Analysis Problem of the Day 6

Today’s problem is a commonly appearing exercise in the theory of $L^p$ spaces.

Problem: Let $X$ be a finite measure space, and let $f \in L^p(X)$ for all $1 \leq p \leq \infty.$ Show that $\lim_{p \to \infty} \|f\|_p = \|f\|_\infty.$

Solution: Let’s first understand intuitively why this might be true. As $p \to \infty,$ the parts where $|f|>1$ start contributing more to the $L^p$ -norm of $f$ (since $a^p$ gets very large for large $p$ and $|a|>1,$ ) while the parts where $|f|<1$ contribute less and less to the norm. Thus, we would expect that in the limit as $p \to \infty,$ the part that will contribute to the norm is the largest value of $f,$ or in other words, the essential supremum $\|f\|_\infty.$ It also suggests that for this problem, instead of thinking horizontally, that it is instead helpful to think vertically, i.e. in terms of the sets $\{x \in X: |f(x)|>\lambda\}.$

Now for a formal proof. We prove the existence and value of the limit by showing both $\liminf_{p \to \infty} \|f\|_p \leq \|f\|_\infty$ and $\limsup_{p \to \infty} \|f\|_p \geq \|f\|_\infty.$ By Hölder’s inequality with $f$ and $1_X,$ we have

$\|f\|_p \leq \mu(X)^{\frac{1}{p}-\frac{1}{q}} \|f\|_q$

for $1 \leq p < q \leq \infty.$ In particular, for $q = \infty,$ we get that $\|f\|_p \leq \mu(X)^{\frac{1}{p}}\|f\|_\infty.$ Taking the limsup on both sides yields $\limsup_{p \to \infty} \|f\|_p \leq \|f\|_\infty.$

For the other direction, we recall the decomposition

$\int |f|^p dx = \int_{|f|>\lambda} |f|^p dx + \int_{|f| \leq \lambda} |f|^p dx \geq \lambda^p \mu(\{x:|f|>\lambda\}).$

Now, the key here is that by definition, $\|f\|_\infty = \sup\{\lambda: \mu(|\{x:|f|>\lambda\}) >0\}.$ Thus, for any $\epsilon>0,$ let $\lambda = \|f\|_\infty - \epsilon.$ Taking $p$ -th roots on both sides yields

$\|f\|_p \geq (\|f\|_\infty-\epsilon)\mu(\{x:|f|>\|f\|_\infty-\epsilon\})^{\frac{1}{p}}.$

Taking the $\liminf$ as $p \to \infty$ on both sides yields $\liminf_{p \to \infty}\|f\|_p \geq \|f\|_\infty - \epsilon,$ and since $\epsilon>0$ is arbitrary, we obtain

$\limsup_{p \to \infty} \|f\|_p \leq \|f\|_\infty \leq \liminf_{p \to \infty} \|f\|_p,$

so we conclude that $\lim_{p \to \infty} \|f\|_p = \|f\|_\infty.$

Remark 1: By interpolation theorems, it suffices to take $f \in L^p \cap L^\infty$ for any $p<\infty$ for the statement of the problem to hold, since then $f \in L^q$ for any $q>p.$

Remark 2: This result still holds true if $\mu(X)=\infty.$ The only issue in our argument is now the application of Hölder’s, which only works on finite measure spaces. In this case, we can instead use the following trick, taking $r$ to be large enough so that $f \in L^r$ :