
Real Analysis Real Analysis Math Forum 
 LinkBack  Thread Tools  Display Modes 
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_\deltaf\_p&=\large\left(\int_{\mathbb R^d}\Bigg\vert\int_{\mathbb R^d}f(xy)K_\delta(y)dyf(x)\Bigg\vert^p dx\right)^{1/p}\\ &=\large\left(\int_{\mathbb R^d}\Bigg\vert\int_{\mathbb R^d}f(xy)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(xy)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}ff\_{p}C\\ &\leq\large 2\f\_{p}C, \end{align*} and noting that $\\tau_{\delta z}ff\_{p}\to0$ 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  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Why is the following inequality correct?  Mathmatizer  Calculus  2  July 12th, 2018 08:24 AM 
Can't figure it out how to use Minkowski Inequality in fractions  lucaslima  Calculus  2  September 13th, 2015 02:18 AM 
Can't figure it out how to use Minkowski Inequality in fractions  lucaslima  Calculus  0  September 12th, 2015 02:09 PM 
is this inequality correct?  cris(c)  Algebra  1  July 19th, 2012 12:46 PM 
is this inequality correct?  cris(c)  Calculus  0  December 31st, 1969 04:00 PM 