# Show that if $f$ is a bounded function on $E$ and $f\in L^{p_1}$ then $f\in L^{p_2}$ for any $p_2>p_1$. [duplicate]

I'm working through Royden & Fitzpatrick's Real Analysis, and one of the questions in the introductory chapters of $L^p$ spaces reads:

Show that if $f$ is a bounded function on $E$ and $f\in L^{p_1}$ then $f\in L^{p_2}$ for any $p_2>p_1$.

There is a theorem in the section that reads:

If $E$ is a measurable set, $m(E)<\infty$ and $1\leq p_1<p_2\leq \infty$, then $L^{p_2}\subseteq L^{p_1}$

Naturally, I would use a similar approach as the one they use to prove this theorem, with Holder's inequality, but the confusing thing is that what I'm after is the opposite. $f\in L^{p_1} => f\in L^{p_2}$ means that $L^{p_1}\subseteq L^{p_2}$, instead of the other way around. So I'm at a loss, either I misunderstand the statement of the theorem against the question, or I have to use something other than Holder's inequality, and this section isn't giving me all that many tools. Any help is appreciated!

## marked as duplicate by Martin R, Community♦Dec 5 '15 at 14:23

Hint: Use $$|f(x)|^{p_2} = |f(x)|^{p_2-p_1} \cdot |f(x)|^{p_1} \leq \|f\|_{\infty}^{p_2-p_1} |f(x)|^{p_1}.$$
• Thanks a lot, So, $\int_E|f|^{p_2}\leq ||f||_\infty^{p_2-p_1}\int_E|f|^{p_1}<\infty$ Should work, long as I'm allowed to pull out the $||f||_\infty^{p_2-p_1}$ from the integral. – Mike Dec 5 '15 at 8:45