# Tensor norms on infinite Banach space

Given two Banach spaces $V$ and $W$ and its tensor product $V \otimes W$.In Ryan's book "Introduction into Tensor product of Banach space", he said that it is natural to choose a norm for elementary tensors as $$\|v \otimes w\|_{V\otimes W} \leqslant \|v \|_{V}\| w \|_W$$

Now I don't understand why it is natural to require the above-mentioned inequality for a tensor norm?

• You mean besides making tensoring a continuous operation? – minimalrho Aug 21 '18 at 0:08
• @minimalrho Yes, I think there are many ways to require the tensoring continuous, (i.e. other inequalities could also make the tensoring continuous). I want to know what is so special about this one. – quallenjäger Aug 21 '18 at 0:11

## 1 Answer

Intuitively, this requirement ensures that V $$\otimes$$ W, combined with the norm $$\|v \otimes w\|_{V \otimes W}$$, is a Banach space, as long as V and W are finite-dimensional.

Banach spaces are normed vector spaces that are closed under limit. In other words, if you take any list of vectors from some vector space $$V: \{v_1, v_2, v_3,..\}$$ that gets arbitrarily close to some other vector $$l$$ (defining distance with the norm $$\|v\|_V$$), then $$l$$ will be in $$V$$. We can formalize this by saying as $$i$$ goes to infinity, $$\|l - v_i\|_V$$ gets closer and closer to 0: $$\lim_{i \to \infty} \|l - v_i\|_V = 0$$

Say we have two such spaces (we'll call them $$V$$ and $$W$$.) If we have a converging sequence $$\{v_1, v_2, v_3,..\}$$ in $$V$$-space, it is natural to assume that the sequence $$\{v_1 \otimes w, v_2 \otimes w, v_3 \otimes w,..\}$$ should also converge (for some arbitrary $$w$$ from $$\textbf{W}$$.)$$^{}$$ Moreso, it should converge to $$l \otimes w$$. Using our above notation, we can write this as $$\lim_{i \to \infty} \|l \otimes w - v_i \otimes w\|_{V \otimes W} = 0$$ However, no metric exists by default on the tensor products $$v \otimes w$$, so this doesn't have to be true!

Let's take another look at your inequality: $$\|v \otimes w\|_{V \otimes W} \leq \|v\|_V \|w\|_W$$ Take any Cauchy sequence $$\{v_i\}$$ from Banach space $$\textbf{V}$$ and replace $$v$$ with $$l - v_i$$. That gives us $$\|(l-v_i) \otimes w\|_{V \otimes W} \leq \|l-v_i\|_V \|w\|_W$$ or, using the bilinearity of the tensor product, $$\|l \otimes w -v_i \otimes w\|_{V \otimes W} \leq \|l-v_i\|_V \|w\|_W$$ As $$i\to\infty$$, the right side goes to 0. Since norms are necessarily non-negative, this means the left side must also go to zero. But this is $$\textit{exactly the definition}$$ of what it means for $$\{v_1 \otimes w, v_2 \otimes w, v_3 \otimes w,..\}$$ to converge to $$l \otimes w$$!

Furthermore, since $$l \in V$$ (by the definition of a Banach space), we know $$l \otimes w$$ must be a vector in $$V \otimes W$$. Since we picked $$\{v_i\}$$ arbitrarily (it could be any Cauchy sequence in $$V$$), any converging sequence in $$V$$ gives a corresponding converging sequence in $$V \otimes W$$, and the limit $$l \otimes w$$ exists in $$V \otimes W$$. In other words, $$V \otimes W$$ is a Banach space. This holds iff$$^{}$$ the distance norm we choose fits the inequality.

$$\\$$ $$\\$$

$$$$: The flipside is also true: a converging sequence $$\{w_1, w_2, w_3,..\}$$ in $$W$$ should imply the sequence $$\{v \otimes w_1, v \otimes w_2, v \otimes w_3,..\}$$ converges to $$v \otimes l_w$$ for any $$v\in\textbf{V}$$. Otherwise we could just write your inequality as$$\|v \otimes w\|_{V \otimes W} \leq \|v\|_V$$.

$$$$: Necessity comes from considering the inverse. If we are allowed to choose a metric s.t. $$\|v \otimes w\|_{V \otimes W} > \|v\|_V \|w\|_W$$, then not only does $$\{v_1 \otimes w, v_2 \otimes w, v_3 \otimes w,..\}$$ not converge to $$l \otimes w$$, it might converge to something that isn't even in $$V \otimes W$$ -- which would be bad news for our hopes of a Banach space.

EDIT: Just realized I forgot to address the general Cauchy sequence in $$V \otimes W$$. Since every vector $$x\in V \otimes W$$ can be written as $$c v \otimes \frac{1}{c^*} w$$ for nonzero $$c\in \mathbb{C}$$, we can say that $$\{v_1 \otimes w_1, v_2 \otimes w_2, v_3 \otimes w_3,..\}$$ should converge to $$l_v \otimes l_w$$. This leads to $$\lim_{i\to\infty} \|l_v \otimes l_w - v_i \otimes w_i \|_{V \otimes W} = 0$$ The given inequality proves the equivalent statement $$\lim_{i\to\infty} \|v_i \otimes w_i - l_v \otimes l_w \|_{V \otimes W} = 0$$ when we consider the expansion of two Cauchy sequences: $$\{v_1, v_2, v_3,..\}$$ in $$V$$ and $$\{w_1, w_2, w_3,..\}$$ in $$W$$. Using the inequality: $$\|(l_v-v_i) \otimes (l_w - w_i)\|_{V \otimes W} \leq \|l_v-v_i\|_V \|l_w - w_i\|_W$$ We can expand the tensor product to $$\|l_v \otimes l_w - l_v \otimes w_i - v_i \otimes l_w + v_i \otimes w_i\|_{V \otimes W} \leq \|l_v-v_i\|_V \|l_w - w_i\|_W$$ By the case shown in the main answer, we know the middle two tensor products converge to $$l_v \otimes l_w$$. The left side then becomes the "equivalent statement" above, and the limit $$i\to\infty$$ goes to 0 with similar logic (non-negative and less or equal to a statement whose limit is 0.)

• Thank you for your answer. Are you sure that it would induce $V \otimes W$ is a banach space? Because your argument would only work if you can represent your tensor algebra as linear combination of $e_i \otimes e_j$, where $e_i,e_j$ are the basis of finite dimensional space $V$ and $W$. You will definitely have a problem if $V$ and $W$ are assumed to be infinite dimensional. – quallenjäger Aug 28 '18 at 23:42
• I'm not sure. This answer seems to suggest so, but I can't think of any definite reason it would. Updated my answer accordingly. – Verta Aug 29 '18 at 0:00