I am interested in whether the uniform symbol spaces $S^m_\infty$ from the theory of pseudo-differential operators have the Heine-Borel property (i.e. every bounded set is relatively compact).
To make the question a bit easier, let's instead a consider similar Fréchet space $E$, defined as follows: Let's say a function $f\in C^\infty(\mathbb{R})$ lies in $E$, if $p_k(f):=\Vert x^kf^{(k)}\Vert_{L^\infty}<\infty$ for all $k\in \mathbb{Z}_{\ge 0}$. Then $E$, together with the semi-norms $(p_k)$, is a Fréchet space and the inclusion into $C^\infty(\mathbb{R})$ is continuous.
Question. Does every bounded sequence in $E$ have a convergent subsequence?
- Since $C^\infty(\mathbb{R})$ has the Heine-Borel property, any bounded sequence $(f_n)\subset E$ converges to some $f$ in the $C^\infty$-topology. The only thing we have to prove is thus that $f\in E$ and $p_k(f-f_n)\rightarrow 0$ for all $k$.
$E$ does not satisfy the Heine-Borel property, here is a counterexample: Take $0\neq \varphi \in C^\infty(\mathbb{R})$ with support compactly contained in $(0,1)$ and define $f_n(x)=\varphi(x/n)$.
Let's check that the sequence is bounded: $f_n^{(k)}(x)=n^{-k}\varphi^{(k)}(x/n)$ is supported in $(0,n)$ and hence $\vert x^kf_n^{(k)}(x)\vert \le \vert \varphi^{(k)}(x/n)\vert$. We conclude that $p_k(f_n)\le \Vert \varphi^{(k)}\Vert_{L^\infty}$, which implies boundedness.
But it does not have a convergent subsequence: Note that $f_n\rightarrow 0$ in $C(\mathbb{R})$ (i.e. uniformly on every compact set). Hence the only possible accumulation point in $E$ is also $0$. But $p_0(f_n)=\Vert \varphi\Vert_{L^\infty}$ does not converge to $0$. We conclude that $(f_n)$ has no convergent subsequence in $E$.