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Is it true that if $\{x_{n}\}_{n=1}^{\infty}$ is a finite union of Riesz sequences in a Hilbert space H, then $\{x_{n}\}$ itself will be a Riesz sequence? What about Frames and Bessel seuences, do we have the same result? (In case of Bessel sequences I beleive it is true).

Note: A sequence $\{x_{n}\}$ is Riesz sequence in H iff there exits $A,B>0$ such that for any scalar sequence $\{a_{n}\}$

$$ A\sum_{n} |a_{n}|^2 \leq \|\sum_{n} a_{n}x_{n}\|^2 \leq B\sum_{n} |a_{n}|^2 $$

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1 Answer 1

No. Observe that a sequence can only be a Riesz sequence if it is linearly independent. This allows to construct a counterexample to your statement:

Let $H=\mathbb C$. Then $\{1\}$ and $\{-1\}$ are Riesz "sequences" but the union $\{-1,1\}$ is not. To see that $\{-1,1\}$ is not a Riesz sequence, consider $a_1=a_2=1$. Then $a_1x_1+a_2x_2=0$ but $\lvert a_1\rvert^2+\lvert a_1\rvert^2>0$.

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