption $\int k_\varepsilon = 0$ anywhere. What it allows us to do is to subtract off a constant term from $f$ in the convolution integral:

$$ \int_{y\in\mathbb{R}^n}k_{\varepsilon}(y)f(x-y)~dy = \int_{y\in \mathbb{R}^n}k_{\varepsilon}(y)(f(x-y) - f(x))~dy $$

What we gain is that $(f(x-y) - f(x))$ will tend to be small when $y$ is small. Of course $f$ isn't assumed continuous, but there is a kind of continuity for $L^p$ functions: $$ \lim_{|y|\to 0} \|f(\cdot - y) - f\|_{L^p} = 0 $$ See Translation operator and continuity.

Thus, by choosing sufficiently small $\alpha$ we can have $\|f(\cdot - y) - f\|_{L^p} < \gamma$, for whatever small number $\gamma$ you want, and then
$$ \left|\int_{|y|\le\alpha}k_{\varepsilon}(y)(f(x-y) - f(x))~dy \right| \le \gamma \int |k_\varepsilon| \le \gamma c $$

The part $|y|>\alpha$ is dealt with as in your proof, using the smallness of $k_\varepsilon$.

Notice how in both cases the product is small, but for a different reason: this is typical for such proofs.