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April 7th, 2019, 01:55 PM   #1
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Math Focus: Measure Theory & Integration
Is Application of Minkowski's Integral Inequality Correct?

>$\textbf{The Problem:}$ 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 $\varepsilon>0$ we have $$\color{blue}{\large\lim\limits_{\delta\to0^{+}}\ int_{\vert x\vert\geq\varepsilon}\vert K_{\delta}(x)\vert dx=0}.$$
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 $\delta\to0^{+}.$

$\textbf{My Toughts:}$ 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, letting $y=\delta z$ at the right time,
\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-\delta z)-f(x)\vert^{p}dx\right)^{1/p}\vert \delta^{d}K_{\delta}(\delta z)\vert dy\\
&\leq\large\|\tau_{\delta z}f-f\|_{p}C\\
&\leq\large 2\|f\|_{p}C,
\end{align*}
and noting that $\|\tau_{\delta z}f-f\|_{p}\to0$ as $\delta\to0^{+}$ we have what we needed using the Dominated Convergence Theorem. The latter follows by a density argument with compactly supported continuous functions and Minkowski's inequality.
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Do you agree with the proof presented above? My main concern is that I did not use the hypothesis in $\color{blue}{\text{blue.}}$

Any feedback is much appreciated.

Thank you for your time.
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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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