Tensor product of algebra 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 $L^1(X)\otimes L^1(Y)=L^1(X\times Y)$
 A: Sorry I'm a little late. 


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*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.

*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).
