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 $Tu_i \to 0$ in $\sC$ , and if $K$ is a compact set in $U$ , and $\{u_i\}_{i=1}^\infty$ is a sequence in $\cD_K$ such that for any multi-index $\alpha$ , we have $$ D^\alpha u_i \to 0$$ in the supremum norm $\lVert\cdot\rVert_\infty$ as $i\to \infty$ . For a contradiction, suppose there is a compact set $K$ in $U$ such that for all constants $C>0$ and $k\in\{0, 1,2,\ldots\}$ there exists a function $u\in \cD_K$ such that $$|T(u)|> C\sum_{|\alpha|\le k} ||D^\alpha u||_\infty.$$ Then, for $C=k=1,2,\ldots$ we obtain functions $u_1,u_2,\ldots$ in $\cD(K)$ such that $ |T(u_i)| > i\sum_{|\alpha|\le i} ||D^\alpha u_i||_\infty.$ Thus $|T(u_i)|>0$ for all $i$ , so for $v_i=u_i/|T(u_i)|$ , we have $$ 1> i\sum_{|\alpha|\le i} ||D^\alpha v_i||_\infty.$$ It follows that $||D^\alpha u_i||_\infty< 1/i$ for any multi-index $\alpha$ with $|\alpha|\le i$ . Thus $\{v_i\}_{i=1}^\infty$ satisfies our assumption, whence $T(v_i)$ should tend to $0$ . However, for all $i$ , we have $T(v_i)= 1$ . This contradiction completes the proof.