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Let $(X,\mathfrak{X},\mu)$ be a finite measure space and let $f \in L^p(\mu)$. Use Holder's inequality to show that: \begin{equation} f \in L^r(\mu) \end{equation}

for $1 \leq r \leq p < \infty$.

Applying Holder's inequality show that: \begin{equation} ||f||_r \leq ||f||_p \mu(X)^s \end{equation}

where $s = (1/r) - (1/p)$. Thus if $\mu(X) = 1$ then $||f||_r \leq ||f||_p$.

A finite measure space satisfies $\mu(X) < \infty$.

Furthermore, Holder's inequality states that: \begin{equation*} \int |fg| \,d\mu \leq \left( \int |f|^p \,d\mu \right)^{1/p} \left(\int |g|^q \,d\mu \right)^{1/q} \end{equation*}

with $f,g$ measurable and $ 1 < p < \infty$ and $q = \frac{p}{p-1}$

Since $f \in L^p(\mu)$ we know that:

\begin{equation} \left( \int |f|^p \,d\mu \right)^{1/p} < \infty \end{equation}

However, I do not see how to apply it in this case, as there are restrictions on $q$, which do not appear in the result I am trying to prove. Any help is much appreciated. I am stuck.

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Use $g\equiv 1$ and appropriate values for $s,r,p$. –  Dirk Oct 1 '12 at 6:18

2 Answers 2

up vote 2 down vote accepted

A little bit of motivation: we are going to use Holder's inequality on $|f|^r$ multiplied by $g \equiv 1$. The trick here is that we want to get rid of the $r$ and replace it with a $p$, so we will manufacture a power that precisely produces this. In particular, notice that $|f|^p = (|f|^r)^{p/r}$. Let $s$ be such that $1/s + r/p = 1$. Then, using Holder, $$ \int |f|^r \, d\mu \le \left(\int (|f|^r)^{p/r} \, d\mu\right)^{r/p}\left( \int \, d\mu \right)^{1/s} = \|f\|_{p}^r \mu(X)^{1/s} < \infty.$$ I suppose, ultimately, the sneaky thing in this problem is that the $p,q$ chosen for Holder's inequality are not the same as the given $p$ in the setup of the problem.

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In general given $f\in L^p(\mu)$ you can't say anything about inclusions like $f\in L_r(\mu)$ with $r\in[1,p)$.

Moreover, for any $p\in[1,+\infty)$ there exist function $f\in L_p(\mathbb{R}_+)$, such that $f\notin L_r(\mathbb{R}_+)$ for all $r\in[1,+\infty)\setminus\{p\}$. For details see the question Is it possible for a function to be in $L^p$ for only one $p$?

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