Does $f'(x)$ always remain close to $f(x)/x$ as $x \rightarrow 0$?

Consider a function $f: (0,1): \to (0,1)$ such that

• $\lim_{x \rightarrow 0+} f(x) = 0$, $\lim_{x \rightarrow 1-} f(x) = 1$;
• $f$ has a power series expansion around $x=1$ that converges in $(0,1)$: $$f(x) = 1 - \sum_{k=1}^\infty c_k(1-x)^k.$$ with $c_k \geq 0$ for all $k$. Note that the condition $\lim_{x \rightarrow 0+} f(x) = 0$ implies $\sum_{k=1}^\infty c_k = 1$.

A typical example may be $f(x) = x^{2/3}$. Note that the derivative of $f$ may tend to $\infty$ as $x$ approaches $0$ from the right, as in this example.

I would like to prove (or find a counter-example) that the derivative $f'(x)$ remains asymptotically close to $f(x)/x$ as $x \rightarrow 0$. By that I mean any of the following:

• the limit $$\lim_{x \rightarrow 0} f'(x)x/f(x)$$ exists and is finite and nonzero;
• or more generally that $f'(x) = \Theta(f(x)/x)$ as $p \rightarrow 0$, where $\Theta$ means "bounded below and above asymptically" (see definition here);
• or any similar result that asserts that $f'(x)$ does not "deviate too much" from $f(x)/x$ as $x \rightarrow 0$.

For the example function this is true. In fact the limit exists and equals $2/3$. This seems to be the case (with other values) for all functions I am testing, but how to prove it/disprove it?

• Hint: Consider the function $f(x) = 0$ for $x\leq 0$ and $f(x) = e^{-1/x}$ for $x>0$. – Chris Janjigian Jun 28 '18 at 17:26
• @ChrisJanjigian I didn't explain it well, sorry again. They are alternating in (x-1), but not in (1-x) – Luis Mendo Jun 28 '18 at 17:37
• What type of functions? Smooth is clearly too rich for that small amount of conditions on the Taylor coefficients at two points to yield that asymptotic, which ca be made fail by just prescribing a pair of Taylor coefficients at a sequence of points that tend to $x=0$. – user569098 Jun 28 '18 at 18:02
• Is the power series convergence radius required to be $1$ (or larger)? – alex.jordan Jun 28 '18 at 18:34
• @alex.jordan Yes, it converges on (0,1). Edited – Luis Mendo Jun 28 '18 at 18:40

Here is a counterexample: Define

$$\tag 1 f(x) = \sum_{k=1}^{\infty}\frac{x^{1/k}}{2^k},$$

the series converging uniformly on $[0,1].$ Each $x^{1/k}$ can be written

$$x^{1/k} = 1-\sum_{n=1}^{\infty}c_k(n)(1-x)^n,$$

where $0\le c_k(n)\le 1$ for all $k,n.$ Thus

$$f(x) = 1 - \sum_{k=1}^{\infty}\frac{1}{2^k}\sum_{n=1}^{\infty}c_k(n)(1-x)^n = 1 - \sum_{n=1}^{\infty}\left (\sum_{k=1}^{\infty}\frac{c_k(n)}{2^k}\right )(1-x)^n.$$

So we see $f$ has the desired form.

Claim: $\lim_{x\to 0^+}\dfrac{xf'(x)}{f(x)} = 0.$

Proof: Note that differenitaing termwise in $(1)$ gives a unifomly convergent series on $[a,1]$ for any $a\in (0,1].$ Thus

$$f'(x) = \sum_{k=1}^{\infty}\frac{x^{1/k-1}}{k2^k}$$

for $x\in (0,1].$ Therefore

$$\frac{xf'(x)}{f(x)} = \frac{ \sum_{k=1}^{\infty}x^{1/k}/(k2^k)}{ \sum_{k=1}^{\infty}x^{1/k}/2^k}.$$

Let $N\in \mathbb N.$ Define $S_N(x) = \sum_{k=1}^{N}\dfrac{x^{1/k}}{2^k},$ $T_N(x) = \sum_{k=N+1}^{\infty}\dfrac{x^{1/k}}{2^k}.$ Then

$$\tag 2\frac{xf'(x)}{f(x)} \le \frac{S_N(x) +T_N(x)/N}{S_N(x) + T_N(x)} \le \frac{S_N(x)}{x^{1/(N+1)}/2^{N+1}} + \frac{1}{N}.$$

As $x\to 0^+,$ the right side of $(2) \to 0+1/N = 1/N.$ Because $N$ was arbitrary, the $\limsup_{x\to 0^+}$ of the left side of $(2)$ is $0,$ proving the claim.

Not quite a counterexample (though I initially thought it was, and why it fails might be of interest): Let $$f(x) = \frac{x}{\sin 1} \sin \left( \frac{1}{x} \right).$$ We obviously have $\lim_{x \to 1} f(x) = 1$; and in the limit $x \to 0$, we note that $$-\frac{x}{\sin 1} \leq f(x) \leq \frac{x}{\sin 1},$$ so $\lim_{x \to 0} f(x) = 0$ by the squeeze theorem.

This is a holomorphic function in the complex plane, and so its radius of convergence in a series about $x = 1$ reaches to the pole nearest to $x = 1$. But the nearest (and only) pole of this function is at $x = 0$. So the power series converges on the ball $(0,2)$.

However, we also have $$f'(x) = \frac{1}{\sin 1} \left[\sin \left( \frac{1}{x} \right) - \frac{1}{x} \cos \left( \frac{1}{x} \right) \right],$$ and $$\frac{f'(x) x}{f(x)} = \frac{x \sin (1/x) - \cos (1/x)}{x \sin(1/x)} = 1 - \frac{1}{x \tan (1/x)}.$$ This function does not have a well-behaved limit as $x \to 0$.

However, this is not quite a counterexample since the power series coefficients are not all positive. In fact, the signs of the $c_k$ for this series seem to generally be alternating.

• I gave a (deleted) similar answer, the problem is that the coefficients of the taylor expansion will not be positive. – The way of life Jun 29 '18 at 18:17

What follows does not answer the original question. However, it does show that even if $f'(x)$ can diverge to infinity for $x\to 0$, with the constraints above $xf'(x)$ does not; so to prove any counterexample, one must exploit $f(x)\to 0$.

$|xf'(x)|\leq 1$ for any $f(x)$ satisfying the constraints above and any $x\in(0,1)$. Note that $\forall x\in(0,1)$:

$|xf'(x)|=\left|x\sum_{k=1}^{\infty}\left(c_k k(1-x)^{k-1}\right)\right|\leq \sum_{k=1}^{\infty}\left| c_k k \max_{x\in(0,1)}\left(x(1-x)^{k-1}\right)\right|$.

Since $c_k\geq 0$ and $\sum_{k=1}^\infty c_k = 1$, observing that $x(1-x)^{k-1}$ is maximized by $x=1/k$ as its derivative equals $(1-x)^{k-1}+x(k-1)(1-x)^{k-2}(-1)=(1-x)^{k-2}(1-kx)$, one immediately obtains:

$|xf'(x)| \leq c_1 + \sum_{k=2}^{\infty} c_k \left(1-\frac{1}{k}\right)^{k-1} = c_1+\sum_{k=2}^{\infty} c_k \frac{k}{k-1} \left(1-\frac{1}{k}\right)^{k} \leq c_1 + \sum_{k=2}^{\infty} (c_k 2e^{-1}) \leq 1$.