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Suppose $1, then show $L^{p_1}[a,b] \supset L^{p_2}[a,b]$.

I was able to show $\|f\|_{p_1} \le \|f\|_{p_2} (b-a)^{1/p_1 - 1/p_2}$ but I'm not sure how to proceed from here.

  • 1
    Hint: Jensen's inequality.2011-02-11
  • 10
    Well, your inequality shows that if $\|f\|_2$ is finite, then so is $\|f\|_1$, what else do you want?2011-02-11
  • 0
    Sorry, you must have already used something like Jensen's inequality to get what you have so far. But aren't you almost there? If you have an inequality then that helps you compare when the norms are finite.2011-02-11
  • 0
    oh haha sorry that was silly of me.2011-02-11

1 Answers 1

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As Jonas T commented, your inequality shows the inclusion. If you use Jensen or Hölder to get a general inequality that shows $\|f\|_{p_1}\leq C\cdot\|f\|_{p_2}$ for a constant $C$ depending only on $b-a$, $p_1$, and $p_2$, then in particular you know that $\|f\|_{p_1}$ is finite whenever $\|f\|_{p_2}$ is.

Here is another way to see the containment. Because you're integrating over a set of finite measure, the only thing that might cause a power of $|f|$ to not be integrable is that $|f|$ is too big. But when $|f|$ is big, the larger powers of $|f|$ are bigger than smaller powers.

More formally, suppose $f$ is in $L^{p_2}$. Define $A=\{x\in [a,b]: |f(x)|\leq 1\}$ and $B=\{x\in[a,b]: |f(x)|>1\}$. Then $$\int_a^b|f|^{p_1} = \int_A|f|^{p_1}+\int_B|f|^{p_1}\leq \int_A 1 + \int_B |f|^{p_2}\leq (b-a) +\int_a^b|f|^{p_2}<\infty,$$ so $f$ is in $L^{p_1}$.