Functional Analysis – Growth of L^p Norms as p Approaches Infinity

ca.classical-analysis-and-odesfa.functional-analysislebesgue-measuremeasure-theoryreference-request

Let $f$ be a non-negative function defined on the unit interval. It is well known that $N(p) := \left(\int_0^1 f^p(t) dt\right)^{\frac{1}{p}} $ converges to $\operatorname{esssup}_{[0,1]} f$ when $p \to \infty$.

I am interested in the case when $\operatorname{esssup}_{[0,1]} f =+\infty$ but $f \in L^p$ for all $p \in [1,\infty)$. One can show that the function $p \to N(p)$ is increasing and continuous, using the Hölder inequality.

Examples suggest to me that the growth of $N(p)$ is polynomial as $p \to \infty$ (subexpontential would suffice for me). For example, if $f(x) = \lvert\ln(x)\rvert$ then $N(p) \sim p e$, where $e$ is the Euler constant.

Of course this is hard to believe, but I was wondering if there are results/theory available about the growth of $N(p)$ as $p \to \infty$.

Any chunk of information is appreciated.

Best Answer

$N(p)$ can grow arbitrarily quickly. Given a sequence $a_m \downarrow 0$ with $a_0=1$ and $a_m<a_{m-1}/2$ for all $m$, define $f(x)=x^{-1/m}$ for all $x \in (a_m,a_{m-1}]$ and $m \ge 1$. Then $N(p)<\infty$ for all $p<\infty$, but for $p>2m$, we have $$\int_0^1 f^p \,dx \ge \int_{a_m}^{2a_m} x^{-p/m} \ge \frac{1}{2a_m} \,.$$

Given a function $\psi(p) \uparrow \infty$, choose $a_m<\frac12 \psi(3m)^{-3m}$. For $p>6$, find an integer $m \in (p/3,p/2)$ to see that $N(p)>\psi(p)$.

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