Prove that $\sum_{k=0}^{\infty}a_k e^{kt} = 0$ implies each $a_k$ is zero

Question

Let $t$ to be the variable over $R$. Let $\{a_k\}_{k=0}^{\infty}$ to be a real number sequence. Consider a seires of exponential function $e^{kt}$, where $k = 0, 1, 2,...$

How to prove that $\sum_{k=0}^{\infty}a_k e^{kt} = 0$ (the zero function) implies each $a_k$ is zero?

I know $\{1, e^t, e^{2t},...,e^{nt}\}$ is a linear independent list in the vector space consists of all continuous function over $R$ if $n$ is a finite positive integer. And I am curious about is there any theroy telling us something like "linearly independent list with infinite lenth"?

Answer 1

First show that the series $\sum_{k=0}^\infty a_k$ is absolutely convergent: For $t = 1$ we get that the series $\sum_{k=0}^\infty a_k e^k$ is convergent, so that $a_k e^k \to 0$ and therefore $|a_k| \le C e^{-k}$ for some constant $C$.

Now assume that not all $a_k$ are zero and let $m$ be the index of the first non-zero element in $(a_k)$. Then $$ -a_m = \sum_{k=m+1}^\infty a_k e^{(k-m)t} \, . $$ For $t < 0$ and $k -m \ge 1$ is $e^{(k-m)t} \le e^t$, so that $$ |a_m| \le \sum_{k=m+1}^\infty |a_k| e^{(k-m)t} \le e^{t}\sum_{k=m+1}^\infty |a_k| \le e^{t}\sum_{k=0}^\infty |a_k| \, . $$ But for $t \to -\infty$ the right-hand side converges to zero. It follows that that $a_m = 0$, in contradiction to the assumption that $a_m \ne 0$.

It follows that all $a_k$ are zero.

Answer 2

We could let $t = \ln(x)$ and note that the hypothesis that $\sum_{k=0}^\infty a_ke^{kt} = 0$ for all $t \in \mathbb{R}$ now implies that $\sum_{k=0}^\infty a_kx^k = 0$ for all $x > 0$. Some theorems (cf. chapter 6 in Abbott's Understanding Analysis) about power series now imply that $\sum_{k=0}^\infty a_kx^k = 0$ for all $x \in \mathbb{R}$, and that the $a_k$ are related to the derivatives of the function $x \mapsto 0$. In particular, $a_k = 0$ for all $k$.