In reading my textbook, the author give a lemma as follows:
Let $X\subset \mathbb{R}^{n}$ be an open set, and let $u\in \mathcal{E}'(X)$ have order $N$. Then $\langle u,\phi \rangle=0$ for all $\phi$ such that $\partial^{\alpha}\phi(x)=0$ when $x\in supp(u)$ and $|\alpha|\le N$.
The author gives a counter-example stating that it is generally wrong to apply this to the case $K=supp(u)$ and assert that if a sequence of functions $\phi_{k}$ converges uniformly to 0 on $supp(u)$, and all its derivatives do so as well, then $\langle u,\phi_{k}\rangle \rightarrow 0$ as $k\rightarrow \infty$. The counter-example is as follows:
Let $X=\mathbb{R}$, $\langle u,\phi \rangle=\lim_{m\rightarrow \infty}\left(\sum^{m}_{k=1}\phi\left(\frac{1}{k}\right)-m\phi(0)-\phi'(0)\log(m)\right)$ Then $supp (u)$ is $\left\{0,1,\frac{1}{2},\frac{1}{3}..\right\}$ We can construct $\phi_{k}\in C^{\infty}_{c}(\mathbb{R})$ such that $\phi_{k}=k^{-\frac{1}{2}}$ for $x\ge \frac{1}{k}$ and $\phi_{k}=0$ for $x\le \frac{1}{k+1}$. Then we have $\phi_{k}$ converge to $0$ uniformly as $k\rightarrow \infty$, and all their derivatives vanish on $supp (u)$. However we have $\langle u,\phi_{k}\rangle=kk^{-\frac{1}{2}}=k^{\frac{1}{2}}\rightarrow \infty$ as $k\rightarrow \infty$.
My question is - is $u$ in here of finite order? If not why the lemma failed? I feel confused because the author claimed when the boundary of $supp u$ is nice enough then the above claim makes sense, but it feels (nice enough or not) is not related as it is not used at all in the proof of the lemma.