How to show $\lim_{p\to \infty} x_p=1$ Let $f$ be nonnegative, continuous, and strictly increasing on $[0,1]$. For $p >0$, let $x_p$ be the number in $(0,1)$ which satisfies $f^p(x_p) = \int^1_0 f^p(x)dx$. Find $\lim_{p\to \infty} x_p=?$ I tried letting $f(x)=x$, $x^2$, $x^n$, etc. The limit seems converging to 1. However, for any other functions, how can I show this limit is 1? 
Thanks a lot,
-Belen
 A: This is a particular case of a classical real analysis problem. But in this setting it can actually be solved in an easier way.
Let $c$ be any point in $(0,1)$, then (since $f$ is increasing)
$$
\int_0^1 f^p \geq \int_x^1 f^p \geq (1-c)f(c)^p,
$$
so
$$
\biggl( \int_0^1 f^p \biggr)^{1/p} \geq (1-c)^{1/p} f(c).
$$
Taking limits as $p \to \infty$ we see that
$$
\limsup_{p \to \infty} \biggl( \int_0^1 f^p \biggr)^{1/p} \geq f(c)
$$
for all $c \in (0,1)$, and so by continuity
$$
\limsup_{p \to \infty} \biggl( \int_0^1 f^p \biggr)^{1/p} \geq \max_{[0,1]}f.
$$
On the other hand
$$
\int_0^1 f^p \leq \max_{[0,1]}f^p
$$
(since $f$ is increasing). Putting these together we find that
$$
\lim_{p \to \infty} \biggl( \int_0^1 f^p \biggr)^{1/p} = \max_{[0,1]}f.
$$
Notice that $\max_{[0,1]}f = f(1)$ since $f$ is increasing, so
$$
f(x_p) = \biggl( \int_0^1 f^p \biggr)^{1/p} \to f(1)
$$
as $p \to \infty$. But now, since $f$ is strictly increasing and continuous, it has a continuous inverse we can apply it and obtain
$$
x_p \to 1.
$$
