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Let $X$ be a Banach space and $T \colon \ell^2\rightarrow \ell^2$ be a bounded linear map. Suppose that the linear map $T\otimes Id_ {X}:\ell^2\otimes X\rightarrow \ell^2\otimes X$ which maps $e_i \otimes x$ to $e_i\otimes T(x) $ is not bounded when we use the norm on $\ell^2\otimes X$ induced by the Banach space $\ell^2(X)$.

Does it exist a sequence $(X_n)$ of finite dimensional subspaces of $X$ such that $$ ||T \otimes Id_{X_n}||_{ \ell^2(X_n)\to \ell^2(X_n) } \xrightarrow[n \to +\infty]{}+\infty\ ? $$

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I think you mean $T(e_i)\otimes x$. – Matthew Daws Oct 18 '11 at 19:24
up vote 0 down vote accepted

Yes. For any $m>0$ we can find a finite-rank tensor $\tau=\sum_{i=1}^n \xi_i\otimes x_i \in \ell^2 \otimes X$ with $\|\tau\|_{\ell^2(X)} = 1$ but with $$\Big\| \sum_i T(\xi_i)\otimes x_i \Big\|_{\ell^2(X)} > m.$$Let $Y$ be the linear span of the $x_i$, so $Y$ is a finite-dimensional subspace of $X$. Clearly the inclusion map $\ell^2(Y) \rightarrow \ell^2(X)$ is an isometry, and observe that $\tau$ and $(T\otimes I_X)\tau$ both live in $\ell^2(Y)$. Thus $\|T\otimes I_Y\|>m$; let $m$ vary and you get your sequence $(X_n)$.

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