# List of tensor product spaces with uniform crossnorms

Let $H^{(j)}$ and $G^{(j)}$ be Banach spaces for $j\in\{1,\dots,n\}$. Call norms $\|\cdot\|_{H}$ and $\|\cdot\|_{G}$ on the algebraic tensor products $H:=\bigotimes_{j=1}^n H^{(j)}$ and $G:=\bigotimes_{j=1}^n G^{(j)}$ uniform if the operator norm satisfies $$\|\bigotimes_{j=1}^n A^{(j)}\|_{H\to G}=\prod_{j=1}^n \|A^{(j)}\|_{H^{(j)}\to G^{(j)}}.$$

Is there an extensive list of pairs of spaces with uniform crossnorms? (Preferably with a focus on function spaces; Lebesgue, Sobolev, Hoelder would already be great)

Of course, the results for tensor products of well-known spaces depend on the norms that we equip these tensor products with. For example, it would be good to know if $C(\Omega_1)\otimes C(\Omega_2)$ equipped with the $C(\Omega_1\times\Omega_2)$ norm and $H^1(\Omega_3)\otimes H^1(\Omega_4)$ equipped with the $H^{1}_{\text{mix}}(\Omega_3\times\Omega_4)$ norm are uniform, and if these norms actually turn the algebraic tensor products into the spaces $C(\Omega_1\times\Omega_2)$ and $H^{1}_{\text{mix}}(\Omega_3\times\Omega_4)$, respectively (by closure).

Posting any specific results instead of a reference would be appreciated too; I will keep track in the list below:

• If $H^{(j)}$ and $G^{(j)}$ are Hilbert spaces, then equipping $H$ and $G$ with the induced Hilbert space structure yields uniform crossnorms. The induced scalar product on $H$ is the unique bilinear extension of $$\langle \otimes_{j=1}^n f^{(j)}_1 ,\otimes_{j=1}^n f^{(j)}_2\rangle_{H}=\prod_{j=1}^n \langle f^{(j)}_1,f^{(j)}_2\rangle_{H^{(j)}}$$ (Proposition 4.127 in W. Hackbusch, "Tensor spaces and numerical tensor calculus". Springer, 2012)
• Equipping $H$ with the projective norm and $G$ with any crossnorm (that is, a norm that is multiplicative w.r.t the tensor product) yields uniform crossnorms. (Have no reference)

• Equipping $G$ with the injective norm and $H$ with any crossnorm yields uniform crossnorms (Have no reference)

• Operator ideal property is what you want, but with $\le$ instead of $=$. Equality is easy on decomposable tensors. In the second reference it is the functorial property. – Peter Michor Mar 30 '16 at 9:55