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let $(\Omega ,A,\mu)$ be a measure space and let $1<p<q<r<\infty$. show that $$L^q(\Omega)\subset L^p(\Omega)+L^r(\Omega)$$

answer:Iknow that I can assume that $f\in L^q(\Omega)$,$E=\{\omega \in\Omega : \mid f(\omega )\mid >1\}$ and decompose fby $f1_E+f1_(\Omega$\E) but I donot know how?

moj
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Suggestion: $E$ is a set where $f$ is "large". When $f > 1$, $|f|^p$ increases and decreases in direct relation to $p$. So if you know $f$ integrates at the $q$th power, what other types of powers will be integrable on this set?

Likewise when $|f| < 1$ we are talking about the "tail" of the integral if $\Omega$ has infinite measure, or simply the region where $f$ is small on a finite $\Omega$. Here, increasing the exponent increases integrability. For a hopefully helpful mental example, consider $\int_1^{\infty} \frac{1}{x} \, dx$ vs $\int_1^{\infty} \frac{1}{x^2} \, dx$.

Hopefully you can use this idea to show that $f 1_E \in L^p$ and $f 1_{E^c} \in L^r$. But $f = f 1_E + f 1_{E^c}$, so $f$ is a sum of a $L^p$ function and a $L^r$ function. Since $f$ was a general member of $L^q$, do you see how we have shown the desired inclusion?

Jason Knapp
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  • but I can not understand still :( how can I solve it?should I from minkowski theory solve it?? – moj Oct 30 '14 at 11:46
  • Just use direct inequalities. I'll give more detail on $E$: $\int_{E} |f|^p d\mu \leq \int_{E} |f|^q , d\mu$ because for every $x \in E$, $|f(x)|^p \leq |f(x)|^q$, since $|f(x)| > 1$ on $E$ and $p < q$. Since the integral on the right is finite, the integral on the left is as well. Does this make sense, and then can you extent this to show the same fact using $r > q$ on $E^c$? – Jason Knapp Oct 30 '14 at 12:27
  • thank u:).but for $E^{c} $ how can I say the same? because 1<q<r .and from this that $f\in L^q$ I can not conclude that $f\in L^r$. would u plz show me how I can prove it because I have to prove it on the board and I must know the details!! – moj Oct 30 '14 at 12:39
  • When $0 < t < 1$, $t^r < t^q$ since $r > q > 1$. Thus on $E^c$, $|f(x)|^r < |f(x)|^q$. Yes? – Jason Knapp Oct 30 '14 at 13:15
  • yes,and $\int_{E^c}\mid f(\omega)\mid ^r+\int_E \mid f(\omega)\mid^p$<$\int_\Omega\mid f(\omega)\mid^q$ and from this conclude we that $L^q(\omega)\in L^p(\omega)+L^r(\omega)$?? – moj Oct 30 '14 at 15:14
  • Answer updated. – Jason Knapp Oct 30 '14 at 15:18
  • and if r=$\infty$?? – moj Nov 02 '14 at 13:09