5

Let $$f(x) = \sum_{k=0}^\infty a_kx^k$$ be a power series mapping reals to reals, with radius of convergence $1$. Suppose $f'(x_0)$ exists in $(-1,1]$ (take the one-sided limit if $x_0 = 1$). Also suppose $a_k \geq 0$ for all $k$. Then is it always true that $$f'(x_0) = \sum_{k=0}^\infty ka_kx^{k-1}?$$

This clearly holds if $x_0 \in (-1,1)$. But what about $x_0 = 1$?

Wakaka
  • 1,353

1 Answers1

1

The answer to the question is yes. This follows from Abel's theorem, since you assume the $a_k$ (and hence $k a_k$) are positive. If you assume that $\sum_k k a_k$ converges, then it is an immedeate consequence of Abel's theorem that $\sum_k k a_kx^{k-1}$ converges for $x\rightarrow 1^- $ to $\sum_k k a_k$.

If, on the other hand, the sum does not converge, it will diverge to $\infty$ (since all terms are nonnegative real numbers), so the generalized Abel theorem (see the link supplied above under remarks) applies again, which states that in this case the function $g(z)=\sum_k k a_kx^{k-1}$ goes off to $\infty$ as well when $x\rightarrow 1$. Since you assume that this is not true you are back in the first case.

What you are requesting here is a special case of what is known as 'Tauberian therom', which you could also refer to. I wonder, though, whether it's possible to give an elementary proof in this case....

Thomas
  • 22,361
  • How do we know that $f'(1)$ exists $\implies$ $\lim_{x\to 1^-} \sum_k ka_kx^{k-1} < \infty$? – Wakaka Oct 11 '15 at 08:28
  • This is the second paragraph in my response. Assume the rhs does not exist. The theorem I cited then implies that the lhs also does not exist. – Thomas Oct 11 '15 at 08:31
  • I mean, you mentioned that I assume "$g(z)$ goes off to $\infty$ when $x \to 1$" is not true. I'm guessing this comes from the assumption that $f'(1)$ exists? But I don't get why... – Wakaka Oct 11 '15 at 09:26
  • You wrote "Suppose $f^\prime(x_0)$ exists in $(-1,1]$ (take the one-sided limit in $x_0=1$. Note that $g=f^\prime$ (except for a typo I'll correct in a minute). – Thomas Oct 11 '15 at 09:45
  • But how do we know $f'=g$ at 1? It requires a condition that f' is continuous from the left at 1, doesn't it? – Wakaka Oct 11 '15 at 18:42
  • You assumed that $f^\prime$ is continuous from the left at $1$ by defining it's value as the limit. That the power series attains that same value is a consequence of the reasoning above. – Thomas Oct 11 '15 at 18:54
  • By assuming $f'(1)$ exists, I'm assuming $\lim_{x \to 1^-} \frac{f(1)-f(x)}{1-x}$ exists. But saying that $f'$ is continuous from the left at $1$ is a different thing; this is saying that $\lim_{x \to 1^-} f'(x)$ exists and is equal to $f'(1)$. That is, this is saying that $f$ is continuously differentiable. Is it obvious that this true? – Wakaka Oct 11 '15 at 19:15
  • In your question you formulated this differently. But yes, by the mean value theorem, $\frac{f(1)-f(x)}{1-x}=f^\prime(\xi)$ for some $\xi\in (x,1)$ so if $ \lim_{x\rightarrow 1}\frac{f(1)-f(x)}{1-x}$ exists then so does $\lim_{x\rightarrow 1} f^\prime(x)$ and it converges to the same value. – Thomas Oct 11 '15 at 19:23
  • But $\lim_{x \to 1} f'(x) = f(1)$ means that $f'(x)$ is close to $f(1)$ for all $x$ that is close to $1$. The mean value theorem only gives us closeness for specific $\xi$. In fact this shows that not all differentiable functions have continuous derivatives. Is there a special property about power series that we are using here? – Wakaka Oct 11 '15 at 19:48
  • I do not understand what you are asking here. You did either assume convergence of $f^\prime(x)$ for $x\rightarrow 1$ or of $\frac{f(1)-f(x)}{1-x} $ for $x\rightarrow 1$. The second implies the first by the mean value theorem as I tried to explain. There is nothing specific for the $\xi$ involved apart from the fact that it is contained in $(x,1)$. The fact that there are differentiable functions which do not admit continuous derivatives is not affected by this. If you have any doubts like the ones you detailed in the comments here consider opening a new question. – Thomas Oct 11 '15 at 19:58
  • For example, consider $f(x) = (x-1)^2\sin(\frac{1}{x-1})$ for $x \neq 1$ and $f(1) = 0$. We have $f'(1) = 0$, so indeed $\frac{f(1)-f(x)}{1-x}$ converges for $x \to 1$. But for $x \neq 1$, $f'(x) = 2(x-1)\sin(\frac{1}{x-1}) - \cos(\frac{1}{x-1})$ does not converge as $x \to 1$. So the second doesn't seem to imply the first in this case. – Wakaka Oct 11 '15 at 21:22
  • Sorry, my fault, should not answer in a hurry. For the conclusion I mentioned you need to know that $f^\prime(x)$ converges as $x\rightarrow 1$. But since you assume $a_n\ge 0$ for all $n$ and $x>0$, this is clear, isn't it? This implies that $f$ and all derivatives are nonnegative (for positive x), hence $f^\prime$ is monotonically increasing. – Thomas Oct 12 '15 at 16:28