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Consider $f$ being a measurable function on $R^n$ such that $$\int_{E} e^{|f|}=1$$ ($E$ measurable) and $f$ vanishes outside $E$ . Then $f\in L^p(R^n)$ for all $p\in (0,\infty)$.

I tried using that measure of $E$ cannot be bigger than $1$ and the formulae

$$\int|f|^p=p\int_0^\infty \alpha^{p-1}\omega(\alpha)d\alpha$$

where $\omega(\alpha)=\{x \in R^n: |f(x)|>\alpha\}$.

3 Answers 3

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You can in fact prove more. By Beppo-Leví we have $$\int e^{|f|}d\mu=\sum_{k=0}^{\infty}\frac{1}{k!}\int|f|^{k}d\mu\geq\frac{||f||_{L^{k}}^{k}}{k!},\; k\in\mathbb{N}$$ hence $f\in L^{p},\: p\in\mathbb{N}$ and so by interpolation $f\in L^{p},\: p\geq1$

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    Good answer but it doesnt give the best constant.2012-08-25
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    I didnt read the entire question I thought it said $f\in L^{1}$. Anyway you can also prove that $f\in L^{p},\: p\in(0,1)$ by noting that $e^{x}\geq x,\: x\geq0$, what constant are you reffering to btw?2012-08-25
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For any $p \in (0,\infty)$ there is $c$ such that $e^x \ge c x^p$ for all $x > 0$.

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For all $x\ge0$, $x. Therefore, substituting $x\mapsto x/p$ and raising both sides to the $p^{\text{th}}$ power yields $$ \left(\frac{x}{p}\right)^p