Let $(E,\langle \cdot,\cdot\rangle_1)$, $(F,\langle \cdot,\cdot\rangle_2)$ be two complex Hilbert spaces. We recall $$E \otimes F:=\left\{\xi=\sum_{i=1}^dv_i\otimes w_i:\;d\in \mathbb{N},\;\;v_i\in E,\;\;w_i\in F \right\}.$$

We endow $E \otimes F$, with the following inner product $$ \langle \xi,\eta\rangle=\sum_{i=1}^n\sum_{j=1}^m \langle x_i,z_j\rangle_1\langle y_i ,t_j\rangle_2, $$ for $\xi=\displaystyle\sum_{i=1}^nx_i\otimes y_i\in E \otimes F$ and $\eta=\displaystyle\sum_{j=1}^mz_j\otimes w_j\in E \otimes F$.

Let $(x_n)_n\subset E$ and $(y_n)_n\subset F$ such that $\displaystyle\lim_{n\to+\infty}x_n=x$ and $\displaystyle\lim_{n\to+\infty}y_n=y$. Why $$\displaystyle\lim_{n\to+\infty}x_n\otimes y_n=x\otimes y\;?$$

Thank you.


The map $T:E\times F\to E\otimes F$ mapping $(x,y)\to x\otimes y$ is bilinear and bounded. Hence, it is continuous. Note that $\lim_{n\to\infty} (x_n,y_n) = (x,y)\in E\times F$. Since $T$ is continuous, your statement follows.

We could also write $x_n=x+p_n$ and $y_n=y+q_n$ for two nullsequences $p_n$ and $q_n$, then \begin{align*} \|(x_n\otimes y_n)&-(x\otimes y)\|^2 = \langle (x_n\otimes y_n)-(x\otimes y), (x_n\otimes y_n)-(x\otimes y)\rangle \\&= \|x_n\|_1\|y_n\|_2 -2\langle x_n, x\rangle\langle y_n,y\rangle + \|x\|_1\|y\|_2 \\ &=\|x_n\|_1\|y_n\|_2 -2\langle x+p_n, x\rangle\langle y+q_n,y\rangle + \|x\|_1\|y\|_2 \\ &=\Bigl(\|x_n\|_1\|y_n\|_2 - \|x\|_1\|y\|_2\Bigr) - \langle p_n, x\rangle\langle q_n,y\rangle - \langle p_n, x\rangle\langle y,y\rangle - \langle x, x\rangle\langle y,q_n\rangle \end{align*} is a nullsequence because each summand is one and therefore $x_n\otimes y_n$ converges to $x\otimes y$.

  • $\begingroup$ Thank you for your answer but wht $T$ is continuous? because $E$ and $F$ are not necessary finite dimensional $\endgroup$ – Student Jan 20 '18 at 13:55
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    $\begingroup$ @Student, continuity of $T$ follows from the definition of the inner product on $E \otimes F$. $\endgroup$ – fourierwho Jan 20 '18 at 13:58
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    $\begingroup$ Because we have $\|x\otimes y\|=\|x\|_1\|y\|_2$? $\endgroup$ – Student Jan 20 '18 at 14:00
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    $\begingroup$ @Student, yes that's exactly it. You could also use that to show the statement directly, of course. You'll have to use that the product of two convergent sequences converges to the product of the limit points. $\endgroup$ – Jesko Hüttenhain Jan 20 '18 at 14:12
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    $\begingroup$ Why does continuity follow from bilinearity? $\endgroup$ – Mariano Suárez-Álvarez Jan 30 '18 at 21:31

You can prove it directly using Squeeze theorem for sequences.

$$\begin{split} 0 \leqslant &\|x_n\otimes y_n - x\otimes y\| \hspace{110pt}(\text{subtract and add }x_n\otimes y)\\ = & \|x_n\otimes y_n - x_n \otimes y + x_n\otimes y - x\otimes y\| \hspace{28pt}(\text{triangle inequality})\\ \leqslant&\|x_n\otimes y_n - x_n \otimes y\| + \|x_n\otimes y - x\otimes y\| \hspace{18pt}(\text{grouping})\\ = &\|x_n\otimes (y_n - y)\| + \|(x_n - x)\otimes y\| \hspace{44pt}(\text{cross norm } \|x\otimes y\|=\|x\|_1\|y\|_2)\\ = &\|x_n\|_1\|y_n-y\|_2+\|x_n-x\|_1\|y\|_2\rightarrow 0 \end{split}$$

  • $\begingroup$ Thank you for your answer. What is Squeeze theorem for sequences? $\endgroup$ – Student Jan 31 '18 at 7:35
  • $\begingroup$ @Student math.stackexchange.com/questions/1135350/… In your case: $x_n =0$, $y_n = \|x_n\otimes y_n - x\otimes y\|$, $z_n = \|x_n\|_1\|y_n-y\|_2+\|x_n-x\|_1\|y\|_2$. Both $x_n, z_y\rightarrow 0$ (the notation is little unpleasent, because we have the same symbols on LHS and RHS, but I bet you get the point) $\endgroup$ – Fallen Apart Jan 31 '18 at 10:30

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