Proving $\liminf |f'(x)| = 0$ when $\lim f(x)/x = 0$

Question

Suppose $f$ is differentiable on $[0,\infty)$ and $\displaystyle \lim_{x \to \infty} \frac{f(x)}{x} = 0$. Show that $\displaystyle \liminf_{x \to \infty}|f'(x)| = 0$.

What I have tried is to apply the mean value theorem to $\frac{f(y)}{y} - \frac{f(x)}{x}$ with $0 < x < y$. There exists $c_x \in (x,y)$ such that $\dfrac{\frac{f(y)}{y} - \frac{f(x)}{x}}{y-x} = \frac{f'(c_x)}{c_x}- \frac{f(c_x)}{c_x^2}$. From here we get $$|f'(c_x)| \leq \frac{c_x}{y-x}\left|\frac{f(y)}{y} - \frac{f(x)}{x} \right| + \left|\frac{f(c_x)}{c_x} \right| $$

Now I want to set $y = 2x$ and take $\liminf$ of both sides to get $0$, but I am unclear on:

1. If $\liminf_{x \to \infty} |f'(c_x)| = 0$, then $\liminf_{x \to \infty} |f'(x)| = 0$. I think this is true?

2. How to handle the factor $c_x/(y-x) = c_x/x$ when taking $\liminf$?

Answer 1

It seems easier to prove this using the contrapositive - here‘s a sketch.

Suppose that $\liminf |f‘(x)|=k > 0$, so that the infimum is not equal to zero. It must then be the case that either $f’(x)>k$ or $f’(x)<-k$ for all $x\in\mathbb R^+$. This means that either $f(x) >kx+C$ or $f(x)<-kx+C$ for all $x$, for some constant $C$. This implies that the limit of $f(x)/x$ as $x\to\infty$ cannot equal zero.

Answer 2

Let $\epsilon>0$ and choose some $M>0$ such that $\left|\dfrac{f(x)}{x}\right|<\epsilon$ for all $x\geq M$.

Now we choose by Mean Value Theorem some $c_{x}\in[x,2x]$ such that \begin{align*} \left|\dfrac{f(2x)-f(x)}{2x-x}\right|=|f'(c_{x})|, \end{align*} then for all $x\geq M$, we have \begin{align*} |f'(c_{x})|\leq 2\left|\dfrac{f(2x)}{2x}\right|+\left|\dfrac{f(x)}{x}\right|\leq 2\epsilon+\epsilon=3\epsilon. \end{align*} Now realize those $x$ to $M,3M,9M,...$ successively, so $c_{M}\in[M,2M]$, $c_{3M}\in[3M,6M],...$, so we look at the sequence $(\eta_{n})$ defined by $\eta_{n}=c_{3^{n}M}$, and of course $\eta_{n}\rightarrow\infty$ with $|f'(\eta_{n})|\leq 3\epsilon$, so $\liminf_{n\rightarrow\infty}|f'(\eta_{n})|\leq 3\epsilon$. Note that the $M$ may depend on $\epsilon$, so the construction of the whole $f(\eta_{n})$ is somehow depending on $\epsilon>0$ as well, but this does not matter.

Finally, note that $\liminf_{x\rightarrow\infty}|f'(x)|=\inf\{\liminf_{n\rightarrow\infty}|f'(y_{n})|: y_{n}\rightarrow\infty\}$, and hence $\liminf_{x\rightarrow\infty}|f'(x)|\leq\liminf_{n\rightarrow\infty}|f'(\eta_{n})|\leq 3\epsilon$, so by arbitririness of $\epsilon>0$, we deduce that $\liminf_{x\rightarrow\infty}|f'(x)|=0$.

Answer 3

What you have is for $y=2x$ and $c_x\in (x,2x)$ that $$|f'(c_x)|\leq \frac{c_x}x\left|\frac{f(2x)}{2x}-\frac{f(x)}x\right|+\left|\frac{f(c_x)}{c_x}\right|,$$ where $1<\frac {c_x}x<2.$ Since $\frac{c_x}x$ is bounded and $\lim_{x\rightarrow \infty}\frac{f(x)}x=0$, one has $$\lim_{x\rightarrow\infty}|f'(c_x)|=0,$$ which implies that $$\liminf_{x\rightarrow\infty}|f'(x)|=0,$$ as required.