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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 $(u_2, v_1) + (u_1, v_2) + (u_2, v_2) = (u_1 + 2.u_2, v_1 + 2.v_2)$.

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    $\begingroup$ 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 $0$. $\endgroup$ – Alex Ortiz Sep 18 at 17:07
  • $\begingroup$ @TomCollinge:John Hughes answer gives you the obvious additional and scalar multiplication and also a suggestion for the norm: $\|(u_1,u_2)\|_{E_1\times E_2}:=\|u_1\|_{X_1}+\|u_2\|_{X_2}$. $\endgroup$ – Oliver Diaz Sep 18 at 17:32
  • $\begingroup$ 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? $\endgroup$ – Joonas Ilmavirta Sep 18 at 17:43
  • $\begingroup$ @AlexOrtiz Thanks, I made an edit to clarify this. $\endgroup$ – Tom Collinge Sep 18 at 17:47
  • $\begingroup$ @JoonasIlmavirta Thanks, I made an edit to clarify this. $\endgroup$ – Tom Collinge Sep 18 at 17:48
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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.

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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.

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  • $\begingroup$ 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 ?) $\endgroup$ – Tom Collinge Sep 18 at 17:29
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    $\begingroup$ 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. $\endgroup$ – John Hughes Sep 18 at 17:58
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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.

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