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.

Suppose $f\in L^p$, $f=f \chi_E+f\chi_{\tilde E}$ where $E=\{x:|f|> 1\}$ and $E$ has finite measure. Find an inequality for $\|f\|_p$ in terms of $\|f \chi_E\|_r$ and $\|f \chi_{\tilde E}\|_s$ where $r\le p$ and $s\ge p$. Assume the Lebesgue measure on $\mathbb R$ and ${\tilde E}=\mathbb R \setminus E$.

Since $E$ has finite measure, by Jensen's inequality, $\|f\chi_E\|_r \le k \|f\|_p$ for all $r\le p$ and $k=m(E)^{{1 \over r} - {1 \over p}}$. Since $|f\chi_{\tilde E}|\le 1$ and $s \ge p$, $\int |f\chi_{\tilde E}|^s \le \int |f|^p$ for all $s \ge p$. From this I don't see how to relate $\|f \chi_{\tilde E}\|_s$ and $\|f\|_p$.

share|improve this question
    
What's $\tilde{E}$? –  Rudy the Reindeer Apr 10 '12 at 14:04
    
@MattN. Sorry, it's $\mathbb R \setminus E$. –  Nicolas Essis-Breton Apr 10 '12 at 14:07
    
Thanks : ) That helps. –  Rudy the Reindeer Apr 10 '12 at 14:15
    
What kind of inequality are you looking for? –  Davide Giraudo Apr 10 '12 at 16:32
    
@DavideGiraudo This point is not clear from the question. With the work I did, I think the inequality should be of the form $\|f \chi_E\|_r+ \|f\chi_{\tilde E}\|_s \le c \|f\|_p$ for some $c>0$. –  Nicolas Essis-Breton Apr 10 '12 at 17:04

1 Answer 1

up vote 2 down vote accepted

We have, if $r\leq p$ and $|f(x)|\geq 1$, that $$|f(x)|^p=\exp(p\log |f(x)|)\geq \exp(r\log |f(x)|)=|f(x)|^r,$$ and as $p\leq s$, if $|f(x)<1$, then $$|f(x)|^p=\exp(p\log |f(x)|)\geq |f(x)|^s.$$ So we got the following inequalities: $|f|^p\chi_E\geq |f|^r\chi_E$ and $|f|^p\chi_{E^c}\geq |f|^s\chi_{E^c}$. This gives $$\lVert f\rVert_p^p\geq \lVert f\chi_E\rVert_r^r+\lVert f\chi_{E^c}\rVert_s^s.$$

share|improve this answer

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.