Does a power series vanish on the circle of convergence imply that the power series equals to zero?

Let $f(z)=\sum_{n=0}^{\infty} a_n z^n$ be a power series, $a_n, z\in \mathbb{C}$. Suppose the radius of convergence of $f$ is $1$, and $f$ is convergent at every point of the unit circle.

Question:If $f(z)=0$ for every $|z|=1$, then can we draw the conclusion that $a_n=0$ for all nonnegative integer $n$?

I think the answer is yes, but I failed to prove it. My approach is concerning about the function $F_\lambda(z):=f(\lambda z)$ for $0\leq\lambda\leq 1$, $|z|=1$. Abel's theorem shows that $F_\lambda$ converge to $F_1$ pointwisely as $\lambda\rightarrow 1$ on the unit circle. If I have the property that $f$ is bounded in the unit disk, then I can apply Lebesgue's dominated convergence theorem to prove $a_0=0$, and by induction I can prove $a_n=0$ for all $n$. However, I cannot prove $f(z) $ is bounded in the unit disk.

Any answers or comments are welcome. I'll really appreciate your help.

Solution 1:

It seems to me that this is a particular case of an old Theorem from Cantor (1870), called Cantor's uniqueness theorem. The theorem says that if, for every real $x$, $$\lim_{N \rightarrow \infty} \sum_{n=-N}^N c_n e^{inx}=0,$$ then all the complex numbers $c_n$'s are zero.

You can google "Uniqueness of Representation by Trigonometric Series" for more information. See e.g. this document for a proof and some history of the result.