An inequality about $L^p$ norm?

Question

If $f\in L^p$ for some $0<p<\infty$, and every set of positive measure in $X$ has measure at least $m$, show that for all $p<q<\infty$, with $\|f\|_{L^q}\leq m^{\frac{1}{q}-\frac{1}{p}}\|f\|_{L^p}$?

I can prove it by starting with simple functions. By homogeneity, one can assume $\|f\|_{L^p}=m^{1/p}$, then imitate the proof that $\|x\|_{q}\leq \|x\|_p$ for $0<p<q<\infty$ in $\ell^p(\mathbf{N})$， the proof can be found here. But I think the proof is not very good, can we prove it using Holder inequality? Since $q>p$, we can not use Holder inequality directly.

Answer 1

The statement is not true if $\|f\|_\infty=\infty$ since $\|f\|_q\xrightarrow{q\rightarrow\infty}\infty$. (here is a proof of that)

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Suppose $\|f\|_\infty<\infty$ and that $f\in L_p$ for some $p>0$.

For $0<\alpha<\|f\|_\infty$, an application of Markov-Chebyshev's inequality gives

$$0<\alpha\Big(\mu(|f|>\alpha\Big)^{1/p}\leq\|f\|_p$$ and so,

$$0<\alpha m^{1/p}\leq\|f\|_p$$ which in turn implies $$\|f\|_\infty\leq\|f\|_pm^{-1/p}$$

For $q>p$, \begin{aligned} \|f\|^q_q&=\int|f|^q\,d\mu=\int|f|^{q-p}|f|^p\,d\mu\leq\|f\|^{q-p}_{\infty}\|f\|^p_p\leq \big(\|f\|_pm^{-1/p}\big)^{q-p}\|f\|^p_p \end{aligned}

Putting things together gives

$$ \|f\|_q\leq m^{\tfrac{1}{q}-\tfrac{1}{p}}\|f\|_p $$

Answer 2

I happened to find this exercise in Terry's notes and I was thinking about the same question. But I now believe application of Hölder is not feasible, because it gives upper bounds in terms of norms of higher exponents (correct me if I am wrong). My argument basically follows the case of $\ell^p$.

Assume every set in $X$ of positive measure has measure at least $m$. Let $f\in L^{p}$. We need to find a bound for the essential supremum in terms of $L^{p}$-norm (the essential supremum cannot be $+ \infty$, since otherwise the function is unbounded on a set of measure at least $m$ and thus not in $L^p$). Let $x\in X$ be such that $$ |f(x)|+\epsilon>\sup|f(x)|. $$ And let $E$ be a set of finite positive measure where each $x\in E$ satisfies the above for some $\epsilon$ small enough. Then

\begin{eqnarray*} \int_{X\backslash E}|f(x)|^{p}d\mu+\int_{E}\left(|f(x)|+\epsilon\right)^{p}d\mu &\geq & \int_{E}\left(|f(x)|+\epsilon\right)^{p}d\mu \\ &>&\int_{E}\sup|f(x)|d\mu \\ &\geq& m\cdot\sup|f(x)|.\\ \end{eqnarray*} Sending $\epsilon\to0$, we have $$\|f\|_{p}^{p}\geq\int_{E}\sup|f(x)|d\mu\geq m\cdot\sup|f(x)|.$$ So we have for $p_{0}\leq p_{1}$ \begin{eqnarray*} \left(\int_{X}|f|^{p_{1}}\right)^{1/p_{1}} & = & \left(\int_{X}|f|^{p_{0}}|f|^{p_{1}-p_{0}}\right)^{1/p_{1}}\\ & \leq & \left(\int|f|^{p_{0}}\right)^{1/p_{1}}\left(\sup|f|^{p_{1}-p_{0}}\right)^{1/p_{1}}\\ & \leq & \left(\int|f|^{p_{0}}\right)^{1/p_{1}}\left(\int|f|^{p}\right)^{\frac{1}{p_{0}p_{1}}(p_{1}-p_{0})}m^{\frac{1}{p_{0}p_{1}}(p_{1}-p_{0})}\\ & = & \left(\int|f|^{p_{0}}\right)^{1/p_{0}}m^{\frac{1}{p_{0}}-\frac{1}{p_{1}}}. \end{eqnarray*}

This type of argument, I find, is typical to use in proving the kind of inequalities in reverse order of Hölder's.