On page 156 of Rudin's Functional Analysis, he gives the following condition for a linear functional over the test functions $D(\Omega)$ to be continuous:
A linear functional $\Lambda$ on $D(\Omega)$ is continuous if and only if for each compact subset $K\subset \Omega$ there is an $N$ and a $C$ such that $|\Lambda\phi|\le C\|\phi\|_N$ for all $\phi\in D(K)$.
Here $\|\phi\|_N:=\operatorname{max}_{|\alpha|\le N, x\in\Omega}|D^{\alpha}(\phi)|$.
I am having trouble to see why continuity of $\Lambda$ implies the existence of such $C$ and $N$. In general, for a fixed $n$, sets of the form \begin{equation} \{\phi\in D(K):\|\phi\|_n\le 1\} \end{equation} is not bounded in $D(K)$, right?
Can someone give a hint why the continuity condition is true?
Thanks!
A distribution $\Lambda$ is continuous, iff the restriction to each $D(K)$ is continuous. Now a continuous $\Lambda$ implies for $\varepsilon = 1$ a neighbourhood $U$ in $D(K)$ represented by a seminorm $||.||_n$ and a $\delta>0$ with $$ U = \{\phi\in D(K); ||\phi||_n < \delta\} \,\text{and}\ |\Lambda(\phi)| < \varepsilon \,\forall\phi\in U. $$ So for a non-zero $\phi$ set $\psi=\frac{\delta}{||\phi||_n} \phi$. Then $|\Lambda(\psi)|<\varepsilon$ and so with $C=\varepsilon/\delta = 1/\delta$ $$ |\Lambda(\phi)| = \big|\frac{||\phi||_n}{\delta} \Lambda(\psi)\big| < \frac{\varepsilon}{\delta} ||\phi||_n = C ||\phi||_n.$$