proof of convergence theorem

Let us show the equivalence of (2) and (3).First, the proof that (3) implies (2) is a direct calculation.Next, let us show that (2) implies (3):Suppose $T{u}_i\to 0$ in $ℂ$, and if $K$ is a compact set in $U$, and${\{u_i\}}_{i=1}^{\mathrm{\infty }}$ is a sequence in ${𝒟}_K$ such thatfor any multi-index $\alpha$, we have

$$D^{\alpha }{u}_i\to 0$$

in the supremum norm ${\parallel \cdot \parallel }_{\mathrm{\infty }}$ as $i\to \mathrm{\infty }$.For a contradiction, suppose there is a compact set $K$ in $U$such that for all constants $C>0$ and $k\in \{0,1,2,\mathrm{\dots }\}$ there existsa function $u\in {𝒟}_K$ such that

$$|T(u)|>C\sum _{|\alpha |\le k}{||D^{\alpha }u||}_{\mathrm{\infty }}.$$

Then, for $C=k=1,2,\mathrm{\dots }$ we obtain functions $u_1,u_2,\mathrm{\dots }$ in $𝒟(K)$such that$|T(u_i)|>i{\sum }_{|\alpha |\le i}{||D^{\alpha }{u}_i||}_{\mathrm{\infty }}.$Thus $|T(u_i)|>0$ for all $i$, so for $v_i=u_i/|T(u_i)|$, we have

$$1>i\sum _{|\alpha |\le i}{||D^{\alpha }{v}_i||}_{\mathrm{\infty }}.$$

It follows that for any multi-index $\alpha$ with $|\alpha |\le i$.Thus ${\{v_i\}}_{i=1}^{\mathrm{\infty }}$ satisfies our assumption, whence $T(v_i)$ should tend to $0$.However, for all $i$, we have $T(v_i)=1$. This contradictioncompletes the proof.