How to show that $\|f\|_{L^p} \to \|f\|_{L^\infty}$ as $p \to \infty$ if $f \in L^\infty \cap L^{p_0}$ for some $0 < p_0 < \infty$?

Question

This problem is from Terrence Tao's blog: Exercise 4

If $f \in L^\infty \cap L^{p_0}$ for some $0 < p_0 < \infty$, show that $\|f\|_{L^p} \to \|f\|_{L^\infty}$ as $p \to \infty$. (Hint: use the Monotone Convergence Theorem.)

I have worked this problem out using the Chebychev's inequality (or Markov's inequality by some authors) for integrals and the squeeze theorem, but I cannot work it out by the given hint using the Monotone Convergence Theorem since I have no idea in seeing or constructing a monotone sequence.

EDIT: I once had an "answer" given below, using the Simple Approximation Lemma which works well for any measurable function but there is a gap already pointed out by someone that if we write the process out, we need to justify the interchanging of limits $$\lim_{p\to \infty}\lim_{n\to \infty}||f_n||_p=\lim_{n\to \infty}\lim_{p\to \infty}||f_n||_p$$

And I asked Professor Tao on his blog and he gave a comment: "…as Holder’s inequality will give enough quantitative continuity to interchange limits. …" But I do not figure out in what sense does it reveal the continuity and how to see the interchangeability of limits by Holder's inequality? Also the condition $f \in L^{p_0}$ has not been used yet. (I have put foward another question but he has not replied yet)

I would be appreciate it if anyone has an idea or an answer on how to rigorously work it out.

Answer 1

Let $(X,\mathcal{F},\mu)$ be a measure space and let $f:X\rightarrow\mathbb{R}$ be a measurable function such that $||f||_{p_{0}}<\infty$ for some $0<p_{0}<\infty$ and $||f||_{\infty}<\infty$. If $||f||_{p_{0}}=0$, then $f=0$ a.e. and we are done. Therefore, we consider the case $||f||_{p_{0}}>0$ only.

Clearly, by replacing $f$ with $|f|$, without loss of generality, we may assume that $f\geq0$. Notice that in what we are going to prove $\lim_{p\rightarrow\infty}||f||_{p}=||f||_{\infty}$, the LHS and the RHS are homogeneous. By replacing $f$ with $f/||f||_{\infty}$, we may further assume that $||f||_{\infty}=1$. (Note that $||f||_{\infty}\neq0$, because it is false that $f=0$ a.e.).

Let $p\geq p_{0}$, then we have that \begin{eqnarray*} ||f||_{p}^{p} & = & \int f^{p}d\mu\\ & \leq & \int f^{p_{0}}d\mu\\ & = & ||f||_{p_{0}}^{p_{0}}. \end{eqnarray*} Therefore, $||f||_{p}\leq||f||_{p_{0}}^{\frac{p_{0}}{p}}\rightarrow1$ as $p\rightarrow\infty$. It follows that $\limsup_{p\rightarrow\infty}||f||_{p}\leq1$. (In the above, we also show that $||f||_{p}<\infty$ for all $p\geq p_{0}$. This fact will be used at later time.)

On the other hand, let $\alpha\in(0,1)$ be arbitrary and define $A_{\alpha}=\{x\in X\mid f(x)>\alpha\}$. Since $\alpha$ is not an essential upper bound of $f$, $\mu(A_{\alpha})>0$. Let $p\geq p_{0}$. We have that \begin{eqnarray*} ||f||_{p}^{p} & \geq & \int_{A_{\alpha}}f^{p}d\mu\\ & \geq & \alpha^{p}\mu(A_{\alpha}). \end{eqnarray*} Therefore $\infty>||f||_{p}\geq\alpha\mu(A_{\alpha})^{\frac{1}{p}}.$ It follows that $\liminf_{p\rightarrow\infty}||f||_{p}\geq\alpha\lim_{p\rightarrow\infty}\mu(A_{\alpha})^{\frac{1}{p}}=\alpha$. Since $\alpha\in(0,1)$ is arbitrary, we further have $\liminf_{p\rightarrow\infty}||f||_{p}\geq 1.$

Hence, $\lim_{p\rightarrow\infty}||f||_{p}=1=||f||_{\infty}.$

Answer 2

I think I may figure out what did Tao mean, and I sketch the proof as follows:

By taking the positive part and negative part respectively we may assume that $f$ is an unsigned (nonnegative) measurable function.

First suppose that $f$ is a simple function: $f = \sum^{n}_{i=1}a_i \chi_{E_{i}}$. Then we have an estimation: $$||f||_\infty \cdot \mu(\{f = ||f||_\infty\})^{1/p} \leq ||f||_p \leq ||f||_\infty\cdot \mu(\bigcup_{i=1}^n E_i)^{1/p}$$ Then by Squeese Theorem we conclude that $\lim_{p\rightarrow\infty}||f||_{p} = ||f||_{\infty}$.

For general (unsigned) measurable function we may apply the Simple Approximation Lemma and the Monotone Convergence Theorem to finish the proof.