Riesz representation for products?

Question

Given is a continuous linear functional $T:C_c^0(\mathbb{R})\otimes C_c^0(\mathbb{R}) \rightarrow \mathbb{R}$ where $C_c^0(\mathbb{R})$ is the space of continuos functions with compact support. Since $C_c^0(\mathbb{R})\otimes C_c^0(\mathbb{R}) \cong C_c^0(\mathbb{R}^2)$ via the Riesz representation theorem i can find a Radon measure $\nu$ on $\mathbb{R}^2$ such that

$$T(\psi)=\int_{\mathbb{R}^2}\psi(x)\nu(dx) \; \; \; \forall \psi \in C_c^0(\mathbb{R}^2) $$

where here $T$ is understood to be a continuous linear functional from $C_c^0(\mathbb{R}^2)$ to $\mathbb{R}$.

Is it true then that for $T:C_c^0(\mathbb{R})\otimes C_c^0(\mathbb{R}) \rightarrow \mathbb{R}$ there exists a Radon measure $\mu$ on $\mathbb{R}$ such that

$$T(\phi_1 \otimes \phi_2)=\int_\mathbb{R}\phi_1(\xi)\phi_2(\xi)\mu(d\xi) \; \; \; \forall \phi_1,\phi_2 \in C_c^0(\mathbb{R}) ?$$

Answer 1

To complement the answer above: There are several subtleties that you are missing.

1. To justify the identity $C_c^\infty(\mathbb{R}) \otimes C_c^\infty(\mathbb{R}) = C_c^\infty(\mathbb{R}^2)$ you need to define tensor products in the category of topological locally convex vector spaces. In that category there are several tensor products available (the projective and injective being the minimal and maximal objects). In the case of some very special spaces, called nuclear spaces both coincide. You are working with a nuclear space.
2. You can identify the space of bi-continuous bilinear forms $B: X \times Y \to \mathbb{C}$ with the dual of the projective tensor product of $X$ and $Y$. I.e: $$ B_{bi}(X \times Y \to \mathbb{C}) = (X \otimes_{\pi} Y)^\ast, $$ where $\otimes_\pi$ is the projective tensor product- In your case $X = Y = C_c^\infty(\mathbb{R})$ and so $X \otimes_\pi Y = C_c^\infty(\mathbb{R}^2)$. It dual is the space of (non tempered) distribution over $\mathbb{R}^2$
3. What you are really asking is which bilinear forms $B: C_c^\infty(\mathbb{R}^2) \times C_c^\infty(\mathbb{R}^2) \to \mathbb{C}$ factor through the multiplication map. I.e. there is a $T: C_c^\infty(\mathbb{R}^2) \to \mathbb{C}$ such that $B = T \circ m$, where $m: C_c^\infty(\mathbb{R}^2) \times C_c^\infty(\mathbb{R}^2) \to C_c^\infty(\mathbb{R}^2)$ is the multiplication. Those forms are the ones with support inside the diagonal.

Answer 2

This is not true. For example, let $T = \delta_x \otimes \delta_y$ where $x \neq y$ so that $$T(\phi \otimes \psi) := \phi(x) \psi(y)$$ Since $x \neq y$, we can find $\phi_x$ and $\phi_y$ with disjoint support such that $\phi_x(x) = 1$ and $\phi_y(y) = 1$. This implies that for any Radon measure $\mu$ on $\mathbb{R}$ we have $$\int_{\mathbb{R}} \phi_x(\xi) \phi_y(\xi) \mu(d \xi) = 0$$ but $$T(\phi_x \otimes \phi_y) = \phi_x(x) \phi_y(y) = 1$$