function $f\notin L^{\infty}(\Omega)$, but $f\in L^p(\Omega)$ for all $1\le p <\infty$

Asked Nov 13 '15 at 00:27

Active Nov 13 '15 at 00:32

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I'm searching for a function $f\notin L^{\infty}(\Omega)$, but $f\in L^p(\Omega)$ for all $1\le p <\infty$. And it has to be $|\Omega |<\infty$. I tried $f(x)=\frac{1}{2\sqrt{x}}$ and $\Omega= (0,1)$ and something like that, but it is $f\notin L^p(\Omega)$ for all $1\le p <\infty$. Could you help me?

edited Nov 13 '15 at 00:32

asked Nov 13 '15 at 00:27

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Try logarithm.. – Nov 13 '15 at 00:29
All standard real analysis questions were already asked and answered; searching for a while is highly recommended. – Nov 13 '15 at 00:31

function $f\notin L^{\infty}(\Omega)$, but $f\in L^p(\Omega)$ for all $1\le p <\infty$

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