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 $H$.
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 $Te_n=f_n.$