Lim Sup of n-th root of integrable functions

Question

I am working on some analysis problems for a qualifying exam and came across this one which has given me some problems:

Let $f_{n}$ be $\textbf{nonnegative}$ measurable functions on a measure space $(X,\mathcal{M},\mu)$ satisfying $\int f_{n}\,\mathrm{d}\mu=1$ for all $n=1,2,...$ Prove that \begin{align} \mathrm{lim\,sup}\,(f_{n}(x))^{1/n}\leq1. \end{align}

My intuition tells me that, since $\int f_{n}\,\mathrm{d}\mu=1$ for each $n$, we know $f_{n}(x)<\infty$ a.e. (Right?) Initially, then, I thought to then say that there is some finite $M>0$ so that $\,f_{n}(x)\leq M$ a.e., so then $(f_{n}(x))^{1/n}\leq(M)^{1/n}$, and $\mathrm{lim\,sup}\,(f_{n}(x))^{1/n}\leq1$.

But I am guessing that there is not necessarily such an $M$; in particular, if I consider $f_{n}(x)=(2\sqrt{x})^{-1}$ on the interval $(0,1)$ and 0 otherwise, then $||f_{n}||_{1}=1$, but $||f_{n}||_{\infty}=\infty$, so I think no such $M$ exists. Is there a less-common measure theory theorem I'm missing out on using here?

Answer 1

Consider $\mu \{x:f_n(x) >(1+1/\sqrt n) ^{n}\}$. This is $ \leq (1+1/\sqrt n) ^{-n}$ because $\int f_n d\mu=1$. Check that $\sum_n (1+1/\sqrt n) ^{-n} <\infty$. This implies that limsup of the sets $\{x:f_n(x) >(1+ 1/\sqrt n) ^{n}\}$ has measure $0$. Hence $f_n^{1/n} \leq 1+1/\sqrt n$ for $n$ sufficiently large, for almost all $x$. Hence $\lim \sup f_n^{1/n} \leq 1$ almost everywhere.

[ I have used two elmentary facts from measure theory: a) $\mu \{f>t\} \leq \frac 1 t \int f$ if $f\geq 0$ is measurable and $t>0$. b) If $\sum_n \mu(E_n) <\infty$ then $\mu (\lim \sup E_n) =0$ where $\lim \sup E_n$ is the set of points that belong to infinitely many of the sets $E_n$].

Answer 2

Consider for $0\leq r<1$, the function $$ F_r(x) = \sum_{j=1}^\infty r^jf_j(x). $$ Then we have $$ \int F_r d\mu = \frac{r}{1-r}<\infty $$ for all $0\leq r<1$. Hence it holds that for each $r$, $$ F_r(x) <\infty $$ for $\mu$-almost every $x$. This implies $$ \bigcap_{j\in \mathbb{N}}\{x\;|\;F_{1-\frac{1}{j}}(x)<\infty\} $$ has a full measure. Finally observe that if $x$ satisfies $$ F_{1-\frac{1}{j}}(x)<\infty, \quad\forall j\geq 1, $$ then the radius of convergence $R(x)$ of power series $\sum_{n\geq 1} r^nf_n(x)$ satisfies $$ R(x) = \limsup_n( f_n(x))^{-\frac{1}{n}}\geq 1, $$ as desired.

$\textbf{Note:}$ It can also be proved by brute force method. Assume to the contrary that $$ \limsup_n( f_n(x))^{\frac{1}{n}}\geq 1+\epsilon $$ for $x\in E$ with $\mu (E)>0$. Then for each $x\in E$, there is $N$ such that $$ n\geq N \Rightarrow f_n(x) \geq (1+\frac{\epsilon}{2})^n\quad\cdots(*). $$ If we choose minimal $N$ such that $(*)$ holds, $x\in E\mapsto N(x)$ is measurable and finite for almost every $x\in E$. Choose $M$ such that $$ \mu\{x\in E\;|\;N(x)\leq M\} >\frac{\mu (E)}{2}>0. $$ If we integrate on $F=\{x\in E\;|\;N(x)\leq M\}$, for any $j\geq 1$, we have $$ 1\geq \int f_{M+j}d\mu \geq \int_F f_{M+j}d\mu\geq \mu(F)(1+\frac{\epsilon}{2})^{M+j} \to \infty, $$ leading to contradiction.