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.

  • $\begingroup$ i am thinking define E1={x: f(x)≤ 1} and E2={{x: f(x) ≥1} $\endgroup$ – Biswa Nov 24 '15 at 19:06
  • 1
    $\begingroup$ 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

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 $f \in L^{\infty}(E)$.

  • $\begingroup$ 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

Hint: If $|f|\le 1,$ then $|f|^{p_2} \le |f|^{p_1}.$

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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