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Suppose that $f$ is a continuous function on $[0,1]$. Moreover, $f(0)=0$, $f(1)=1$. Calculate the following limit $$\lim_{n\rightarrow +\infty}n\int_0^1 f(x)x^{2n}dx.$$

From mean value theorem, there would be a $\zeta_n\in[0,1]$ such that $\int_0^1f(x)x^{2n}dx=f(0)\int_0^{\zeta_n}x^{2n}dx+f(1)\int_{\zeta_n}^1x^{2n}dx=\int_{\zeta_n}^1x^{2n}dx=\frac{1-(\zeta_n)^{(2n+1)}}{2n+1}$.

So the answer would be 1/2 if we could prove that $\zeta_n$ has a upper bound less than $1$.

Any one has some better ideas to prove this?

user26857
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Intoks Liobein
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  • I don't think the hypothesis $f(0)=0$ figures into the answer. – zhw. Jul 20 '17 at 05:38
  • I don't agree that the mean value theorem grants you the statement you've written, considering that we can always choose $f$ such that $f(0) = 0$, $f(1) = 1$, and $\int_0^1 f(x)x^{2n},\mathrm{d}x < 0$ (for a particular $n$). This would imply that there is no $\zeta_n\in [0, 1]$ such that $\int_0^1 f(x)x^{2n},\mathrm{d}x = \int_{\zeta_n}^1 x^{2n},\mathrm{d}x$. For example, what is $\zeta_1$ for $f(x) = 6x^2-5x$? – Michael L. Jul 20 '17 at 08:56
  • Yes, it needs $f$ to be monotonic as a prerequisite. – Intoks Liobein Jul 20 '17 at 09:00
  • The limit holds even if $f$ is not monotonic, though, which is nice. – Michael L. Jul 20 '17 at 09:01
  • mean value theorem states that $\zeta_n\in(0,1)$ thus $\zeta_n^{2n+1}\to 0$ as $n\to\infty$ – Raffaele Jul 20 '17 at 10:42
  • @Raffaele That reasoning is incorrect. What if $\zeta_n = \sqrt[2n+1]{c}$ for some $c\in (0, 1)$? Then $\zeta_n^{2n+1} = c$ for all $n$. – Michael L. Jul 20 '17 at 18:06

2 Answers2

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Hint: for any $t \in (0,1)$, $\lim_{n \to \infty} n \int_0^t x^{2n}\; dx = 0$ while $\lim_{n \to \infty} n \int_t^1 x^{2n}\; dx = 1/2$. Choose $t$ so that $1-\epsilon < f(x) < 1 + \epsilon$ for $t \le x \le 1$, $B$ so that $|f(x)| \le B$ for $0 \le x \le 1$, and estimate $n \int_0^t f(x) x^{2n}\; dx$ and $n \int_{t}^1 f(x) x^{2n}\; dx$.

Robert Israel
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For the sake of exercise, I decided to approach this a different way. Let's prove a bit of a technical (but very helpful) lemma first.

Lemma 1: For any continuous function $f : [0, 1]\to \mathbb{R}$ such that $f(0) = f(1) = 0$ and any $\epsilon > 0$, there is a function $f_-(x) = ax^2-ax-\epsilon$ and a function $f_+(x) = bx^2-bx+\epsilon$ with $a, b\in \mathbb{R}$ such that $f_-\leq f\leq f_+$ on $[0, 1]$.

Proof: First, we note that $f$ is uniformly continuous on $[0, 1]$ (as this is a compact space). Therefore, for any $\epsilon > 0$, there is a $\delta > 0$ such that $\lvert x-y\rvert < \delta$ for any $x, y\in [0, 1]$ implies $\lvert f(x)-f(y)\rvert < \epsilon$. We also note that $f$ is bounded on $[0, 1]$. We let $\lvert f\rvert\leq M$. Then, for $0 < \epsilon < M$ we choose $a = \frac{M-\epsilon}{\delta (1-\delta)} > 0$ (where $\delta < 1$ satisfies the uniform continuity condition for $\epsilon$). This gives us the following facts:

(1) $f_-(0) = f_-(1) = -\epsilon$ and $f_-(x) < -\epsilon$ for $x\in (0, 1)$ by the strict convexity of $f_-$

(2) $f_-(\delta) = f_-(1-\delta) = -M$ and $f_-(x) < -M$ for $x\in (\delta, 1-\delta)$.

Combined, these tell us that $f_-\leq f$ on $[0, 1]$: $f_-(x)\leq -\epsilon\leq f(x)$ for $x\in [0, \delta]\cup [1-\delta, 1]$, and $f_-(x)\leq -M\leq f(x)$ for $x\in [\delta, 1-\delta]$. Similarly, we can let $b = -a$ to get $f\leq f_+$ on $[0, 1]$. Therefore, the lemma follows. We could easily show something similar with a different class of continuous convex functions with a minimum in $(0, 1)$ (such as absolute value functions).

Problem 1: For any continuous function $f : [0, 1]\to \mathbb{R}$ such that $f(0) = 0$ and $f(1) = 1$, $$\lim_{n\to \infty} n\int_0^1 f(x)x^{2n}\,\mathrm{d}x = \frac{1}{2}$$

Proof: Let $a, b\in \mathbb{R}$ such that for $\epsilon > 0$, $f_-(x) = ax^2-ax-\epsilon$ and $f_+(x) = bx^2-bx+\epsilon$ satisfy $f_-\leq (f-x)\leq f_+$ as in Lemma 1. Then, let $f_-^*(x) = ax^2+(1-a)x-\epsilon$ and $f_+^*(x) = bx^2+(1-b)x+\epsilon$ so that $f_-^*\leq f\leq f_+^*$. Then, we will have that $$n\int_0^1 f_-^*(x)x^{2n}\,\mathrm{d}x\leq n\int_0^1 f(x)x^{2n}\,\mathrm{d}x\leq n\int_0^1 f_+^*(x)x^{2n}\,\mathrm{d}x$$ for all $n$. Note that \begin{align*} \int_0^1 (cx^2+(1-c)x\pm \epsilon)x^{2n}\,\mathrm{d}x &= \int_0^1 cx^{2n+2}+(1-c)x^{2n+1}\pm \epsilon x^{2n}\,\mathrm{d}x \\ &= \frac{c}{2n+3}+\frac{1-c}{2n+2}\pm \frac{\epsilon}{2n+1} \\ &= \frac{2n+(3-c)}{2(n+1)(2n+3)}\pm \frac{\epsilon}{2n+1} \end{align*} Therefore, as $\lim_{n\to \infty} \frac{n(2n+(3-c))}{2(n+1)(2n+3)}\pm \frac{n\epsilon}{2n+1} = \frac{1\pm \epsilon}{2}$ for any $c\in \mathbb{R}$, we have that $$\lim_{n\to \infty} n\int_0^1 f_{\pm}^*(x)x^{2n}\,\mathrm{d}x = \frac{1\pm \epsilon}{2}$$ so therefore $$\liminf_{n\to \infty} n\int_0^1 f(x)x^{2n}\,\mathrm{d}x\geq \frac{1-\epsilon}{2}$$ and $$\limsup_{n\to \infty} n\int_0^1 f(x)x^{2n}\,\mathrm{d}x\leq \frac{1+\epsilon}{2}$$ As these hold true for any $\epsilon > 0$, we have that $\liminf_{n\to \infty} n\int_0^1 f(x)x^{2n}\,\mathrm{d}x = \limsup_{n\to \infty} n\int_0^1 f(x)x^{2n}\,\mathrm{d}x = \frac{1}{2}$, so therefore, $$\lim_{n\to \infty} n\int_0^1 f(x)x^{2n}\,\mathrm{d}x = \frac{1}{2}$$

Michael L.
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