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?

  • $\begingroup$ You mean besides making tensoring a continuous operation? $\endgroup$ – minimalrho Aug 21 '18 at 0:08
  • $\begingroup$ @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. $\endgroup$ – quallenjäger Aug 21 '18 at 0:11

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 $\textbf{W}$.)$^{[1]}$ 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$^{[2]}$ the distance norm we choose fits the inequality.

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$[1]$: 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$\|v \otimes w\|_{V \otimes W} \leq \|v\|_V$.

$[2]$: 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.)

  • $\begingroup$ 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. $\endgroup$ – quallenjäger Aug 28 '18 at 23:42
  • $\begingroup$ 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. $\endgroup$ – Verta Aug 29 '18 at 0:00

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