Find $\lim\limits_{p\to+\infty}x_p.$ (a problem in mathematical analysis)

Question

Please give me some hints not a whole solution to solve this problem.

Problem Let $f$ be a continuous positive strictly monotone function on the segment $[0,1].$ For all $p > 0$ consider the point $x_p$ such that $$[f(x_p)]^p = \int_0^1[f(x)]^p dx.$$ Find $\lim\limits_{p\to+\infty}x_p.$

Answer 1

Proposition: Let $C\left([0,1]\right)$ be the vector space of all real-valued continuous functions defined on $[0,1]$. For each $p\in[1,\infty)$, define $||f||_{p}=\left(\int_{0}^{1}|f(x)|^{p}\,dx\right)^{\frac{1}{p}}$. Define $||f||_{\infty}=\sup_{x\in[0,1]}|f(x)|$. Then for each $f\in C\left([0,1]\right)$, we have $||f||_{\infty}=\lim_{p\rightarrow\infty}||f||_{p}$.

Proof: Let $f\in C\left([0,1]\right)$. If $||f||_{\infty}=0$, we are done because $||f||_{p}=||f||_{\infty}=0$ for all $p\in[1,\infty)$. Suppose that $||f||_{\infty}>0$. By direct verification, for each $p\in[1,\infty)$, we have $||f||_{p}\leq||f||_{\infty}$. In particular, $\limsup_{p\rightarrow\infty}||f||_{p}\leq||f||_{\infty}.$ On the other hand, choose $x_{0}\in[0,1]$ such that $|f(x_{0})|=||f||_{\infty}.$ Let $0<\alpha<||f||_{\infty}$ be aribtrary. Since $|f|$ is continuous at $x_{0}$, there exists $\delta>0$ such that $|f(x)|>\alpha$ whenever $x\in(x_{0}-\delta,x_{0}+\delta)$. (If $x_{0}$ is an end-point of $[0,1]$, the interval is adjusted in an obvious way.) Now $$ \int_{0}^{1}|f(x)|^{p}\,dx\geq\int_{x_{0}-\delta}^{x_{0}+\delta}|f(x)|^{p}\,dx\geq\alpha^{p}\cdot2\delta. $$ Therefore $||f||_{p}\geq\alpha\cdot(2\delta)^{\frac{1}{p}}.$ In particular, $\liminf_{p\rightarrow\infty}||f||_{p}\geq\liminf_{p\rightarrow\infty}\alpha\cdot(2\delta)^{\frac{1}{p}}=\alpha$. Since $\alpha\in(0,||f||_{\infty})$ is arbitrary, we have $\liminf_{p\rightarrow\infty}||f||_{p}\geq||f||_{\infty}$. Now, $\limsup_{p\rightarrow}||f||_{p}\leq||f||_{\infty}\leq\liminf_{p\rightarrow\infty}||f||_{p}$. It follows that $\lim_{p\rightarrow\infty}||f||_{p}$ exists and equals to $||f||_{\infty}$.

///////////////////////////////////////////////////////////

We consider the strictly increasing case.

Let $f:[0,1]\rightarrow(0,\infty)$ be a strictly increasing function.

Firstly, we show that for each $p\in[1,\infty)$, there exists a unique $x_{p}\in[0,1]$ such that $f(x_{p})=||f||_{p}$. Let $p\in[1,\infty)$. Note that $$ f(0)^{p}=\int_{0}^{1}f(0)^{p}\,dx\leq\int_{0}^{1}f(x)^{p}\,dx\leq\int_{0}^{1}f(1)^{p}\,dx=f(1)^{p}. $$ Taking $p$-th root, we have $$ f(0)\leq||f||_{p}\leq f(1). $$ By Intermediate Value Theorem, there exists $x_{p}\in[0,1]$ such that $f(x_{p})=||f||_{p}$. Since $f$ is strictly increasing, $x_{p}$ is unique.

Next, it is well-known that the inverse function $f^{-1}:[f(0),f(1)]\rightarrow[0,1]$ of $f$ exists and is continuous. Therefore, $$ x_{p}=f^{-1}\left(f(x_{p})\right)=f^{-1}\left(||f||_{p}\right). $$ By continuity of $f^{-1}$ and the above Proposition, we have $$ \lim_{p\rightarrow\infty}x_{p}=f^{-1}\left(\lim_{p\rightarrow\infty}||f||_{p}\right)=f^{-1}\left(||f||_{\infty}\right)=f^{-1}\left(f(1)\right)=1. $$

Answer 2

Hint:

1. Since $f$ is a positive continuous function, it admits a maximum $M>0$. Moreover, $f$ is strictly monotone, so the $M$ is achieved only at $x=1$ if $f$ is increasing, or only at $x=0$ if $f$ is decreasing. Assume, just to fix the ideas, that $f$ is increasing.

2. [Difficult point] You can prove that $$ \left(\int_0^1 [f(x)]^p\, dx\right)^{1/p} \to \max_{[0,1]} f = f(1), \qquad \text{as}\ p\to +\infty. $$

3. By the previous points it holds $$ f(x_p) \to f(1) \quad \text{as}\ p\to +\infty. $$ On the other hand, $f$ is continuous and strictly increasing...