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Assume $(X,\mathcal{M},\mu)$ is a measure space and for some $1\leq p<\infty$, $1\leq q<\infty$, $L^p(\mu)\subset L^q(\mu)$. Prove there is a constant $C>0$ so that $\|f\|_q\leq C\|f\|_p$ for all $f\in L^p(\mu)$.

I need help getting started on this.

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up vote 7 down vote accepted

$L^p(\mu)$ and $L^q(\mu)$ are Banach spaces, so we can apply the closed graph theorem: if $f_n\to f$ in $L^p$ and $f_n\to g$ in $L^q$ then passing to a subsequence $f_{n_k}\to f$ and $g$ almost everywhere. Hence the graph of the identity $\iota\colon L^p\to L^q$ is closed, so $\iota$ is continuous (and linear).

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thank you for your help –  john May 27 '12 at 21:13
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You could argue by contradiction: Suppose there was no such $C>0$. Then we can find a sequence $f_n\in L^p(\mu)$ satisfying

$$\Vert f_n\Vert_q \ge 4^n, \qquad \Vert f_n\Vert_p = 1$$

for all $n\in \mathbb N$. Now look at $$g(x) = \sum_{n=1}^\infty 2^{-n} |f_n(x)|$$ We have $g\in L^p(\mu)$, but $g$ is not in $L^q(\mu)$.

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In proving the last statement, is that just playing around with the triangle inequality? –  john May 27 '12 at 21:15
    
@john: We have $g \ge 2^{-m}|f_m|$ for all $m$. So upon integrating one obtains $$\Vert g\Vert_q \ge 2^{-m} \Vert f_m\Vert_q \ge 2^m$$ for any $m\in \mathbb N$. –  Sam May 27 '12 at 22:02
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