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If $X$ is a Banach space, then I want to know if $X\times X$ is also Banach. What is the norm of that space?

So for example, we know $C^k(\Omega)$ is Banach and I have a vector $v = (u_1, u_2)$ where $u_i \in C^k(\Omega)$ and I want to use properties on this vector.

Thanks

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3 Answers 3

Yes, though the product is more commonly called a direct sum. Many equivalent norms are possible; common choices are $\lVert(u_1,u_2)\rVert=\lVert u_1\rVert+\lVert u_2\rVert$ and $\lVert(u_1,u_2)\rVert=\max(\lVert u_1\rVert,\lVert u_2\rVert)$, or more generally $\lVert(u_1,u_2)\rVert=(\lVert u_1\rVert^p+\lVert u_2\rVert^p)^{1/p}$ for $1\le p<\infty$.

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Wouldn't the direct sum give you the *co*product? –  Rasmus Jul 27 '12 at 13:14
    
@Rasmus: For a finite number of factors, the product and the coproduct in the category of Banach (or linear) spaces are the same. –  Harald Hanche-Olsen Jul 27 '12 at 13:19
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But Rasmus has a point when you restrict to linear maps of norm at most one, where the $\ell^1$-sum is the coproduct and the $\max$-norm is the product norm. –  t.b. Jul 27 '12 at 13:21
    
@t.b.: Good point. –  Harald Hanche-Olsen Jul 27 '12 at 13:22
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If these spaces happen to be Hilbert spaces, you can use this with $p=2$ and the sum/product is again a Hilbert space. –  GEdgar Jul 27 '12 at 14:37

The product of a finite number of Banach spaces can easily made into a Banach space by, e.g., adding the norms or by taking their maximum. There are more choices, but none of them is natural, to my knowledge, or preferred.

A better way to put it is, in my opinion, to say that such a product is, in a natural way, a topological vector space, which admits a complete norm.

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"none of them is natural..." As t.b. points out in a comment on Harald's answer, there are unique choices for the product and coproduct norms when the morphisms are linear contractions. –  Jonas Meyer Jul 27 '12 at 14:20

Yes.

Let $(X, \lVert - \rVert)$ be a Banach space. Then $X \times X$ is a Banach space under the norm $\lVert (x,y) \rVert = \lVert x \rVert + \lVert y \rVert$.

Proof: It's easy to check that this defines a norm, so we just need completeness. Let $(x_n, y_n)$ be a Cauchy sequence in this norm. Let $\varepsilon > 0$ and let $N \in \mathbb{N}$ be such that for all $m,n \ge N$ we have

$$\lVert (x_n, y_n) - (x_m, y_m) \rVert < \varepsilon$$

Then by the definition of our norm we must also have $\lVert x_n - x_m \rVert + \lVert y_n - y_m \rVert < \varepsilon$, and hence $(x_n)$ and $(y_n)$ are Cauchy in $X$, hence convergent, and it's easy to check that if $x_n \to x$ and $y_n \to y$ then $(x_n,y_n) \to (x,y)$. $\square$

(In fact we may choose other norms for $X \times X$.)

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