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Show that if $f$ is a bounded function on $E$ that belongs to $L^{p_1}(E)$ then it belongs to $L^{p_2}(E)$ for any $p_2>p_1$

How can I insert argument about the boundedness of $f$? I can prove $L^{p_2} \subset L^{p_1}$ but I am stuck here. Please help.

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  • $\begingroup$ i am thinking define E1={x: f(x)≤ 1} and E2={{x: f(x) ≥1} $\endgroup$ – Biswa Nov 24 '15 at 19:06
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    $\begingroup$ There's no inclusion between the two spaces unless the measure of $E$ is finite. $\endgroup$ – Silvia Ghinassi Nov 24 '15 at 19:10
  • $\begingroup$ its a problem of royden fitzpatrick sec7.2 problem 13. the only condition is f is a bounded function $\endgroup$ – Biswa Nov 24 '15 at 19:13
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Let $f \in L^{p_1}(E)$ bounded. Then there exists $M\geq 0$ such that $\sup_{x \in E} |f(x)| \leq M$ and $\|f\|_{p_1}^{p_1}=\int_E |f(x)|^{p_1}\,dx < \infty$.


Case 1: $p_2<\infty$

For $p_2 > p_1$, we have \begin{align} \|f\|_{p_2}^{p_2}& =\int_E |f(x)|^{p_2}\,dx \\ &= \int_E |f(x)|^{p_2-p_1}|f(x)|^{p_1}\,dx \\ & \leq \left(\sup_{x \in E} |f(x)|^{p_2-p_1}\right)\int_E |f(x)|^{p_1}\,dx \\ & \leq \left(\sup_{x \in E} |f(x)|\right)^{p_2-p_1}\int_E |f(x)|^{p_1}\,dx \\ & \leq M^{p_2-p_1} \|f\|_{p_1}^{p_1} < \infty \end{align}

so that $\|f\|_{p_2}^{p_2} < \infty$ and hence $f \in L^{p_2}(E)$.

Note that we have used the fact that $p_2-p_1>0$ to claim $\sup_{x \in E} |f(x)|^{p_2-p_1}\leq \left(\sup_{x \in E} |f(x)|\right)^{p_2-p_1}$.


Case 2: $p_2=\infty$

Since $f$ is bounded, $\|f\|_{\infty}=\operatorname{ess sup}_{x \in E} |f(x)| \leq \sup_{x \in E} |f(x)| \leq M< \infty$ by hypothesis, so $f \in L^{\infty}(E)$.

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  • $\begingroup$ thanks so much.such a nice proof it is.that looks alright.but what about if P2=∞? there is no restriction on P2 $\endgroup$ – Biswa Nov 24 '15 at 19:33
  • $\begingroup$ See my edit.$ $ $\endgroup$ – Silvia Ghinassi Nov 24 '15 at 19:38
  • $\begingroup$ thank you so much.My confusion has gone $\endgroup$ – Biswa Nov 24 '15 at 19:41
  • $\begingroup$ You're welcome, @Biswa. Since you are a new user, I thought I should let you know about accepting and/or upvoting answers, see here. $\endgroup$ – Silvia Ghinassi Nov 24 '15 at 19:58
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Hint: If $|f|\le 1,$ then $|f|^{p_2} \le |f|^{p_1}.$

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  • $\begingroup$ The size of $E$ can be unbounded. You probably cannot use that. $\endgroup$ – user398843 Oct 31 '18 at 21:55
  • $\begingroup$ @user398843 We know $f\in L^{p_1}$ and $f$ is bounded. The result follows easily $\endgroup$ – zhw. Oct 31 '18 at 22:20

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