# Tensor product space with projective norm is incomplete

Ryan says in his book "Introduction to Tensor Products of Banach Spaces"(pg. 17) that for Banach spaces $X$ and $Y$, $X\otimes Y$ equipped with projective norm is not complete unless $X$ and $Y$ are finite dimensional.
First I want the example of this.
Second, is there any sources about the proof of this statement? Thanks.

• Can you explain how we guarantee that if one of them is finite dimensional then the projective tensor product space is complete? – Salam Mohammed Oct 7 '17 at 16:56
• If $X$ is of finite dimension $n$, then $X \otimes Y$ is isomorphic to a sum of $n$ copies of $Y$, or equivalently to $Y^n$, which is complete since $Y$ is complete. – Todd Trimble Oct 7 '17 at 18:45

The projective tensor product $\ell_1\widehat{\otimes}X$ is naturally isometrically isomorphic to the $\ell_1$-sum of countably many copies of $X$. The uncompleted tensor product $\ell_1 \odot X$ is then the linear span of elements of the form $(\xi_n x)$, where $(\xi_n)$ is in $\ell_1$ under this identification, which is hardly complete as there exist infinite convergent series. For example, take a linearly independent sequence $(x_n)_{n=1}^\infty$ of unit vectors in $X$ and consider $(n^{-2}x_n)_{n=1}^\infty$; it does not belong to (the image of) $\ell_1\odot X$.

In general the proof goes along the same lines -- it uses the possibility of (non-unique) representation of elements of the projective tensor product as infinite series of simple tensors. You then have to show that if $X$ and $Y$ are infinite-dimensional then there is an infinite series that cannot be truncated to a finite one.

• It is quite correct to say that "The uncompleted tensor product $\ell_1 \odot X$ is then the algebraic direct sum of countably copies of $X$..."? If $f = (2^{-n}) \in \ell_1$ what is $f \otimes x$ in this picture? – Matthew Daws Jul 24 '17 at 8:45
• @MatthewDaws, thanks Matt, corrected. – Tomek Kania Jul 24 '17 at 9:52

It wasn't clear to me, immediately, how to do the general case, but I think you can use Biorthogonal Systems. Indeed, let $(x_n)\subseteq X$ and $(f_n)\subseteq X^*$ satisfy that $f_n(x_m) = \delta_{n,m}$. Let $(y_n) \subseteq Y$ be a linearly independent sequence. Set $$\tau = \sum_{n=1}^\infty \|x_n\|^{-1} \|y_n\|^{-1} 2^{-n} x_n\otimes y_n \in X \widehat\otimes Y.$$ Suppose, towards a contradiction, that $\tau \in X\otimes Y$. Now, $\tau$ induces a bounded linear map $T:X^*\rightarrow Y$, and by our assumption, $T$ is finite rank. However, clearly $T(f_n) = \|x_n\|^{-1} \|y_n\|^{-1} 2^{-n} y_n$ and so $(T(f_n))$ is a linearly independent sequence, contradicting $T$ being finite rank.

To construct $(x_n), (f_n)$ we can follow an old argument due to Markushevich. The argument is not hard, but is a touch long to type out here. The best link I could find was: Biorthogonal Systems in Banach Spaces (Google books). Search for "M-basis".

• I've tried to thinking all day but can not found the series. So thank you very much about the answer and more intuition about that biorthogonal systems(I think it is just as the Gram-Schmidt process?). – CSH Jul 24 '17 at 14:17
• @CSH: Yes, pretty much. You start with $(x_n)$ being just linearly independent, and $(f_n)$ being linearly independent separating the points of the linear span of the $(x_n)$. Then adjust $f_1$ so $f_1(x_1)=1$, adjust $x_2$ so $f_1(x_2)=0$, adjust $f_2$ with $f_2(x_1)=0, f_2(x_2)=1$, and so on.... – Matthew Daws Jul 24 '17 at 14:30
• But it seems that the argument holds only when $X^*$ is separable, isn't it? – CSH Jul 24 '17 at 14:33
• No: you just need that if $X_0$ is the linear span of $(x_n)$ (so $X_0$ is not closed, and it's closure is separable) then $(f_n)$ should separate the points of $X_0$. Hahn-Banach does this. – Matthew Daws Jul 24 '17 at 14:54