| Version current |
Version 6 |
| \newcommand{\cD}[0]{\mathcal{D}} |
\newcommand{\cD}[0]{\mathcal{D}} |
| \newcommand{\scomp}[0]{C^\infty_0} |
\newcommand{\scomp}[0]{C^\infty_0} |
| Let us show the equivalence of (2) and (3). |
Let us show the equivalence of (2) and (3). |
| First, the proof that (3) implies (2) is a direct calculation. |
First, the proof that (3) implies (2) is a direct calculation. |
| Next, let us show that (2) implies (3): |
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 |
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 |
$\{u_i\}_{i=1}^\infty$ is a sequence in $\cD_K$ such that |
| for any multi-index $\alpha$, we have |
for any multi-index $\alpha$, we have |
| $$ D^\alpha u_i \to 0$$ |
$$ D^\alpha u_i \to 0$$ |
|
in the supremum norm $\lVert\cdot\rVert_\infty$ as $i\to \infty$.
|
in the supremum norm $||\cdot||_\infty$ as $i\to \infty$.
|
| For a contradiction, suppose there is a compact set $K$ in $U$ |
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 |
such that for all constants $C>0$ and $k\in\{0, 1,2,\ldots\}$ there exists |
| a function $u\in \cD_K$ such that |
a function $u\in \cD_K$ such that |
| $$|T(u)|> C\sum_{|\alpha|\le k} ||D^\alpha u||_\infty.$$ |
$$|T(u)|> C\sum_{|\alpha|\le k} ||D^\alpha u||_\infty.$$ |
| Then, for $C=k=1,2,\ldots$ we obtain functions $u_1,u_2,\ldots$ in $\cD(K)$ |
Then, for $C=k=1,2,\ldots$ we obtain functions $u_1,u_2,\ldots$ in $\cD(K)$ |
| such that |
such that |
| $ |T(u_i)| > i\sum_{|\alpha|\le i} ||D^\alpha u_i||_\infty.$ |
$ |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 |
Thus $|T(u_i)|>0$ for all $i$, so for $v_i=u_i/|T(u_i)|$, we have |
| $$ 1> i\sum_{|\alpha|\le i} ||D^\alpha v_i||_\infty.$$ |
$$ 1> i\sum_{|\alpha|\le i} ||D^\alpha v_i||_\infty.$$ |
| It follows that $||D^\alpha u_i||_\infty< 1/i$ |
It follows that $||D^\alpha u_i||_\infty< 1/i$ |
| for any multi-index $\alpha$ with $|\alpha|\le i$. |
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 $0$. |
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 |
However, for all $i$, we have $T(v_i)= 1$. This contradiction |
| completes the proof. |
completes the proof. |
