$\textbf{The Problem:}$ Suppose that $(K_{\delta})_{\delta>0}$ is a family of integrable functions such that there exists a constant $C\in(0,\infty)$ such that $\int K_{\delta}=1,\int\vert K_{\delta}\vert\leq C$ for every $\delta>0$ and for every $\eta>0$ we have $$\color{blue}{\large\lim\limits_{\delta\to0^{+}}\int_{\vert x\vert\geq\eta}\vert K_{\delta}(x)\vert dx=0}.$$ Let $p\in[1,\infty)$. Prove that for every $f\in L^p(\mathbb R^d)$ we have $f\ast K_{\delta}\to f$ in $L^p(\mathbb R^d)$ as $\delta\to0^{+}.$

$\textbf{My Thoughts:}$ Here we go. For $p\in[1,\infty)$ we use Minkowski's inequality for integrals, for reference, this is $6.19$ on page $194$ of Folland's Real Analysis, $2$nd Edition. With this in mind we have, let $\varepsilon>0$ be given, then we have that there is $\eta>0$ such that $\|\tau_yf-f\|_{p}<\varepsilon$ for all $\vert y\vert<\eta.$ Putting these together we have \begin{align*}\large\|f\ast K_\delta-f\|_p&=\large\left(\int_{\mathbb R^d}\Bigg\vert\int_{\mathbb R^d}f(x-y)K_\delta(y)dy-f(x)\Bigg\vert^p dx\right)^{1/p}\\ &=\large\left(\int_{\mathbb R^d}\Bigg\vert\int_{\mathbb R^d}f(x-y)K_\delta(y)dy-\int_{\mathbb R^d}f(x)K_\delta(y)dy\Bigg\vert ^p dx\right)^{1/p}\\ &=\large\left(\int_{\mathbb R^d}\Bigg\vert\int_{\mathbb R^d}[f(x-y)-f(x)]K_\delta(y)dy\Bigg\vert ^p dx\right)^{1/p}\\ &\leq\large\int_{\mathbb R^d}\left(\int_{\mathbb R^d}\vert f(x-y)-f(x)\vert^{p}dx\right)^{1/p}\vert K_{\delta}(y)\vert dy\\ &\leq\large\int_{\mathbb R^d}\|\tau_{y}f-f\|_{p}\vert K_{\delta}(y)\vert dy\\ &\leq\large\int_{\vert y\vert\geq\eta}2\|f\|_{p}\vert K_{\delta}(y)\vert dy+\int_{\vert y\vert<\eta}\|\tau_{y}f-f\|_{p}\vert K_\delta(y)\vert dy\\ &\leq\large2\|f\|_{p}\int_{\vert y\vert\geq\eta}\vert K_{\delta}(y)\vert dy+\varepsilon\\ &\large\overset{\delta\to0^{+}}{\longrightarrow\varepsilon.} \end{align*} It follows that $f\ast K_{\delta}\to f$ in $L^p(\mathbb R^d)$ as $\delta\to0^{+}.$

Do you agree with the proof presented above?

Any feedback is much appreciated.

Thank you for your time.

As a reference, here is the statement of Minkowski's Inequality for Integrals; Suppose that $(X,\frak{M},\mu)$ and $(Y,\frak{N},\nu)$ are $\sigma$-finite measure spaces, and let $f$ be an $(\frak{M}\otimes\frak{N})$-measurable function on $X\times Y$. Then if $f\geq0$ and $1\leq p<\infty,$ we have $$\left[\int\left(\int f(x,y)d\nu(y)\right)^p d\mu(x)\right]^{1/p}\leq\int\left[\int f(x,y)^p d\mu(x)\right]^{1/p}d\nu(y).$$

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    $\begingroup$ How are you going from the third line to the fourth line? $\endgroup$ – Dionel Jaime Apr 7 at 17:24
  • $\begingroup$ @DionelJaime I corrected a mistake I had made in that step. Thank you for pointing that out. I will add more details as well. $\endgroup$ – G the Stackman Apr 8 at 1:57
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    $\begingroup$ @GabyAlfonso With your correction, the proof looks correct. $\endgroup$ – Gyu Eun Lee Apr 8 at 8:49

It seems to me, fourth inequality is doubt, though I don't remember Minkowski's inequality and I don't have folland's book.


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