# Homework Help: Application of Lebesgue differentiation theorem

Tags:
1. Mar 21, 2015

### fderingoz

If $f\in L_{p}^{\rm loc}(\mathbb{R}^{n})$ and $1\leq p<\infty$, then a stronger version of Lebesgue differentiation theorem holds: $$\lim\limits_{r\rightarrow 0}\dfrac{\|f\chi_{B(x,r)}\|_{L_{p}(\mathbb{R}^{n})}}{\|\chi_{B(x,r)}\|_{L_{p}(\mathbb{R}^{n})}}=|f(x)|$$ for almost all $x\in\mathbb{R}^{n}$. See, for example, "Grafakos, Classical Fourier Analysis, Third Edition, Page 101-102". Here $\chi_{B(x,r)}$ denotes the characteristic function of the open ball $B(x,r)$. I wonder that is there an analogue of this property in Orlicz spaces, that is, $$\lim\limits_{r\rightarrow0}\dfrac{\|f\chi_{B(x,r)}\|_{L_{\Phi}(\mathbb{R}^{n})}}{\|\chi_{B(x,r)}\|_{L_{\Phi}(\mathbb{R}^{n})}}=|f(x)|,$$ for almost all $x\in\mathbb{R}^{n}$ ?

Where $\Phi:[0,\infty)\to [0,\infty)$ is an increasing, continuous, convex function with $\Phi(0)=0$ and
$$\|f\|_{L_{\Phi}(\mathbb{R}^{n})}:=\inf\{\lambda>0:\int_{\mathbb{R}^{n}}\Phi\left(\frac{|f(x)|}{\lambda}\right)dx\leq 1\}.$$
It is a generalization of $L_p$ norm. Indeed, if we take $\Phi(t)=t^p,\,1\leq p< \infty$ we get $\|f\|_{L_{\Phi}}=\|f\|_{L_{p}}$.

2. Mar 26, 2015

### Staff: Admin

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted