Limit of a certain expression

Question

Let $f$ be a twice differentiable function on $(0,1)$ s.t. $\lim\limits_{x\to0+} f(x)=0$ and estimates $$ |f^{(k)}(x)|\le Cx^{-k},\quad k=0,1,2;\ x\in(0,1), $$ hold.

Is it true that $$\lim_{x\to0+} xf'(x)=0?$$

Or, for function $g(x)=xf(x)$, it is equivalent to $\lim_{x\to0+} g'(x)=0$ since $g'(x)=f(x)+xf'(x)$.

Remark. If to drop the condition on derivatives for $k=2$, i.e. $|f''(x)|\le Cx^{-2}$, the statement is false as the following example shows:
Let $f(x)=x \cos \frac1x$, then$$ x f'(x)=\sin \frac{1}{x}+x \cos \frac{1}{x} $$ does not tend to $0$. In this case,$$ f''(x)=-\frac1{x^{3}}\cos \frac{1}{x}. $$

Answer 1

The statement is true! In order to prove this, we recall following modifacted Taylor-forumla: We have $$f(t) = f(x) + f'(x)(t-x) + \int_x^t (f'(s)-f'(x)) \,\mathrm{d}s$$ for $x,t\in (0,1)$. This formula can be easily checked by integration. Note that we only need that $f$ is continuously differentiable on $(0,1)$.

We apply this formula to $t=(1+\varepsilon)x$ in order to find $$f'(x) x = \frac{1}{ε} \left( f((1+ε)x)-f(x) - \int_x^{(1+ε)x} (f'(s)-f'(x)) \,\mathrm{d}s \right).$$ Now, we know that $f((1+ε)x)-f(x) \rightarrow 0$ for $x \rightarrow 0+$ and fixed $ε >0$. Moreover, we can bound the integrand using$$|f'(s)-f'(x)| \leq |s-x| \sup_{h \in [x,(1+ε)x]} |f''(h)| \leq |s-x| · Cx^{-2} \leq Cεx^{-1}.$$ Thus $$ \frac{1}{ε}\left|\int_x^{(1+ε)x} (f'(s)-f'(x)) \,\mathrm{d}s \right| \leq \int_x^{(1+ε)x} C x^{-1} \,\mathrm{d}s \leq Cε.$$ All togehter implies $$\limsup_{x \rightarrow 0+} |f'(x) x| \leq Cε$$ and, since $ε>0$ is arbitary, we get already the claim.

Inspecting the proof, we see that we only used $\lim\limits_{x \rightarrow 0+} f(x) =0$ and $|f''(x)| \leq C x^{-2}$, and that $f$ is twice differentiable.

Edit: My initial formula was wrong. I have repaired the proof.

Answer 2

Let $m(x)=\sup_{y\in(0,x]}\sqrt{\big|f(y)\big|}$; this function is non-decreasing, $m(x)<\tfrac12$ for sufficiently small $x$, and $\lim_{x\to+0}\frac{f(x)}{m(x)}=0$.

Apply Taylor's formula with center $b=x$ and $a=x(1-m(x))>\tfrac{x}2$ as $$ f(a) = f(b) - f'(b)(b-a) + \frac12f''(\xi)(b-a)^2. \\ f(a) = f(x) - f'(x)\cdot x~m(x) + \frac12f''(x)\cdot (x~m(x))^2 \\ x~f'(x) = \frac{f(x)}{m(x)} - \frac{f(a)}{m(x)} + \frac{m(x)}{2}\cdot (x/\xi)^2 \big(\xi^2f''(\xi)). $$ On the R.H.S.each term converges to $0$: if $x\to+0$ then $a\to0$, $\frac{f(x)}{m(x)}\to0$, $\frac{\big|f(a)\big|}{m(x)} \le \frac{\big|f(a)\big|}{m(a)}\to0$, $\xi^2f''(\xi)$ and $x/\xi$ are bounded and $m(x)\to0$.