Suppose $f$ is a distribution on $\mathbb{R}^1$. Show that $\langle F, \phi \rangle=\langle f, \phi_y \rangle$, for $\phi \in \mathcal{D}(\mathbb{R}^2)$, where $\phi_y(x)=\phi(x,y)$ (here $y$ is fixed), defines a distribution in $\mathbb{R}^2$.
My question lies in showing how the function is continuous in $\mathbb{R}^2$. Really appreciate the help! Thank you :)
OK I will use the heuristic integral notation for the pairing of a distribution with a test function, i.e., what I said I would avoid in https://mathoverflow.net/questions/72450/can-distribution-theory-be-developed-riemann-free
Here $f(x)$ is a generalized function of the single variable $x\in\mathbb{R}$. Now the quantity $\langle F,\phi\rangle$ defined by $\langle f,\phi_y\rangle$ should be $$ \int_{\mathbb{R}} f(x)\phi(x,y)\ dx\ . $$ This is not a number, but a function of the variable $y$ which has not been "integrated over". This function belongs to $\mathscr{D}(\mathbb{R})$, i.e., is a smooth compactly supported function of $y$. This is part of the statement of Fubini's Theorem for distributions, in the $\mathscr{D},\mathscr{D}'$ variant (see my answer to the above MO question for the $\mathscr{S},\mathscr{S}'$ variant).
Now what lcv proposes makes perfect sense but is something different: the application of the distribution $f\otimes 1\in\mathscr{D}'(\mathbb{R}^2)$ on the test function $\phi\in\mathscr{D}(\mathbb{R}^2)$ which indeed produces a number. Heuristically this is $$ \int_{\mathbb{R^2}} f(x)\phi(x,y)\ dx dy $$
where this time $y$ has been integrated over.