Let $E$, $F$ be complex Hilbert spaces and $\mathcal{L}(E)$ (resp. $\mathcal{L}(F)$) be the algebra of all bounded linear operators on $E$ (resp. $F$).
The algebraic tensor product of $E$ and $F$ is given by
$$E \otimes F:=\left\{\xi=\sum_{i=1}^dv_i\otimes w_i:\;d\in \mathbb{N}^*,\;\;v_i\in E,\;\;w_i\in F
\right\}.$$
In $E \otimes F$, we define
$$
\langle \xi,\eta\rangle=\sum_{i=1}^n\sum_{j=1}^m \langle x_i,z_j\rangle_1\langle y_i ,w_j\rangle_2,
$$
for $\xi=\displaystyle\sum_{i=1}^nx_i\otimes y_i\in E \otimes F$ and $\eta=\displaystyle\sum_{j=1}^mz_j\otimes w_j\in E \otimes F$.
The above sesquilinear form is an inner product in $E \otimes F$.
It is well known that $(E \otimes F,\langle\cdot,\cdot\rangle)$ is not a complete space. Let $E \widehat{\otimes} F$ be the completion of $E \otimes F$ under the inner product $\langle\cdot,\cdot\rangle$.
It is true that
$$\|x_1\otimes x_2\|_{E\widehat{\otimes} F}=\|x_1\|_{E}\|x_2\|_{F},$$
for all $x_1\in E$ and $x_2\in F$ ?
It is clear that the norm on $E \otimes F$ is a cross norm. Indeed
$$\|x\otimes y\|_{E\otimes F}=\|x\|_{E}\|y\|_{F},$$
for all $x\in E$ and $y\in F$.
Best Answer
This is true and is a straightforward consequence of the fact that $\| x \otimes y \|_{E \otimes F} = \|x\|_E \|y\|_F$ because $\| \cdot \|_{E \hat \otimes F} = \| \cdot \|_{E \otimes F}$ on $E \otimes F$ by definition of the completion.