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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?

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    $\begingroup$ LaTeX tips: < and > mean "less than" and "greater than", and produce spacing correct for that meaning only. When you want angle brackets, you need to use \langle and \rangle. Also, to get proper norm symbols use \| instead of ||. $\endgroup$ Apr 2, 2013 at 23:16
  • $\begingroup$ Thanks for the correction! $\endgroup$
    – Max
    Apr 2, 2013 at 23:19
  • $\begingroup$ What is $ || \cdot || $? Is this the Frobenius norm? $\endgroup$
    – Duns
    Oct 29, 2020 at 13:03

2 Answers 2

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$\newcommand{\norm}[1]{\left\|{#1}\right\|} \newcommand{\ip}[1]{\left\langle{#1}\right\rangle} \newcommand{\abs}[1]{\left|{#1}\right|}$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.

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  • $\begingroup$ But what about the other inequality? How do I prove that $\|T_1 \otimes T_2\| \leq \|T_1\| \|T_2\|$ $\endgroup$
    – Max
    Apr 3, 2013 at 1:50
  • $\begingroup$ @Max Sorry about that, I clearly misread your question! Now I've answered the question you actually asked. $\endgroup$ Apr 3, 2013 at 7:23
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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.

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