# Does $Tx=\left( T_1x_1, T_2x_2, \cdots \right)$ define an element of $L( \oplus_{n=1}^\infty H_n , \oplus_{n=1}^\infty H_n )$?

Given Hilbert spaces $H_1, H_2, \cdots$ with inner products $\langle\cdot,\cdot\rangle_1,\langle\cdot,\cdot\rangle_2, \cdots$ and norms $\|\cdot\|_1, \|\cdot\|_2, \cdots,$ respectively, consider $$\bigoplus_{n=1}^\infty H_n=\left\{ \underline{x}=(x_1, x_2, \cdots )\in \prod_{n=1}^{\infty}H_n:\sum_{n=1}^{\infty}\|x_n\|_n^2<\infty \right\}.$$ Given $T_n\in L(H_n,H_n)$ for $n=1,2,\cdots ,$ define for each $\underline{x}\in \bigoplus_{n=1}^\infty H_n$ $$T\underline{x}=\left( T_1x_1, T_2x_2, \cdots \right).$$ Does $T$ always define an element of $L\big( \bigoplus_{n=1}^\infty H_n , \bigoplus_{n=1}^\infty H_n \big)$? I want to prove it or give a counterexample. I think the answer to the question is no, but I need help to come up with a counterexample. Anyone?

• Do you mean to ask whether $T$ linear or bounded linear? Apr 14 '17 at 11:02
• @Bernard Wojcik. Bounded linear. Apr 14 '17 at 14:41

The answer is no I think. Take $H_i = \mathbb{R}$ and $T_i$ multiplication with $i$. Each $T_i$ is bounded but $T$ is not bounded if you look norm 1 vectors in $H$ of the form $(0,...0,1,0,...)$
To get a positive result, you need that $K:= \sup_{\mathbb N} \lVert T_n \rVert < \infty$. Then $$\lVert Tx \rVert ^2 = \sum_{n=1}^\infty \lVert T_i x_i \rVert ^2 \leq K^2 \lVert x \rVert^2.$$