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The Proposition 2.8 of Ryan's textbook(Introduction to tensor products of Banach spaces) says: Let $X$ and $Y$ be a Banach space. Let $u \in X\hat\otimes_{\pi} Y$ and $\epsilon >0$. Then there exist bounded sequences $(x_n), (y_n)$ in $X, Y$ respectively such that the series $\sum_{n=1}^{\infty} x_n \otimes y_n$ converges to $u$ and $$ \sum_{n=1}^{\infty} \|x_n\|\|y_n\|<\pi(u)+\epsilon. $$

From this we have that the projective norm $\pi (u)$ is that $$ \pi(u)=\inf\left\{\sum_{n=1}^{\infty} \|x_n\|\|y_n\|:\sum_{n=1}^{\infty} \|x_n\|\|y_n\|<\infty,\, u = \sum_{n=1}^{\infty} x_n \otimes y_n \right\}, $$ the infimum being taken over all the representations of $u$.

Question: Could you explain why $$ \pi(u) = \inf\left\{\sum_{n=1}^{\infty} |\lambda_n| \|x_n\|\|y_n\|: u = \sum_{n=1}^{\infty} \lambda_n x_n \otimes y_n, \, \sum_{n=1}^{\infty} |\lambda_n|<\infty,\, x_n, y_n \rightarrow 0\right\}? $$

I have tried to show this by multiplying $x_n, y_n$ by some values to find the desired representation of $u$ (which is of the form as in the statement in Question), but I couldn't get it.

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  • $\begingroup$ What is $X\hat\otimes_{\pi} Y$? $\endgroup$ – Paul Sinclair Nov 15 '19 at 2:30
  • $\begingroup$ @PaulSinclair It is the completion of the normed space $X\otimes Y$. $\endgroup$ – cdamle Nov 15 '19 at 6:08
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Try showing that if $u = \sum_{n=1}^{\infty} x_n \otimes y_n$ and $\sum_{n=1}^{\infty} |\lambda_n|<\infty$ for some sequence with $\lambda_n \ne 0$. Then if you let $\hat x_n = (1/\lambda_n)x_n$, it must be true that both $\hat x_n \to 0$ and $y_n = 0$.

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  • $\begingroup$ If $\hat{x_n}=(1/\lambda_n) x_n$, how could you see that $\hat{x_n} \rightarrow 0$? $\endgroup$ – cdamle Nov 15 '19 at 6:10

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