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Suppose that a function $u:\Omega \rightarrow \mathbb{R}^n$ is such that $u \in L^2(\Omega)$. Does this imply that $u^p \in L^2(\Omega)$? if not can you give a counterexample?

Here $\Omega$ is an open bounded subset of $\mathbb{R}^n$ and $p > 2$ is a real number.

Giovanni
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kamerove
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    Why would this be true? $|u|^2$ could blow up at a point and barely be integrable. $|u|^p$ will blow up worse and might not make it. – zhw. Dec 08 '15 at 03:34

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It does not. Indeed let $\Omega = (0,1)$, $u(x) = \frac{1}{\sqrt[3]{x}}$. Then $u \in L^2((0,1))$, but $u^3 = \frac{1}{x}$ and $\frac{1}{x}$ is clearly not square integrable.

Giovanni
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