Edit: According to comments of Eric Wofsy and Yemon Choi I edit the question.

For a (compact) topological space $X$, we put $A=\{f:X\to \mathbb{C}\mid f\text{ is bounded}\}$. We define a semi-norm on $A$ with $\parallel f \parallel=\parallel \omega_{f}\parallel_{\infty}$ where $\omega_{f}(x)$ is the oscilation of $f$ at $x$. It defines a norm on $A/C(X)$. The completion of this normed space is denoted by $DC(X)$.

How can $DC(X\times Y)$ be written in term of $DC(X)$ and $DC(Y)$?(An appropriate completion of the algebraic tensor product $DC(X)\otimes DC(Y)$? (Or some thing else?)

Note: We can not expect to have an algebra structure on $DC(X)$, since $C(X)$ is not an ideal. But in the following particular case we may expect that we have a Banach algebra structure: When $X$ is a compact topological group with its Haar measure, $A$ is the algebra of bounded measurable functions with the convolution product. again we consider the oscillation norm. For abelian $X$, it would be interesting to study the Gelfand spectrum of the resulting commutative Banach algebra. In particular, what is the resulting algebra and its Gelfand spectrum in the particular case of $X=S^{1}$?

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    $\begingroup$ Why would you expect this to be true? Oscillation is not multiplicative, and there is no obvious map $A(X)/C(X)\otimes A(Y)/C(Y)\to A(X\times Y)/C(X\times Y)$. $\endgroup$ – Eric Wofsey Dec 28 '14 at 11:28
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    $\begingroup$ You do realize that the projective t.p. of C(X) with C(Y) is usually not isomorphic to $C(X\times Y)$? As @EricWofsey says, why would you expect this result to be true? $\endgroup$ – Yemon Choi Dec 28 '14 at 12:32
  • $\begingroup$ @EricWofsey Thanks for your two comments. so may be I should change the first part to the following: How$ DC(X\times Y)$ can be written in term of DC(X) and DC(Y) $\endgroup$ – Ali Taghavi Dec 28 '14 at 12:45
  • $\begingroup$ @YemonChoi May be my previous comment would be reasonable? $\endgroup$ – Ali Taghavi Dec 28 '14 at 12:48

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