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 April 7th, 2019, 01:55 PM #1 Newbie   Joined: Apr 2019 From: Madison, Wisconsin Posts: 2 Thanks: 0 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. ____ 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. ____ 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 $$\left[\int\left(\int f(x,y)d\nu(y)\right)^p d\mu(x)\right]^{1/p}\leq\int\left[\int f(x,y)^p d\mu(x)\right]^{1/p}d\nu(y).$$ Tags application, correct, inequality, integral, minkowski, real analysis Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Mathmatizer Calculus 2 July 12th, 2018 08:24 AM lucaslima Calculus 2 September 13th, 2015 02:18 AM lucaslima Calculus 0 September 12th, 2015 02:09 PM cris(c) Algebra 1 July 19th, 2012 12:46 PM cris(c) Calculus 0 December 31st, 1969 04:00 PM

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