Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.