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For all $j \geq 1$ vectors $e_j$ forms an orthonormal basis for separable Hilbert space $H$.

Suppose that $f_j \in H$ where $j \geq 1$ such that $\sum_j^\infty\|f_j\|_H^2 < \infty$. Show that there exists an unique bounded linear operator on $H$ such that $Te_j = f_j$ for all $j \geq 1$.

I tried to construct an operator with the fact that we can represent any $x \in H$ as $x = \sum_n^\infty<x,e_n>e_n$, but without any success. Any help is appreciated.

Atleast a case where $e_j = f_j$ isnt possible since then we have that $\sum_j^\infty\|f_j\|_H^2 = \infty$. I think i need to construct some sort of a diagonal operator. Or can i just write $f_j = \lambda_j e_j$ because then we clearly have an unique bounded linear operator on $H$.

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  • $\begingroup$ There is only one possible way to define such an operator. If it is linear, continuous, and maps $e_j$ to $f_j$, then where it must map the element $\sum\limits_{n=1}^{\infty}\langle x,e_n\rangle e_n$? $\endgroup$
    – Mark
    Jan 29 at 18:28
  • $\begingroup$ Are the $e_j$ a Schauder basis for $H$? $\endgroup$
    – copper.hat
    Jan 29 at 18:35
  • $\begingroup$ I think you have an unstated assumption that $\{e_j\}_j$ forms an orthonormal basis for $H$, in the Hilbert space sense. $\endgroup$ Jan 29 at 18:35
  • $\begingroup$ yes, for all $j \geq 1$ vectors $e_j$ forms an orthonormal basis for separable Hilbert space $H$. $\endgroup$
    – voroshilov
    Jan 29 at 18:36

1 Answer 1

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Let $$Tx=\sum_{n=1}^\infty \langle x,e_n\rangle f_n$$ where $x$ belongs to the linear span of finitely many $e_n.$ Then basing on the triangle inequality and the Cauchy-Schwarz inequality we get $$\|Tx\|^2\le \sum_{n=1}^\infty |\langle x,e_n\rangle|^2\sum_{n=1}^\infty \|f_n\|^2\\ =\sum_{n=1}^\infty \|f_n\|^2\,\|x\|^2$$ Therefore the operator is bounded and defined on a dense subspace of $H.$.Hence it extends uniquely to the entire space. Clearly we have $Te_n=f_n.$

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