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Suppose $||f_n||_{L^p(\Omega)} \leq C$, where $\Omega$ is a bounded set in $\mathbb{R}^n$. Moreover, $f_n \geq 0$.

Using weak compactness, we know that there exists a subsequence $\{f_{n_k} \}$ such that $f_{n_k} \rightarrow f$ weakly in $L^p$.

Since $||\sqrt{f_{n_k}}||_{L^{2p}} \leq C$, we similarly obtain $\sqrt{f_{n_k}} \rightarrow \sqrt{g}$ weakly in $L^{2p}$, up to a subsequence.

My question is whether $f=g$. I guess $f=g$ due to the choice of subsequence. If true, how to prove it?

Akira
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inhonorof Lagrange
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1 Answers1

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In short, the answer is no. To produce a counterexample, we can use Riemann-Lebesgue's lemma. Suppose for instance that $\Omega=(0,2\pi)$ and $f_n=\cos(nx)+1$. Then, by Riemann-Lebesgue's lemma, $f_n$ converges weakly in $L^p(\Omega)$ to the constant function

$$f(x)=\frac{1}{2\pi}\int_0^{2\pi}\cos(x)+1\, dx=1,$$

while $\sqrt{f_n}$ weakly converges in $L^{2p}(\Omega)$ to the constant function

$$\sqrt{g(x)}=\frac{1}{2\pi}\int_0^{2\pi}\sqrt{\cos(x)+1}\, dx\neq 1,$$

so that $g\neq f$.

With a similar argument you can produce counterexamples in dimension $d>1$.

Tytiro
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