Growth of $L^p$ norms as $p \to \infty$

Question

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.

Answer 1

$N(p)$ 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 $N(p)>\psi(p)$.

Answer 2

$N(p)$ can grow to $\infty$ however fast. Indeed, let $g\colon(0,\infty)\to(1,\infty)$ be any (say) strictly increasing continuous function, with $g(\infty-)=\infty$. We want to show that for some nonnegative function $f$ such that $f\in L^p$ for all $p>0$ we have
\begin{equation*} N(p)\ge g(p). \tag{1}\label{1} \end{equation*}

Note that
\begin{equation*} G(p):=g(p)^p>1 \end{equation*} is also strictly increasing and continuous in $p$. Let \begin{equation*} h_p:=\frac1{G(p)}\in(0,1) \end{equation*} and \begin{equation*} a(x):=\frac2{G^{-1}(1/x)}; \end{equation*} everywhere here, by default, $x\in(0,1)$. Note that $a(x)\downarrow0$ as $x\downarrow0$. Moreover, the condition that $g$ is increasing means exactly that
\begin{equation*} b(x):=a(x)\ln\tfrac1x \end{equation*} is decreasing in $x$.

Letting now \begin{equation*} f(x):=x^{-a(x)}=e^{b(x)} \end{equation*} and recalling that $a(x)\downarrow0$ as $x\downarrow0$, for each real $p>0$ we have $f(x)^p\le x^{-1/2}$ for all $x$ in a right neighborhood of $0$. Also, $f$ is bounded outside any right neighborhood of $0$. So, $f\in L^p$ for all $p>0$.

On the other hand, recalling that $b$ is decreasing, we have \begin{equation*} N(p)^p\ge \int_0^{h_p} f^p =\int_0^{h_p} e^{pb(x)}\,dx \ge h_p e^{pb(h_p)}=1/h_p=G(p)=g(p)^p, \end{equation*} so that \eqref{1} follows, as desired.