# Find an inequality for $\|f\|_p$ when $f=f \chi_E+f\chi_{\tilde E}$ and $m(E)=m(\{x:|f|> 1\})<\infty$

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 $\|f\|_p$.

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What's $\tilde{E}$? –  Matt N. 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. –  Matt N. 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
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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 $$\lVert f\rVert_p^p\geq \lVert f\chi_E\rVert_r^r+\lVert f\chi_{E^c}\rVert_s^s.$$