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LEt $H_1,H_2$ are two Hilbert spaces. $\{e_1,\ldots,e_n\}\subseteq H_1$ and $\{f_1,\ldots,f_n\}\subseteq H_2$ two orthonormal systems. $\lambda_1,\ldots,\lambda_n\in\mathbb K$. Let $$ U:H_1\to H_2:x\mapsto\sum_{i=1}^n \lambda_i f_i(x,e_i) $$ Now I would like to know how the norm $\Vert U\Vert$ can be calculated. I know that if $U$ is bounded linear functional, then $\Vert U\Vert=\sup\{\Vert U(x)\Vert: \Vert x\Vert=1\}$ but I do not know how to use this here.

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Since $\{f_i:i\in\{1,\ldots,n\}\}\subset H$ is an orthonormal system, then $$ \left\Vert \sum\limits_{i=1}^n\alpha_i f_i\right\Vert^2=\sum\limits_{i=1}^n|\alpha_i|^2 $$ for all $\{\alpha_i:i\in\{1,\ldots,n\}\}\subset \mathbb{C}$, so you can show that $$ \Vert U(x)\Vert^2 =\left\Vert\sum\limits_{i=1}^n\lambda_i\langle x, e_i\rangle f_i\right\Vert^2 =\sum\limits_{i=1}^n|\lambda_i\langle x, e_i\rangle|^2 =\sum\limits_{i=1}^n|\lambda_i|^2|\langle x, e_i\rangle|^2 \leq\max\limits_{i=1,\ldots,n}|\lambda_i|^2\sum\limits_{i=1}^n|\langle x, e_i\rangle|^2 $$ By Bessel's inequality we have $$ \Vert U(x)\Vert^2 \leq\max\limits_{i=1,\ldots,n}|\lambda_i|^2\sum\limits_{i=1}^n|\langle x, e_i\rangle|^2 \leq(\max\limits_{i=1,\ldots,n}|\lambda_i|)^2\Vert x\Vert^2 $$ This gives $\Vert U\Vert\leq \max\limits_{i=1,\ldots,n}|\lambda_i|$. Let $\max\limits_{i=1,\ldots,n}|\lambda_i|=|\lambda_j|$ for some $j$, then $$ \Vert U(e_j)\Vert =\left\Vert\sum\limits_{i=1}^n\lambda_i\langle e_j, e_i\rangle f_i\right\Vert =\left\Vert\sum\limits_{i=1}^n\lambda_i\delta_{ij} f_i\right\Vert =\left\Vert\lambda_j f_j\right\Vert =|\lambda_j| =\max\limits_{i=1,\ldots,n}|\lambda_i|\Vert e_j\Vert $$ So we get $\Vert U\Vert\geq \max\limits_{i=1,\ldots,n}|\lambda_i|$. This two inequalities gives us $$ \Vert U\Vert=\max\limits_{i=1,\ldots,n}|\lambda_i| $$

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  • $\begingroup$ Thank you for your answer, but it is not quite clear to me how you received $||U(x)||\le max|\lambda_i|||x||$. My idea was: $||U(x)||\le\sum |\lambda_i| ||f_i (x,e_i)||\le C||x||$ but I do not see how to get the $max|\lambda_i|$ $\endgroup$ – Alexander Nov 18 '12 at 0:58

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