I was reading one of the answers to this question, and got stuck on the following point.
Let $\sum_{k=1}^\infty a_k x^k$ be a power series that converges uniformly to $0$ on $[0,1]$.
Question: Then is it true that $a_k=0$ for all $k$?
I'm probably missing something obvious, and would appreciate a hint. So far I just see that if we call the partial sums $P_n=\sum_{k=1}^n a_k x^k$, then for every $\varepsilon>0$ there exists an $n_\varepsilon$ such that $$\|P_{n_\varepsilon}-0\|_\infty=\|P_{n_\varepsilon}\|_\infty\leq\varepsilon.$$ But is this enough to conclude that the $a_k$ are all $0$?
Let's denote the power series by $f(x)=\sum_{n=1}^\infty a_kx^k$, because $f(x)$ converge in $x=1$, by a claim i will prove at the end of the answer, it is converging uniformly in $[-r,r]$ for $0\leq r<1$, thus we can conclude that f(x) is differentiable in $0$ infinitely many times. because it is differentiable the derivative is equal to the directional derivative, so we can conclude that $f'(0)=\lim_{x\to0^+}\frac{f(x)-f(0)}{x-0}=\lim_{x\to0^+}\frac{0-0}{x-0}=0$. for every $0<x<1$ we know that $f^{(n)}(x)=0$, so by the same reasoning as before we can conclude that $f^{(n)}(0)=0$. now, $f(x)$ is analytic on $(-1,1)$ we can conclude that it's power representation is it's taylor series, and we will get that the coefficients are $a_k=\frac{f^{(k)}(0)}{k!}=0$.
proof of the claim: i will prove that if a power $\sum_{k=0}^\infty a_kx^k$ is converging in $x_0$ then it is converging uniformly and absolutely in $[-r,r]$ for $0\leq r<|x_0|$.
because the series $\sum_{k=0}^\infty a_kx_0^k$ is converging, then $a_kx_0^k\xrightarrow{k\to\infty}{}0$, thus this sequence is bounded. let us denote $\sup_k|a_kx_0^k|=M$. let $0\leq r<|x_0|$, and $q=\frac{r}{|x_0|}$, then $|q|<1$ and for all $x$ such that $|x|\leq r$ $$ |a_kx^k|=|a_kx_0^k|\cdot|\left(\frac{x}{x_0}\right)^k|\leq Mq^k $$
because the series $\sum_{k=0}^\infty Mq^k$ is convergent, then by weierstrass M test the original power series is converging uniformly and absolutely in $[−r,r]$.