1
$\begingroup$

It is well-known that a continuous map $f:M\to\mathbb{R}^n$ from a Hilbert manifold can be closely approximated by a smooth map $g:M\to\mathbb{R}^n$ which has no critical points.

But, can such a continuous map $f$ also be closely approximated by a map $h:M\to\mathbb{R}^n$ which has infinitely many critical points? Also, are these critical points dense in $M$?

A reference would be much appreciated. Thanks in advance! Cross-posted on MSE.


Edit: As mentioned by Pietro, a $C^0$ perturbation creates infinitely many local minima and maxima. However, do we rigorously show this?

$\endgroup$
  • 3
    $\begingroup$ No difficulty at all in creating infinitely many local minima and maxima by a small $C^0$ perturbation. Of course impossible, in general, by a small $C^1$ perturbation. $\endgroup$ – Pietro Majer Aug 3 '18 at 19:30
  • $\begingroup$ @PietroMajer Would you mind explaining why a $C^0$ perturbation creates infinitely many local minima and maxima? Could you also add this as an answer, if you don't mind? $\endgroup$ – Sergio Charles Aug 3 '18 at 23:39
  • 1
    $\begingroup$ @Multivariablecalculus: Just take the function that you want to approximate itself, but in a small neighborhood make the function constant using a partition of unity. The neighborhood can be made arbitrarly small. $\endgroup$ – Thomas Rot Aug 4 '18 at 14:48
  • $\begingroup$ @ThomasRot Do you mind including this as an answer, for the sake of completeness? $\endgroup$ – Sergio Charles Aug 4 '18 at 16:04
  • 2
    $\begingroup$ Working in local charts. Let $\sigma(t):=\max\big(0,\min(2t-1,1)\big)$ for $t\in\mathbb{R}$. For $\epsilon>0$ define $f_\epsilon(x):=f\big(\sigma (\|x\|/\epsilon)\ x\big)$. Then $f_\epsilon$ is constant in the ball of radius $\epsilon/2$, coincides with $f$ outside the ball of radius $\epsilon$, $\|f-f_\epsilon\|_\infty=o(1)$ for $\epsilon\to0$ $\endgroup$ – Pietro Majer Aug 5 '18 at 17:03
0
$\begingroup$

As suggested by Pietro, a continuous map $f:X\to\mathbb{R}^n$ on a Hilbert manifold $X$ may be approximated to have infinitely many points. That is, a a $C^0$-perturbation creates infinitely many local minima and maxima. However, this is, in general, not true for a $C^1$-perturbation. In particular, in local charts, suppose $\sigma(t):=\max(0,\min(2t-1,1))$ for $t\in\mathbb{R}$. Then for $\varepsilon>0$, define $f_{\epsilon}(x):=f(\sigma(\|x\|/\varepsilon)x)$. As such, $f_{\varepsilon}$ is constant in the ball of radius $\epsilon/2$, it coincides outside the ball of radius $\varepsilon$, and $\|f-f_{\varepsilon}\|_{\infty}=o(1)$ for $\epsilon\to 0$.

$\endgroup$

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.