Norm of a tensor product of operators I have two Hilbert spaces $H_1$ and $H_2$ which are subspaces of a bigger Hilbert space $H$.
I also have two bounded linear functions $T_1:H_1\rightarrow H$ and $T_2:H_2\rightarrow H$.
I define the tensor product space $F=H_1\otimes H_2$, and a linear function on it $T=(T_1\otimes T_2)$. The vector space $F$ has the induced inner product from the Hilbert spaces:
$$\langle\phi_1\otimes \psi_1, \phi_2 \otimes \psi_2\rangle = \langle\phi_1,\phi_2\rangle\langle\psi_1,\psi_2\rangle$$ 
and therefore an induced norm.
I want to show that $\|T\| =\|T_1\| \cdot\|T_2\|$, but I'm stuck since I cant show that
$\|T\| \leq \|T_1\| \cdot\|T_2\|$ (the other inequality I've already shown).
Can anyone help?
 A: Let me add the other direction. Suppose $\varepsilon >0$. Then from the boundedness of operators $T_{1}$ and $T_{2}$ we know that there exist unit vectors in the Hilbert spaces $\psi \in H_{1}$, $\xi \in H_{2}$ such that:
$\left \| T_{1} \psi \right \|> \left \| T_{1} \right \| - \epsilon$, $\left \| T_{2} \xi \right \|> \left \| T_{2} \right \| - \epsilon$
Using this we can use the definition of the tensor product to see that:
$\left \| \left (T_{1}\otimes T_{2}  \right )\left ( \psi\otimes \xi \right ) \right \|_{H_{1}\bigotimes H_{2}}=\left \| \left ( T_{1}\psi\otimes T_{2}\xi \right ) \right \|_{H_{1}\bigotimes H_{2}}=\left \|T_{1}\psi \right \|_{H_{1}}  \left \| T_{2}\xi \right \|_{H_{2}}\geq$
$ \geq\left \| T_{1} \right \|\left \| T_{2} \right \|-\epsilon \left \| T_{1} \right \| - \epsilon \left \| T_{2} \right \| + \epsilon^{2}\geq\left \| T_{1} \right \|\left \| T_{2} \right \|-\epsilon \left \| T_{1} \right \| - \epsilon \left \| T_{2} \right \|$
Sending $\epsilon$ to 0 shows the desired result.
A: $\newcommand{\norm}[1]{\left\|{#1}\right\|}
\newcommand{\ip}[1]{\left\langle{#1}\right\rangle}
\newcommand{\abs}[1]{\left|{#1}\right|}$Let $S \in B(H_1)$, $T \in B(H_2)$; the claim is that $\norm{S \otimes T} \leq \norm{S}\norm{T}$.
Let me first show that $S \otimes I$ is bounded with $\norm{S \otimes I} \leq \norm{S}$; the same proof, mutatis mutandis, will show that $I \otimes T$ is bounded with $\norm{I \otimes T} \leq \norm{T}$. Since the algebraic tensor product $H_1 \odot H_2$ is dense in $H_1 \otimes H_2$, it suffices to show that $\norm{(S \otimes I)v} \leq \norm{S}\norm{v}$ for any $v \in H_1 \odot H_2$. 
So, let $v = \sum_{k=1}^N x_k \otimes y_k \in H_1 \odot H_2$; by performing Gram--Schmidt orthogonalisation on $\{y_k\}$ and expressing the $y_k$ in terms of the resulting orthonormal basis for $\operatorname{span}\{y_k\}$, we may assume without loss of generality that $\{y_k\}$ is orthonormal. On the one hand, it follows that $\{x_k \otimes y_k\}$ is orthogonal, so that
$$
 \norm{v}^2 = \norm{\sum_{k=1}^N x_k \otimes y_k}^2 = \sum_{k=1}^N \norm{x_k \otimes y_k}^2 = \sum_{k=1}^N \norm{x_k}^2.
$$
On the other hand, since $(S \otimes I)(x_k \otimes y_k) = Sx_k \otimes y_k$, it follows that $\{Sx_k \otimes y_k\}$ is also orthogonal, so that by the same computation, mutatis mutandis,
$$
 \norm{(S \otimes I)v}^2 = \sum_{k=1}^N \norm{S x_k}^2 \leq \sum_{k=1}^N \norm{S}^2 \norm{x_k}^2 = \norm{S}^2 \sum_{k=1}^N \norm{x_k}^2 = \norm{S}^2\norm{v}^2.
$$
Thus, $\norm{(S \otimes I)v} \leq \norm{S}\norm{v}$, as required.
Now, observe that since $(S \otimes T) = (S \otimes I)(I \otimes T)$ on $H_1 \odot H_2$, it follows by the boundedness of $S \otimes I$ and $I \otimes T$ that $S \otimes T$ is also bounded with norm
$$
 \norm{S \otimes T} \leq \norm{S \otimes I}\norm{I \otimes T} \leq \norm{S}\norm{T},
$$
as required.
