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

Question

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

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

Answer 1

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 $\beta_n = \frac{1}{n \log^2 n}$.