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Can we find define a norm on tensor product $C(X) \otimes C(Y)$ such that the norm completion of $C(X)\otimes C(Y)=C(X\times Y)$

And can we define a norm on tensor product $L^1(X)\otimes L^1(Y)$ such that the norm completion of $L^1(X)\otimes L^1(Y)=L^1(X\times Y)$

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The best you can hope for is to define a norm such that the completion of the LHS is the RHS. The LHS doesn't have enough elements in it, e.g. $C(\mathbb{R}) \otimes C(\mathbb{R})$ doesn't contain functions like $\sin xy$. – Qiaochu Yuan Mar 12 '13 at 0:07
Sorry, I should make it more clear. By $C(X)\otimes C(Y)$ I mean norm completion of tensor product. What's the norm should looks like? – owen Mar 12 '13 at 1:26
Let $ X $ and $ Y $ be locally compact Hausdorff spaces. Form the algebraic tensor product $ C(X) \otimes C(Y) $, and equip it with a special locally convex topology to obtain the ‘injective tensor product’ $ C(X) \otimes_{\varepsilon} C(Y) $. Then completing $ C(X) \otimes_{\varepsilon} C(Y) $ with respect to its topology yields the completed injective tensor product $ C(X) \widehat{\otimes}_{\varepsilon} C(Y) $. This will be isomorphic to $ C(X \times Y) $. If $ X $ and $ Y $ are further compact, then $ C(X) \widehat{\otimes}_{\varepsilon} C(Y) $ is even norm-isomorphic to $ C(X \times Y) $. – Haskell Curry Mar 14 '13 at 16:24

Sorry I'm a little late.

  1. For you first question note that $C(X)$ are abelian (and thus nuclear) $C^*$-algebras. This means that there is a unique norm you can put on $C(X)\otimes C(Y)$ such that it's completion is a $C^*$-algebra. I claim that the completion is isomorphic to $C(X\times Y)$. Consider the map from $C(X)\otimes C(Y)$ to $C(X\times Y)$ defined on simple tensors by $f\otimes g\mapsto f(x)g(y)$. Then extend linearly. This map is injective. You can use Stone–Weierstrass to show that the image is dense. Thus you can restrict the norm from $C(X\times Y)$ to the image to get a new $C^*$-norm on $C(X)\otimes C(Y)$. Since the norm is unique you get that this must be the original norm. Since the image is dense you get the desired result.

  2. For your second question it is still true but since $L^1$ is not a $C^*$-algebra it is not as easy. In general there are many norm you can put on $L^1(X)\otimes L^1(Y)$. The theory of Topological vector space tensor product norms was some of the earliest work of Grothendieck (before any algebraic geometry). Anyway there is a tensor product called the projective tensor product (i think it is also called the topological tensor product) that gives the correct isomorphism (though I don't know the proof off the top of my head).

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