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In problem 3 we have:

If $f:\mathbb{R} \longrightarrow\mathbb{R}$ is mensurable, $E:=\mathrm{supp}\ f$ and

$$\int_E e^{|f(x)|}dx =1,$$

then $f\in L^p(\mathbb{R})$, for all $p\in(0,\infty)$ and

$$\|f\|_p \leq Cp,$$ where the constant $C$ does not depend on $f$ or $p$. Moreover, there exist $f\notin L^{\infty}(\mathbb{R})$ such that $\int_E e^{|f(x)|}dx =1$.

So, for $p=1$ we have $|f(x)|\leq e^{|f(x)|}$, whence $\|f\|_p \leq 1$.

If $|E|<\infty$ and $0<p<1$, then $\phi(y):=y^{1/p}$ is convex, and, by Jensen's inequality,

$$\bigg(\frac1{|E|}\int_E |f(x)|^p dx\bigg)^{1/p} \leq \frac1{|E|}\int_E |f(x)|dx,$$

there is,

$$\|f(x)\|_p \leq |E|^{-1-1/p}.$$

But this dont solve the problem. I tried to use Jensen's inequality for $\phi(y) = e^{y}$ in case $p\geq 1$, but it don't conclude the result.

Can someone help me?

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up vote 3 down vote accepted

Noe that $\mathrm e^t\geqslant t$ for every $t\geqslant0$ hence $\mathrm e^{pt}=(\mathrm e^t)^p\geqslant t^p$ for every $p\gt0$. Using this for $t=s/p$ yields $s^p\leqslant p^p\mathrm e^s$ for every $s\geqslant0$. Hence, for every $p\gt0$, $|f|^p\leqslant p^p\mathrm e^{|f|}\mathbf 1_E$.

Integrating this pointwise inequality yields $\|f\|_p^p\leqslant cp^p$ with $c=\int\limits_E\mathrm e^{|f|}$. In the present case, $c=1$ hence we proved that $\|f\|_p\leqslant p$.

For the second question, try $f=\mathbb 1_E$ for some suitable $E$.

For the revised second question, try $f=\sum\limits_{n\geqslant1}n\mathbf 1_{E_n}$ where $|E_n|=1/(2\mathrm e)^n$ for every $n\geqslant1$ and the sets $E_n$ are disjoint.

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Thank you very much @Did. Hoewver, unfortunately, I wrote wrong above. The example is find $f\notin L^{\infty}$, but sastisfies $\int_{E} e^{|f(x)|}dx = 1$. Do you know any example? Thank you again! – Kelson Vieira Jan 23 '13 at 15:25
Thank you very much @Did. And stop smoking lol. :) – Kelson Vieira Jan 23 '13 at 15:42

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