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$U$ is a bounded open subset of $R^n$.

If $u\in L^2(U)$ and $u>0$, how to show $u\ln u\in L^2(U)$ ?

Enhao Lan
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1 Answers1

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HINT:

The statement is not true. Let's see why: in general a function $u$ on some measure space $U$ is in $L^2(U)$ if and only if both of the series

$$ \sum_{\ge 1} n^2\cdot \alpha_n \\ \sum_{\ge 1} \frac{1}{n^2}\cdot \alpha'_n $$ are convergent

where $\alpha_n = \mu( \{ x \ | \ n \le |u(x)| < n+1\} ) $ and $\alpha'_n = \mu( \{ x \ | \ \frac{1}{n+1} \le |u(x)| < \frac{1}{n}\} ) $

By hypothesis the measure of $U$ is finite. Because of this, and the fact that $\frac{\log n}{n^2} \to 0$ the second series for $u \log u$ ( and surely for $u$) will be convergent. However, the first might as well not be. For denote by $\beta_n = n^2 \alpha_n$. Then $u^2 \in L^1(U)$ means in fact $\sum \beta_n < \infty$. However, it may as well be that $\sum \log n^2 \beta_n = \infty$. For instance, choose $\beta_n = \frac{1}{n \log^2 n}$.

Svetoslav
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orangeskid
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  • Very nice approach – Svetoslav Feb 23 '16 at 17:18
  • @Svetoslav: Thanks! Now, I really wonder how to get an explicit function in the way you initiated. What should be the growth at $0$? – orangeskid Feb 23 '16 at 22:21
  • I think you can take a function $u\ge 2$ in a ball (with radius which will become clear in a moment) which is piecewise continuous on $A_n$, where $A_n$ is the annulus from radius $r=r_n$ to $r=r_{n+1}$, such that the area $\alpha_n$ of $A_n=\frac{1}{n^3\ln^2{n}}$. And on each annulus, the function will increase from $y=n$ to $y=n+1$. And because $u\ge 2$ your $A_0$ and $A_1$ will be $\emptyset$. So for example $\alpha_2=\frac{1}{2^3\ln^2(2)}$. Because the area of the ball will be the limit of $\sum\limits_{n=1}^{N}{A_n}$ when $N\to \infty$, then the radius of the ball should be well defined. – Svetoslav Feb 24 '16 at 04:26
  • $\alpha_2=\frac{1}{8\ln^2(2)}\Rightarrow r_2=0,, r_3=\sqrt{\frac{\alpha_2}{\pi}}$ – Svetoslav Feb 24 '16 at 04:28