Verapoulous Algebra $C(K) \mathbin{\hat\otimes} C(L)$ is a subalgebra of $C(K\times L)$?

Let $$K$$ and $$L$$ be compact spaces. Consider the Banach algebra $$V(K,L)=C(K)\mathbin{\hat\otimes} C(L)$$ , which is the completion of the $$C(K)\otimes C(L)$$ with respect to the projective tensor norm. It is known that $$V(K,L)$$ is a subalgebra of $$C(K\times L)$$ , not isometrically. I am trying to establish this result . Heres what I have figured out. Due to the universal property of tensor product , there is an algebra homomorphism $$\theta : C(K)\otimes C(L)\to C(K\times L)$$ such that $$\theta(\sum_{i=1}^nf_i\otimes g_i)(x,y)=\sum_{i=1}^nf_i(x)g_i(y)$$

I want to show that this map is injective. Once this is done then we can extend it to its completion.

• But be careful: The extension of an injective uniformly continuous map to the completion needn't be injective! Consider a Banach space $(X,\|\cdot\|)$ and a strictly finer norm $\||\cdot\||$ on it (e.g., $\|x\|+|f(x)|$ for a discontinuous linear functional). Then the identity from $(X,\||\cdot\||)$ to $(X,\|\cdot\|)$ is continuous and injective but its extension to the completion is not injective (because otherwise its inverse would be continuous by the closed graph theorem). – Jochen Jan 15 at 13:47
• @jochen Thanks for reminding , I was actually thinking the same . – Lav Kumar Jan 16 at 13:55

Suppose $$h\in C(K)\otimes C(L)$$ and $$\theta(h)=0$$. Write $$h=\sum_{i=1}^nf_i\otimes g_i$$, where the $$g_i$$ are linearly independent. But then this implies that for all $$x\in K$$, $$0=\theta(h)(x,\cdot)=\sum_{i=1}^nf_i(x)g_i(\cdot),$$ whence $$f_i=0$$ for all $$i$$, and thus $$h=0$$.
• I wouldn't think so, but possibly. You might be able to do it with the $f_i$ linearly independent and obtain a representation with less elementary tensors. – Aweygan Jan 14 at 17:09