1
$\begingroup$

My question is really a basic (and, I dare say, naive) question about the theory of ordinary differential equations and is likely answered in a book on functional analysis or the like, so any pointers to such a book with the answer would be appreciated.

It is well-known that $\mathcal{C}^n(I)$ with the compact-open topology for any open interval $I$ of positive width is homeomorphic to $\mathcal{C}^n((0,1))$ with the compact-open topology, essentially via the unique orientation-preserving affine diffeomorphism of $I$ with $(0,1)$. In this post and this post, it is outlined that $\mathcal{C}^n((0,1))$ with the compact-open topology is metrizable by a metric that is complete, that is to say, that $\mathcal{C}^n((0,1))$ with the compact-open topology is topologically complete.

My questions is, are the elementary functions dense in $\mathcal{C}^n((0,1))$ with the compact-open topology?

If this is true, then given any $n^\text{th}$-order linear IVP with all coefficient functions continuous on $I$ and leading coefficient function $a_n(t)$ having no roots in $I$ has a Cauchy sequence (in any of the metrics outlined above) of elementary functions which converges to the IVP's unique solution $x(t)$ (which may be non-elementary but which will be $n$ times continuously differentiable on $I$ as $\mathcal{C}^n(I)$ with the compact-open topology is topologically complete). This fact would be useful in motivating differential equations students about some of the strategies for "solving", or approximating solutions to, IVPs.

$\endgroup$
2
$\begingroup$

The polynomials are dense by the Stone-Weierstrass theorem.

$\endgroup$
  • 1
    $\begingroup$ Isn't that only true for closed, bounded intervals $I$? en.wikipedia.org/wiki/Stone%E2%80%93Weierstrass_theorem (I had thought of that but seemed to recall the compactness of the domain was an important hypothesis.) $\endgroup$ – Jeffrey Rolland Apr 13 '17 at 21:19
  • $\begingroup$ It is true for open intervals in the compact-open topology; the proof is pretty much follow-your-nose. (researchgate.net/post/… See Rogier Brussee's comment) $\endgroup$ – Jeffrey Rolland Apr 13 '17 at 22:01
  • $\begingroup$ Compact-open topology: so a basic neighbourhood of $f$ is defined in terms of the values on a compact subset $K \subset (0,1)$. And Stone-Weierstrass says the polynomials are dense in the continuous functions on that compact set. $\endgroup$ – Robert Israel Apr 13 '17 at 23:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.