Hello please i have that $w_{\rho}(x)=h_{\rho}(x)w(x)$ where

$$w\in L^{\Phi}(\mathbb{R}^N)=\{u\in L^1(\mathbb{R}^N); \int_{\mathbb{R}^N}\Phi(\frac{|u|}{\lambda})dx<+\infty~\text{for some}~\lambda>0\}$$

and $h_{\rho}\in C^{\infty}(\mathbb{R}^N,[0,1])$ such that $h_{\rho}(x)\equiv 1~\text{on}~B_{\rho}(0),~\text{and}~ supp(h_{\rho})\subset B_{2\rho}(0)$

How to prove that $$ \int_{\mathbb{R}^N}\Phi(|w_{\rho_n}(x)-w(x)|)dx \overset{n\rightarrow+\infty}{\longrightarrow}0 $$

$(\rho_n)$ is a real sequence where $\rho_n\rightarrow +\infty,~\text{whene}~ n\rightarrow+\infty$

where $\Phi$ is a real positive convex function


The convex image of an integrable function need not be integrable. As an example of this, let $\sqrt 2< a< 2$ and consider $w:\mathbb R\to \mathbb R$ given by \begin{equation*} w(x) = \begin{cases} a^j & \text{ if } 2j - 1< x< 2j-1+2^{1- j}\\ 0 & \text{ otherwise}, \end{cases} \end{equation*} where $j= 1, 2, \ldots$. Let $\Phi(r) = r^2$. On one hand, since $a< 2$ we have \begin{equation*} \int_{\mathbb R} w(x)\; dx = \sum_{j = 1}^\infty a^j 2^{1-j}< \infty. \end{equation*} On the other hand, since $a>\sqrt 2$ we have \begin{equation*} \int_{\mathbb R} \Phi(w(x))\; dx = \int_{\mathbb R} w^2(x)\; dx = \sum_{j = 1}^\infty a^{2j}2^{1 - j} = +\infty. \end{equation*}

  • $\begingroup$ It seems that @supinf has provided a much simpler example illustrating that the convex image of an integrable function need not be integrable. $\endgroup$ – BindersFull Jul 21 '17 at 12:05
  • $\begingroup$ see my edit please $\endgroup$ – Vrouvrou Jul 21 '17 at 12:26
  • $\begingroup$ please delete your answer it is false $\endgroup$ – Vrouvrou Jul 21 '17 at 15:48
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    $\begingroup$ @Vrouvrou: why do you downvote this answer? sorry but your comment is rude but also factually wrong. Moreover, you have not explained why BindersFull is wrong as you claim $\endgroup$ – supinf Jul 21 '17 at 15:55
  • $\begingroup$ A less explicit example with the same message: If $w > 0$ is your favourite positive function in $L^1 \setminus L^p$, $p>1$ and $\Phi(r ) = r^p$ then $∫ w = ‖w‖_1$ and $∫ \Phi(w) = ‖w‖_p^p = ∞$ $\endgroup$ – Calvin Khor Jul 21 '17 at 15:59

the statement is wrong.

Consider the constant function $\Phi=1$ which is convex. Then the integral $$ \int_{\mathbb R^n} 1 \,\mathrm dx = \infty $$ and thus does not converge to $0$.

Edit: OP made same changes to the question and added non-obvious assumptions, but i am leaving my original answer up here.

  • $\begingroup$ why it is $\infty$ ? it is impossible , $\endgroup$ – Vrouvrou Jul 21 '17 at 11:41
  • $\begingroup$ i have edited to clarify. Note that the domain is infinite $\endgroup$ – supinf Jul 21 '17 at 11:55
  • $\begingroup$ from where we can have that $\int_{\mathbb{R}^N}$ see that $supp(h)\subset B_{2\rho}(0)$ $\endgroup$ – Vrouvrou Jul 21 '17 at 12:11
  • $\begingroup$ see the edit please $\endgroup$ – Vrouvrou Jul 21 '17 at 12:38
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    $\begingroup$ @Vrouvrou you might do yourself some good to explain why it is "false". (hint: it isn't) $\endgroup$ – Calvin Khor Jul 21 '17 at 15:54

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