# Show that there exists an unique bounded linear operator

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 $$x = \sum_n^\inftye_n$$, 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 $$H$$.

• 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 $\sum\limits_{n=1}^{\infty}\langle x,e_n\rangle e_n$?
– Mark
Jan 29 at 18:28
• Are the $e_j$ a Schauder basis for $H$? Jan 29 at 18:35
• I think you have an unstated assumption that $\{e_j\}_j$ forms an orthonormal basis for $H$, in the Hilbert space sense. Jan 29 at 18:35
• yes, for all $j \geq 1$ vectors $e_j$ forms an orthonormal basis for separable Hilbert space $H$. Jan 29 at 18:36

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 $$Te_n=f_n.$$