For $\varphi\in C_{0}^{\infty}(\mathbb{R}^{3})$ , define $u(\varphi):=\int\partial^{\alpha}\varphi(x,0,0)dx$ for some multiindex $\alpha$ .
It's pretty clear to me that $u$ is a distribution.
But can anyone please help show what the order of $u$ is? It's obviously $\leq\alpha$ by
$\left|\int\partial^{\alpha}\varphi(x,0,0)dx\right|\leq\int_{\mbox{supp}(\varphi)}dx\mbox{sup}\left|\partial^{\alpha}\varphi(x,y,z)\right|,$
but could the inequality be strict? Or how I can one prove that it's not, which I suspect.
Finally, what about the order of the derivatives, e.g
$\frac{\partial}{\partial y}u(\varphi):=\int\partial^{\alpha}(\frac{\partial}{\partial y}\varphi)(x,0,0)dx.$
Many thanks for any help!
In the multiindex $\alpha=(\alpha_1,\alpha_2,\alpha_3)$ not all components are equally important. Since you integrate over the first variable, having $\alpha_1 >0$ makes the distribution identically zero: the integral over $x$ of $x$-partial derivative simplifies by the fundamental theorem of calculus, and turns out to be $0$ because $\varphi$ is compactly supported.
If $\alpha_1=0$, then the order of distribution is $|\alpha|= \alpha_2 + \alpha_3 $. To prove the lower bound, fix a test function $\varphi$ for which $u(\varphi)\ne 0$. (You'll need to argue that such $\varphi$ exists.) Apply scaling $\varphi_t(x,y,z) =t^{\alpha}\varphi(x,ty,tz) $ where $t$ is large. Observe that $u(\varphi_t)=u(\varphi)$. Yet, the derivatives of $\varphi_t$ of orders below $|\alpha|$ tend to zero as $t\to\infty$. So, these derivatives cannot control $u(\varphi_t)$.