$f\in C^2((0,1)),\lim_{x\to 1^{-}}f(x)=0,\exists C\forall x\in (0,1)$,$(1-x)^2|f''(x)|\leqslant C$,then $\lim_{x\to 1^{-}} (1-x)f'(x)=0$

Question

Question:

Suppose that $f\in C^2((0,1))$,$\lim\limits_{x\to 1^{-}}f(x)=0$.

Assume that there exists a constant $C>0$ such that $\forall x\in (0,1)$, $(1-x)^2|f''(x)|\leqslant C$.

Prove that $\lim_\limits{x\to 1^{-}} (1-x)f'(x)=0$.

Attempt:

I've tried to use Taylor expansion and got

$0=f(x)+f'(x)(1-x)+\frac{f''(c)}{2}(1-x)^2,c\in (x,1).$

But that's not enough.

This is similar to the result "if $f(x) \to 0 $ as $x\to\infty $ and $f''(x) $ is bounded then $f' (x) \to 0$ as $x\to\infty $". See https://math.stackexchange.com/q/730411/72031 — Paramanand Singh, Feb 21 '22 at 02:33
@ParamanandSingh: With $g(x) = f(1-x)$ we have $\lim_{x \to 0^+} g(x) = \lim_{x \to 1^-} f(x)= 0$, $\sup |x^2g(x)| = \sup | (1-x)^2 f(x)| \le C$, and $\lim_{x \to 0^+ }xg'(x) = -\lim_{x \to 1^-} (1-x)f'(x)$. Am I missing something? — Martin R, Feb 23 '22 at 13:00

Stephen Montgomery-Smith · Answer 1 · 2022-02-21T05:00:21.147

ParamanandSingh's hint helped enourmously.

Given $\epsilon>0$, there exist $\delta \in (0,1)$ such that if $x > 1 - \delta$, then $|f(x)| < \epsilon$.

For any $0 < x_0 < x < \tfrac12(1+x_0)$, we have $|f''(x)| \le C/(1-x)^2 \le 4C/(1-x_0)^2$. Hence $$ f'(x) = f'(x_0) + \int_{x_0}^x f''(y) \, dy \ge f'(x_0) - 4C\frac{x-x_0}{(1-x_0)^2} .$$ Similarly, $$ f'(x) = f'(x_0) + \int_{x_0}^x f''(y) \, dy \le f'(x_0) + 4C\frac{x-x_0}{(1-x_0)^2} .$$

Now suppose $f'(x_0) \ge L /(1-x_0)$ for some $x_0>1-\delta$, where $L$ will be chosen later. Then for $$ x_0 \le x \le x_1 := \min\left\{\frac{1+x_0}2, x_0 + \frac{L(1-x_0)}{8C}\right\} $$ we have $$ f'(x) \ge \frac L{2(1-x_0)} $$ Hence if $L \le 4C$, we have $$ 2 \epsilon > f(x_1) - f(x_0) = \int_{x_0}^{x_1} f'(x) \, dy \ge \frac L{2(1-x_0)} \cdot \frac{L(1-x_0)}{8C} = \frac{L^2}{16C} .$$ Without loss of generality, we can assume $32 \epsilon < 16C^2$. Hence if we set $ L^2 \in( 32 \epsilon, 16C^2] $, we obtain a contradiction.

Similarly, if $f'(x_0) \le -L/(1-x_0)$, we obtain a similar contradiction.

Hence for any $x_0 > 1 - \delta$, as long as $\epsilon < C^2/2$, we have $$ (1-x_0)|f(x_0)| \le \sqrt{32 \epsilon} .$$

score 0 · Answer 2 · answered Feb 21 '22 at 12:26

By Taylor expension we have:

$f(\delta+(1-\delta)x)=f(x)+\delta(1-x)f'(x)+\frac{\delta^2}{2}\frac{(1-x)^2}{(1-\xi)^2}(1-\xi)^2f''(\xi),\xi\in(x,\delta+(1-\delta)x)$.

Divided by $\delta$ gives that:

$|(1-x)f'(x)|\leqslant|\frac{f(\delta+(1-\delta)x)-f(x)}{\delta}|+|\frac{\delta}{2}\frac{(1-x)^2}{(1-\xi)^2}(1-\xi)^2f''(\xi)|\leqslant|\frac{f(\delta+(1-\delta)x)-f(x)}{\delta}|+|\frac{\delta}{2(1-\delta)^2}M|$.

$\forall \epsilon >0,\exists \delta'>0,\forall 0<\delta<\delta',\forall 0<x<1$,

the second part is less than $\epsilon/2$ ($\delta\to 0$ in short),

then set $x\to 1$ and the first part is also less than $\epsilon/2$.

$f\in C^2((0,1)),\lim_{x\to 1^{-}}f(x)=0,\exists C\forall x\in (0,1)$,$(1-x)^2|f''(x)|\leqslant C$,then $\lim_{x\to 1^{-}} (1-x)f'(x)=0$

2 Answers2