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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?

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  • $\begingroup$ Do you mean to ask whether $T$ linear or bounded linear? $\endgroup$
    – Bernard W
    Apr 14 '17 at 11:02
  • $\begingroup$ @Bernard Wojcik. Bounded linear. $\endgroup$
    – Barbara
    Apr 14 '17 at 14:41
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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 $(0,...0,1,0,...)$

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To get a positive result, you need that $K:= \sup_{\mathbb N} \lVert T_n \rVert < \infty$. Then $$ \lVert Tx \rVert ^2 = \sum_{n=1}^\infty \lVert T_i x_i \rVert ^2 \leq K^2 \lVert x \rVert^2. $$

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