# Why $\displaystyle\lim_{n\to+\infty}x_n\otimes y_n=x\otimes y\;?$

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.

You can prove it directly using Squeeze theorem for sequences.

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

• Thank you for your answer. What is Squeeze theorem for sequences? Commented Jan 31, 2018 at 7:35
• @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) Commented Jan 31, 2018 at 10:30

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 $x\otimes y$.

• Thank you for your answer but wht $T$ is continuous? because $E$ and $F$ are not necessary finite dimensional Commented Jan 20, 2018 at 13:55
• @Student, continuity of $T$ follows from the definition of the inner product on $E \otimes F$.
– user81375
Commented Jan 20, 2018 at 13:58
• Because we have $\|x\otimes y\|=\|x\|_1\|y\|_2$? Commented Jan 20, 2018 at 14:00
• @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. Commented Jan 20, 2018 at 14:12
• Why does continuity follow from bilinearity? Commented Jan 30, 2018 at 21:31