Let $A=\{\sqrt{2}\sin(\sqrt{n^2+a} \pi x)\} _{n=1}^\infty$, where $a$ is a positive real number. Is $A$ a Riesz Basis of $L^2(0,1)$?
1 Answer
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I suppose that you have an extra $\pi$ somewhere: {$\sin \pi nx$} is a Riesz basis on $(0,1)$, not $(0,\pi)$.
After you correct your question, this is not a Riesz basis when $a>0$ is sufficiently large. This follows from characterization of Riesz bases by Pavlov, Sov. Math. Dokl. 20:4, 1979, 655–659, or Semmler, Ann. Acad. Sci. Fenn. Math. 35 (2010), no. 1, 23–46.