Let $\{a_n\} \in \Bbb C$ be a complex sequence such that $\sum_{n=1}^\infty \lvert a_n \rvert ^2=0 $
I would like to know if the following relation is true
$\sum_{n=1}^\infty a_n = 0 $
Thanks.
Square summable sequences of complex numbers is simply summable?
0
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sequences-and-series
complex-analysis
functional-analysis
2 Answers
2
Yes, it is true. If $\sum_{n \in \mathbb N} \lvert a_n \rvert^2 = 0$ then $a_n = 0$ for all $ n\in \mathbb N$. Indeed, the space $$\ell^2(\mathbb C) = \left\{\{c_n\}_{n \in \mathbb N} : c_n \in \mathbb C \text{ for all } n \text{ and } \sum_{n\in\mathbb N} \lvert c_n \rvert^2 < \infty \right\}$$ is a Hilbert space with inner product $$\langle a , b \rangle = \sum_{n\in\mathbb N} a_n b_n$$ where $a = \{a_n \}_{n \in \mathbb N}, b = \{b_n \}_{n \in \mathbb N}$. The fact that $\sum_{n \in \mathbb N} \lvert a_n \rvert^2 = 0 \implies a_n = 0$ for all $ n\in \mathbb N$ is exactly the statement that the norm induced by this inner product is positive definite.
2
Yes, it is true.
Since the series $\sum_{n=1}^\infty |a_n|^2$ converges to zero, the sequence $S_N:=\sum_{n=1}^N |a_n|^2$ must converge to zero. But we have: $S_N\geqslant S_{N-1}\geqslant 0$, because we are summing non-negative terms. Therefore $(S_N)$ is a non-decreasing sequence with non-negative terms, and $(S_N)$ converges to zero. But the limit of a non-decreasing sequence is greater or equal to any element in the sequence. The only way that can be true for $(S_N)$ is when $S_N=0$ for all $N\in\mathbb N$. On the other hand, every $S_N$ is a finite sum of non-negative terms; therefore $S_N$ can be zero only if all these terms are zero. Therefore, $|a_n|=0$ for all $n\in \mathbb N$, which implies $a_n=0$ for all $n\in \mathbb N$. Then $\sum_{n=1}^\infty a_n$ is a null-series.