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Prove or disprove: A power series that converges on $\left[0,1\right]$ is uniquely determined by any sequence of points converging to $1$.

More precisely, let $f, g:\left[0,1\right]\rightarrow\left[0,\infty\right)$ by expandable into power series $$\begin{align}\forall x\in\left[0,c\right] &\bullet f\left(x\right)=\sum_{i=0}^\infty a_ix^i\\ \forall x\in\left[0,c\right] &\bullet g\left(x\right)=\sum_{i=0}^\infty b_i x^j\end{align}$$ with non-negative coefficients that sum to $1$: $$\left\{\left.a_i, b_i\space:\right|i\in\mathbb{N}_0\right\}\subseteq\left[0,\infty\right)$$ $$\sum_{i=0}^\infty a_i = 1 = \sum_{i=0}^\infty b_i$$

Let $\left(s_j\right)_{j=0}^\infty$ be a strictly ascending sequence of points in $\left(0,1\right)$ that converges to $1$.

Is it true that $$\left(\forall j\in\mathbb{N}_0\bullet f\left(s_j\right)=g\left(s_j\right)\right)\implies\left(\forall i\in\mathbb{N}_0\bullet a_i=b_i\right)$$


Notes

  1. If the radii of convergence $R_f, R_g$ were both $>1$, or if $\left( s_j\right)_{j=0}^\infty$ converged to any number $\in\left[0,1\right)$, the claim would be an immediate consequence of Baby Rudin's Theorem 8.5:

    Suppose the series $\sum a_n x^n$ and $\sum b_n x^n$ converge in the segment $S=\left(-R, R\right)$. Let $E$ be the set of all $x\in S$ at which $$\sum_{n=0}^\infty a_n x^n = \sum_{n=0}^\infty b_n x^n$$ If $E$ has a limit point in $S$, then $a_n=b_n$ for $n=0,1,2,\dots$

  2. The claim derives from Klenke, where it is stated as true without proof. (Theorem 3.2(iii))

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1 Answer 1

up vote 3 down vote accepted

It's not true. Consider the following series: $$ \eqalign{f(x) &= \sum_{j=1}^\infty \frac{3}{2^{2j}} x^{n_{2j-1}}\cr g(x) &= \frac{1}{2} + \sum_{j=1}^\infty \frac{3}{2^{2j+1}} x^{n_{2j}}\cr}$$ where $n_j$ is an increasing sequence of positive integers to be chosen. Thus $$h(x) = g(x) - f(x) = \frac{1}{2} + \frac{3}{2}\sum_{k=1}^\infty (-1)^k \frac{x^{n_k}}{2^k}$$ Let $$h_N(x) = \frac{1}{2} + \frac{3}{2}\sum_{k=1}^N (-1)^k \frac{x^{n_k}}{2^k}$$ Note that $(-1)^N h_N(1) = (1/2)^{N+1} > 0$. I will choose the $n_j$ and an increasing sequence $t_j \in [0,1)$ inductively so that $(-1)^j h(t_j) > 0$. This can be done as follows. We start with $t_0 = 0$, so $h(t_0) = 1/2 > 0$. Given $N$, $t_{N-1}$ and $n_1, \ldots, n_N$, take $t_N \in (t_{N-1},1)$ so close to $1$ that $(-1)^N h_N(t_N) > 0$. Since $|h(t) - h_N(t)| < 3 t^{n_{N+1}}/2^{N+1}$, by taking
$n_{N+1}$ large enough we can ensure that $(-1)^N h(t_N) > 0$.

Now note that by the Intermediate Value Theorem, for each $N$ there is $s_N$ with $t_N < s_N < t_{N+1}$ such that $h(s_N) = 0$.

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Thank you! It's a beautiful counter-example. I should have been more careful and posed my question as: "Prove or disprove..." –  Evan Aad Sep 5 '12 at 7:15

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