# H. Cartan - Differential Calculus. Query?

In H. Cartan - Differential Calculus (1971) p. 29 he investigates differentiating a bi-linear function $$f: E_1 \times E_2 \to F$$ where $$E_1, E_2, F$$ are Banach spaces and $$E_1 \times E_2$$ the product (presumably Cartesian).

He claims $$E_1 \times E_2$$ to be a Banach space with the obvious rules of addition and scalar multiplication.
I think this might be OK if $$E_1, E_2$$ are one-dimensional, but not otherwise. To be algebraically complete mustn't he instead use the tensor product $$E_1 \otimes E_2$$ ?
Since for $$E_1, E_2$$ at least two dimensional with bases $$\{u_1, u_2\}, \{v_1, v_2\}$$ there is a very clear counterexample....
$$(u_1, v_1), (u_2, v_1), (u_1, v_2), (u_2, v_2)$$ are elements of $$E_1 \times E_2$$. But then $$(u_2, v_1) + (u_1, v_2) + (u_2, v_2)$$ is not of the form $$(u, v)$$ and so not in $$E_1 \times E_2$$, i.e. $$E_1 \times E_2$$ is not algebraically closed under addition.

I may have mixed up some concepts in the above.
It seems that $$E_1 \times E_2$$ with addition and scalar multiplication as noted by @JohnHughes is the direct sum of $$E_1, E_2$$ and nothing to do with the tensor product.
And then as noted by @JoonasIlmavirta $$(u_2, v_1) + (u_1, v_2) + (u_2, v_2) = (u_1 + 2.u_2, v_1 + 2.v_2)$$.

• What do you mean by "algebraically complete?" Also, I am not familiar with the notation $0.u_1\otimes v_1$. If $0.u_1\otimes v_1 = (0u_1)\otimes v_1$, then this is just $0$. – Alex Ortiz Sep 18 at 17:07
• @TomCollinge:John Hughes answer gives you the obvious additional and scalar multiplication and also a suggestion for the norm: $\|(u_1,u_2)\|_{E_1\times E_2}:=\|u_1\|_{X_1}+\|u_2\|_{X_2}$. – Oliver Diaz Sep 18 at 17:32
• There is no condition of algebraic completeness, whatever that might mean. (What do you mean by that?) In a Banach space you can't multiply elements together. Perhaps you're thinking of a Banach algebra? – Joonas Ilmavirta Sep 18 at 17:43
• @AlexOrtiz Thanks, I made an edit to clarify this. – Tom Collinge Sep 18 at 17:47
• @JoonasIlmavirta Thanks, I made an edit to clarify this. – Tom Collinge Sep 18 at 17:48

Since for $$E_1, E_2$$ at least two dimensional with bases $$\{u_1, u_2\}, \{v_1, v_2\}$$ there is a very clear counterexample....
$$(u_1, v_1), (u_2, v_1), (u_1, v_2), (u_2, v_2)$$ are elements of $$E_1 \times E_2$$. But then $$(u_2, v_1) + (u_1, v_2) + (u_2, v_2)$$ is not of the form $$(u, v)$$ and so not in $$E_1 \times E_2$$, i.e. $$E_1 \times E_2$$ is not algebraically closed under addition.

The sum you gave is $$(u_2, v_1) + (u_1, v_2) + (u_2, v_2) = (u_2+u_1+u_2,v_1+v_2+v_2)$$ and this is an element of $$E_1\times E_2$$ because $$u_2+u_1+u_2\in E_1$$ and $$v_1+v_2+v_2\in E_2$$. The vector is indeed of the form $$(u,v)$$, where $$u=u_2+u_1+u_2$$ and $$v=v_1+v_2+v_2$$.

This has nothing do to with Banach spaces. This is all about the concept of a product of two vector spaces as another vector space.

I suspect that the definition of addition is that $$(u_1, v_1) + (u_2, v_2) = (u_1 + u_2, v_1 + v_2),$$ where the first addition (on the right-hand side) is the addition on $$E_1$$, and the second is addition on $$E_2$$.

For scalar multiplication, I'd guess the definition is $$r(u, v) = (ru, rv)$$ where the first entry uses scalar multiplication from $$E_1$$, and the second uses scalar multiply from $$E_2$$.

Presumably the norm on $$E_1 \times E_2$$ is just something like the sum of the individual norms. Completeness then follows from something like the triangle inequality: if $$\{(u_i, v_i)\}_i$$ is a Cauchy sequence, then each of $$\{u_i\}_i$$ and $$\{v_i\}_i$$ is too, so both converge, hence the paired-sequence converges to the paired limits.

• In terms of it being a metric space it can be topologically complete (is in fact), but if it is not algebraically complete it can't be a vector space and hence not a Banach space (I think ?) – Tom Collinge Sep 18 at 17:29
• See @JoonasIlmarvirta's answer for this issue of completeness. Perhaps this all would have been clearer if the text had used some notion like "direct sum" rather than the cartesian-product symbol, but as Joonas points out, there's nothing subtle happening here: it's just a product of vector spaces, with everything done termwise. – John Hughes Sep 18 at 17:58

A Banach space is a normed space which is complete, i.e. every Cauchy sequence w.r.t. the space norm is convergent. The fact that there exist bilinear forms on $$E_1 \times E_2$$ which are not pure tensors has nothing to do with $$E_1 \times E_2$$ being a Banach space.