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Let $X,Y$ be $\mathbb R$-Banach spaces and $X\:\hat\otimes_\pi\:Y$ denote the completion of the tensor product $X\otimes Y$ with respect to the projective norm $\pi$. I know that if $u\in X\:\hat\otimes_\pi\:Y$ and $\epsilon>0$, then there are $(x_n)_{n\in\mathbb N}\in\ell^\infty(X)$ and $(y_n)_{n\in\mathbb N}\in\ell^\infty(Y)$ with $$\sum_{n=1}^\infty\left\|x_n\right\|_X\left\|y_n\right\|_Y<\pi(u)+\epsilon\tag1$$ and $$u=\sum_{n=1}^\infty x_n\otimes y_n\;.\tag2$$

Why can we choose $(x_n)_{n\in\mathbb N}$ and $(y_n)_{n\in\mathbb N}$ such that $x_n\xrightarrow{n\to\infty}0$ and $0<\sum_{n\in\mathbb N}\left\|y_n\right\|_Y<\infty$?

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  • $\begingroup$ hmmmm I don't really understand the question... could you elaborate... by the way if your question is if you can pick $(x_n),(y_n)$ such that equations 1 and 2 hold and also $x_n\to \infty$ and $\sum |y_n|<\infty$ then this clearly can be done by a simple redifinition of the sequences $\endgroup$ – David Jaramillo Jan 27 '18 at 23:18
  • $\begingroup$ @DavidJaramillo Could you please explain how this redefinition is done? $\endgroup$ – 0xbadf00d Jan 28 '18 at 15:09
  • $\begingroup$ hmmm I am sorry I am mistaken, couldn't do it.... I might be wrong... but I think it can be done. I put some insight in answers because it doesn't fit here. $\endgroup$ – David Jaramillo Jan 29 '18 at 23:03
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Well it is a little trickier than I thought. But you can define $\tilde x_n=(x_n/|x_n|u_n)$. and $\tilde y_n=|x_n|u_ny_n$, where $(u_n)$ is a slowly divergent sequence (to be fixed after) with positive coefficients. Then $$|x_n||y_n|=|\tilde x_n||\tilde y_n|$$ and $$x_n\otimes y_n=\tilde x_n\otimes \tilde y_n.$$ Also $$|\tilde x_n|\to 0$$.

And using partial sums $$\sum_{n=1}^{N} |\tilde y_n|=\sum_{n=1}^N|y_n|u_n|x_n|= u_1\sum_{n=1}^N|x_n| |y_n|-\sum_{n=1}^{N-1}(u_{n+1}-u_n)\sum_{k=n+1}^{N}|x_k||y_k|$$

Now if both of these series are convergent then $\sum|\tilde y_n|$ is convergent. The first is convergent, for the second it is a little tricker to find $u_n$ which makes it convergent..... I don't know if it's possible I am higly inclined to think it is but I am not sure.

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