Show that $\langle f,g \rangle:=\int_{\mathbb{R}} \frac{\partial g (t,t^2) }{\partial x} dt$ for $g(x,y) \in \mathcal{D}(\mathbb{R}^2)$
defines a Distribution $f \in \mathcal{D}'(\mathbb{R}^2)$
As far as I know I need to show linearity and continuity.
But before I start I am a little bit confused by the expression.
What does $g(t,t^2)$ mean? Do I just handle $t$ and $t^2$ like two seperate variables? And if so since $g$ depends only on $t$ and $t^2$ all partial derivatives with respect to $x$ are equal zero. This sounds very wrong to me.
Is there some typo? And If so could someone write down the correct expression (or give me a link or something).
If it's not a typo, it would be nice if someone could tell what I am understanding wrong.
Edit:
Now my calculations/thoughts Showing convergence: Let $g_n$ be a testfunction which converges to $g$
Since $D^\alpha g_n \rightrightarrows D^\alpha g$ for $\alpha \in \mathbb{N}^2$
One can interpret $\frac{\partial g_n (t,t^2) }{\partial x}$ being dependent on two variables $t$ and $t^2$ Because $D^\alpha g_n \rightrightarrows D^\alpha g$ for $\alpha \in \mathbb{N}^2$, this implies that $\frac{\partial^m g_n (t,t^2) }{\partial x^m}$ $\rightrightarrows$ $\frac{\partial^m g (t,t^2) }{\partial x^m}$.
Considering now that $\frac{d}{dt} \int_{\mathbb{R}} \frac{\partial g_n (t,t^2) }{\partial x} dt= \frac{\partial g_n (t,t^2) }{\partial x}$
The only thing left to show is $\int_{\mathbb{R}} \frac{\partial g_n (t,t^2) }{\partial x} dt \rightrightarrows \int_{\mathbb{R}} \frac{\partial g (t,t^2) }{\partial x}$. This holds because $g_n$ is uniformly convergent, thus we can interchange the limits.
Would be very thankful if someone could tell me, if my calculations are correct? (If not I would probably need some held)