From Chapter 1, Exercise 8 and 9, Strichartz's book: Distribution theory & fourier transforms
Suppose $f$ is a distribution on $\mathbb{R}^1$. Show that $<F,\phi> = <f,\phi_y>$, for $\phi \in \mathcal{D} (\mathbb{R}^2)$. where$\phi_y(x) = \phi(x,y)$ (here y is any fixed value), defines a distribution on $\mathbb{R}^2$.
Suppose f is a distribution on $\mathbb{R}^1$. Show that $<G,\phi> = \int_{-\infty}^\infty<f,\phi_y>dy$ for $\phi\in\mathcal{D}(\mathbb{R}^2)$ defines a distribution on $\mathbb{R}^2$. Is $G$ the same as as $F$ in problem 8.
In problem 8 isn't the functional on $\mathcal{D}(\mathbb{R}^1)$ dependent on the value $y$. So how does it give a the same value for all such test functions $\phi\in\mathcal{D}(\mathbb{R}^2)$ irrespective of the value of $y$.
Problem 9 does recognize that the value is dependent upon $y$ therefore, the variable of the integrand is $y$. This is confusing to me.
Your understanding is correct: in problem 8, $F$ depends on $y$, so it should better be called $F_y$, or similar. But, since $y$ is assumed fixed, the author dropped it. In problem 9 $y$ is not fixed anylonger, so the index makes sense now: $G=\int \limits _{-\infty} ^\infty <F_y, \phi> \Bbb d y$. (In particular, this shows that $F$ and $G$ are not the same, because $F$ is a distribution in dimension $2$, while $G$ is in dimension $1$; but maybe you are asked whether $G$ and $f$ are the same?)