Functional Analysis – Projective Tensor Norm on Banach Spaces

banach-spacesfunctional-analysishilbert-spacesinner-productstensor-products

As presented in the answer of this post, the projective tensor norm on the algebraic tensor product of two Banach spaces $X$ and $Y$ is given by
\[
\Vert \omega\Vert_{\pi} = \inf\left\{\sum \lVert x_{i}\rVert_X \,\lVert y_{i}\rVert_ Y\,:\,\omega = \sum_{i=1}^{n} x_{i} \otimes y_{i}\right\}
\]
with respect to which we complete the algebraic tensor product to obtain the tensor product Banach space $X \otimes Y$

For Hilbert spaces $H_1$ and $H_2$, as shown in this link, the inner product on the algebraic tensor product is given by linear extension of the formula
\begin{equation}
\langle \phi_1 \otimes \phi_2, \psi_1 \otimes \psi_2 \rangle = \langle \phi_1, \psi_1 \rangle_{H_1} \langle \phi_2, \psi_2 \rangle_{H_2}
\end{equation}
for $\phi_1, \psi_1 \in H_1$ and $\phi_2, \psi_2 \in H_2$. Next, the tensor product Hilbert space $H_1 \otimes H_2$ is defined to be completion under this inner product.

Now, my question is

Does the projective tensor norm on general Banach spaces, as defined above for $X$ and $Y$, imply the inner product $\langle \phi_1 \otimes \phi_2, \psi_1 \otimes \psi_2 \rangle$ when $X$ and $Y$ are both Hilbert spaces?

I think this must be the case, but it does not seem obvious at all. Could anyone please provide any clarification?

Best Answer

The Hilbert tensor product is in general not equal to the projective tensor product:

If $H$ is a Hilbert space and $H^*$ its dual space, then

  • $H \hat \otimes_\pi H^*$ (the projective tensor product) is (isometrically isomorphic to) the trace class (nuclear) operators with the trace norm
  • $H \hat \otimes_\epsilon H^*$ (the injective tensor product) is (isometrically isomorphic to) the compact operators with the operator norm
  • $H \hat \otimes_h H^*$ (the Hilbert tensor product, which is a Hilbert space again) is (isometrically isomorphic to) the Hilbert-Schmidt operators with the Hilbert-Schmidt norm

Since the spaces of Hilbert-Schmidt, compact and trace class operators are, in infinite dimensions, never the same it follows that the tensor norms can not be the same either.

To show that they are never the same in infinite dimensions:

Let $(e_n)_{n \in \mathbb{N}}$ be an orthonormal system in $H$. Let $(x_n)_{n \in \mathbb{N}}$ be any sequence of complex numbers that converges to 0, but whose absolute value is not square summable. For example $x_n =1/\sqrt{n}$. Define a linear operator $T: H\to H$ by $Ty = \sum_{n \in \mathbb{N}} x_n \langle e_n, y \rangle e_n$. Then $T$ is compact, but not Hilbert-Schmidt.

Let $(s_n)_{n \in \mathbb{N}}$ be any sequence of complex numbers whose absolute value is square summable, but not summable. For example $s_n := 1/n$. Define a linear operator $S : H \to H$ by $Sy = \sum_{n \in \mathbb{N}} s_n \langle e_n, y \rangle e_n$. Then $S$ is Hilbert-Schmidt, but not trace class.

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