Show that the order of $\delta'$ is one

Question

I try to show that the order of $\delta'$ is one. Clearly $|(\delta', \phi)| = |\phi'(0)| \leq \sup \phi'(x)$ but if I got the definition of order right I have to show that $|\phi'(0)| \leq \sup \phi(x)$ as well. How do I do that?

Thanks in advance!

Answer 1

Distribution $d$ being of order $k$ means that $|(d, \phi)|$ is bounded by an expression of the form $c_0 \sup \phi + c_1 \sup \phi' + \ldots + c_k \sup \phi^{(k)}$ with nonnegative constants $c_i$. With this definition you get that $\delta'$ is of order $1$ by taking $c_0 = 0$ and $c_1 = 1$.

Answer 2

Construct a sequence $\{\phi_n\}$ of test functions with support contained in $[-1,1]$ such that $\phi_n'(0)\geqslant n$ but $\sup_x|\phi_n(x)|=1$. For example, regularize the polygonal line $(-1,0),(-1/2,1),(0,1),(n^{-1},0)$.