= Smooth convergence on a compact interval
{title2=$p_k(f_n-f)\to0\text{ for every }k$}
= Smooth convergence
{synonym}
Smooth convergence on a compact <interval> requires <uniform convergence> of the <functions> and of their <derivatives> of every order separately, as measured by the <derivative seminorms>. The need for several scales is visible in
$$
f_n(t)=\frac{\sin(nt)}n,\qquad p_0(f_n)\leq\frac1n\longrightarrow0,
\qquad f_n'(t)=\cos(nt),\qquad p_1(f_n)=1.
$$
For $n\geq2$, $p_0(f_n)=1/n$ because $t=\pi/(2n)$ lies in $[0,1]$. The <derivative> has <supremum norm> one for every $n$, already at $t=0$. Thus $f_n$ <converges uniformly> to zero, but fails the test for <smooth convergence> to zero. https://www.buttenschoen.ca/MATH725/introduction/ck-spaces/[Andreas Buttenschoen, Applied Functional Analysis, spaces of continuous and differentiable functions] discusses this example and the family of smoothness measurements.
\Image[/smooth-function-seminorm-convergence.png]
{title=Shrinking function values coexist with derivative errors of size one}
More generally, for any fixed $K\geq0$, $h_n(t)=\sin(nt)/n^{K+1}$ has $p_K(h_n)\leq1/n$ but $p_{K+1}(h_n)=1$ for $n\geq2$. If a <norm> $N$ induced the smooth <topology>, its <continuity> would give $N(f)\leq C p_K(f)$ for some finite $K$, while <continuity> of $p_{K+1}$ for $N$ would give $p_{K+1}(f)\leq C'N(f)$. Substituting $h_n$ contradicts these two bounds. <Smooth convergence> therefore cannot be reduced to one <norm>. Its beauty is that every finite order can be controlled without fixing a final order in advance.
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