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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.

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    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 $L^{p_1}$.