Krdytkyu

A function $f$ defined on interval $(0,1)$ with a continuous twice derivation $(fin{C^2(0,1)})$ satisfies $lim_{xto0^+}f(x)=0$ and $|x^2f''(x)|leq{C}$ where $C$ is a fixed positive real number.

Prove $lim_{xto0^+}xf'(x)=0$ (or disprove it!)

I've tried several ways like calculating $yf'(y)-xf'(x)=f(y)-f(x)+int_{x}^{y}tf''(t),dt$ to prove the limitation exists (and failed).

I also think it is similer to L'Hospital rule $lim_{xto0^+}frac{f'(x)}{frac{1}{x}}=lim_{xto0^+}frac{f''(x)}{-frac{1}{x^2}}$,but it's only a sufficient condition and probably wrong.

For $0 < y < x < 1$, by Taylor's theorem there exists $theta in (0,1)$ such that

$$f(y) = f(x) + f'(x)(y-x) + frac{1}{2} f''(x - theta(x-y)) (y-x)^2$$

Taking $y = (1-eta) x$ where $0 < eta < 1/2$ we have

$$f(y) - f(x) = -eta xf'(x) +frac{eta^2}{2}x^2f''(x(1 -thetaeta)) $$

and since $y to 0+$ as $x to 0+$,

$$0 = lim_{x to 0+}frac{f(y) - f(x)}{eta} = lim_{x to 0+}left(-xf'(x)+ frac{eta}{2}frac{1}{(1 - thetaeta)^2} [x(1-thetaeta)]^2 f''(x(1 -thetaeta)) right) $$

Since the limit on the RHS is $0$, for any $epsilon > 0$ if $0 < x < delta$ we have

$$tag{*}|x f'(x)| leqslant epsilon + frac{eta}{2(1 - thetaeta)^2} [x(1-thetaeta)]^2 |f''(x(1 -thetaeta))| leqslant epsilon + frac{eta}{2(1 - thetaeta)^2}C,$$

Since $0 < eta < 1/2$, we have

$$frac{eta}{(1- theta eta)^2} < frac{eta}{(1 - theta/2)^2} < 4eta,$$

and the last term on the RHS of (*) can be made arbitrarily small.

It follows that

$$lim_{x to 0+} x f'(x) = 0$$

Let $g(x)=f(1/x)$ for $xin (1,infty)$. The hypotheses on $f(x)$ imply that
$$lim_{x rightarrow infty} g(x)=0~mbox{and}~|(x^2g'(x))'|<C,$$
since $(x^2g'(x))'=x^2 g''(x)+2x g'(x)=frac{1}{x^2}f''(1/x)$ which is bounded in absolute value by $C$ by the assumption.

Define $h(x)=xg'(x)$, and so we need to show that $lim_{x rightarrow infty}h(x)=0$, since $frac{1}{x}f'(1/x)=-xg'(x)=-h(x)$. We first show that $h(x)$ is bounded on $[2,infty)$. On the contrary, suppose $h(x)$ is unbounded and so there exists a sequence $x_i rightarrow infty$ such that $|h(x_i)| >i$ for all $i$. Also, wlog we can assume that $h(x_i)$ are all positive or all negative. WLOG, suppose that $h(x_i)>0$ for all $i$, since the proof is similar in the other case. For each $i$ large enough, choose $y_i in [2,x_i]$ such that $h(y_i)>i$ and $h'(y_i)>0$. This is possible, since for $i$ large enough $h(x_i)>h(2)$ and so $y_i=inf_{tin [2,x_i]}{h(t)=h(x_i)}$ works. But, one has $$|h(x)+xh'(x)|=|(xh(x))'|=|(x^2g'(x))' |<C,$$
and replacing $x$ by $y_i$ and letting $i rightarrow infty$ results in a contradiction. So $h(x)$ is bounded on $[2,infty)$.

Next, we show that $lim_{xrightarrow infty}h(x)$ exist. Let $alpha=limsup_{xrightarrow infty} h(x)$ and $beta=liminf_{x rightarrow infty}h(x)$. Then $alpha, beta in mathbb{R}$, and we need to show that $alpha=beta$. On the contrary, suppose $alpha>beta$. Assume that $alpha > 0$. Let $x_i rightarrow infty$ such that $h(x_i) rightarrow alpha$. Choose $theta>0$ such that $beta<theta alpha<alpha$.
By the intermediate-value theorem, for $i$ large enough, there exists $t_i>0$ such that $h(x_i+t_i)=theta h(x_i)$. In addition, we can choose $t_i$ such that $h(x)>theta alpha/2$ for all $xin [x_i, x_i+t_i]$.

From $|(xh(x))'|<C$ and the mean-value theorem, we have
$$|x(h(x+t)-h(x))+th(x+t)|=|(x+t)h(x+t)-xh(x)|<Ct$$
for all $tgeq 0$ and $x>1$. It follows that
$$x|h(x+t)-h(x)|<At,$$
for some $A>0$ and all $tgeq 0$ and all $x>1$. Replacing $x$ with $x_i$ and $t$ by $t_i$ gives
$$frac{t_i}{x_i}>frac{1}{A}|h(x_i+t_i)-h(x_i)|>frac{1}{A}(1-theta)h(x_i)$$
for $i$ large enough. It follows that
$$frac{x_i+t_i}{x_i}=1+frac{t_i}{x_i}>D,$$
for some $D>1$ and all $i$ large enough. Now, one has

$$g'(x)= frac{h(x)}{x} >frac{theta alpha}{2x},$$
for all $xin [x_i,x_i+t_i]$. It follows that
$$|g(x_i+t_i)-g(x_i)|geq int_{x_i}^{x_i+t_i} frac{theta alpha}{2x}dxgeq frac{theta alpha}{2} log left (frac{x_i+t_i}{x_i} right) geq frac{theta alpha}{2}log D,$$
for all $i$ large enough. This is a contradiction, since $lim_{x rightarrow infty}g(x)=0$. The proof in the case of $beta<0$ is similar. Therefore, $alpha=beta$ and so $L=lim_{xrightarrow infty}h(x)$ exists.

Finally, we show that $L=0$. Suppose $L>0$. Then for $x$ large enough $|g'(x)|=|h(x)/x|>L/(2x)$ which gives $g(x)>g(x_0)+ (L/2) log(x)$ for some $x_0$ and all $x$ large enough, a contradiction. The case where $L<0$ is similar. Therefore, $L=0$ and the proof is completed.