0
$\begingroup$

Consider two smooth functions $f,g\in C^\infty(\Omega)$ with $\partial \Omega$ smooth and $\Omega\subset \mathbb{R}^3$. Assume that $f=g$ on $\partial \Omega$. For any given $\varepsilon>0$, how to perturb $f$ to another smooth function $\tilde{f}$ such that $\|f-\tilde{f}\|_{L^2}<\varepsilon$ and, if $g(x)\neq g(y)$, then $\tilde{f}(x) \neq \tilde{f}(y)$ for any $x,y\in(\Omega)$? Or a counter example?

$\endgroup$
5
  • $\begingroup$ Usually I've seen perturbation questions asked so that it's just a matter of local closeness, i.e. pointwise $|(f-\stackrel{\sim}{f})(x)|<\epsilon$, do you have a reason for wanting the $L^2$ difference being bounded, or was that more an arbitrary choice for a measure of closeness? I ask mostly because you don't seem to require that $f$ itself be in $L^2$, so closeness in a Banach space norm like $L^2$ doesn't seem natural to me. $\endgroup$ Feb 19, 2014 at 5:25
  • $\begingroup$ There's no way this is true. You're asking that $\tilde{f}$ have the same level sets as $g$. Having $f = g$ on the boundary tells you almost nothing about the level sets of $f$, so you're basically asking for $L^2$-small perturbations of $f$ which have any prescribed level sets. $\endgroup$
    – Nik Weaver
    Feb 19, 2014 at 5:46
  • $\begingroup$ I encounter this problem when I try to estimate the $L^2$ inner product of $f$ and a function of $g$. I work with a compact domain $\Omega$. If the statement holds true with $|f-\tilde{f}|_{C_0(\Omega)}<\varepsilon$, then we have $\|f-\tilde{f}\|_{L^2(\Omega)} <(\varepsilon|\Omega)|)^{1/2}$. Note that this is not a local problem because one needa to perturb $f$ to separate points with avoiding to add new inseparable points. $\endgroup$
    – Lingyun
    Feb 19, 2014 at 5:48
  • 1
    $\begingroup$ @NikWeaver, a nitpick: aren't we just asking that each level set of $\tilde{f}$ is contained in a level set of $g$? $\endgroup$
    – Vectornaut
    Feb 19, 2014 at 5:52
  • $\begingroup$ @NikWeaver I don't see why this is not true. Could you please provide a counter example? I only know, if $f$ is a constant, then $f + t g$ gives the desired perturbation with $t$ small enough. $\endgroup$
    – Lingyun
    Feb 19, 2014 at 6:09

1 Answer 1

3
$\begingroup$

Okay, here's a counterexample. Let $\Omega \subset {\bf R}^3$ be the ball of radius 2 about the origin and observe that $\Omega$ contains the unit cube $[0,1]^3$. Define $f(x,y,z) = x$ and $g(x,y,z) = y$ on the unit cube and extend them to smooth functions on $\Omega$ which agree on $\partial \Omega$.

Now let $\tilde{f}$ be any smooth function on $\Omega$ which satisfies $g(x,y,z) \neq g(x',y',z') \Rightarrow \tilde{f}(x,y,z) \neq \tilde{f}(x',y',z')$. Then on the unit cube we have $y \neq y' \Rightarrow \tilde{f}(x,y,z) \neq \tilde{f}(x',y',z')$. This implies that $\tilde{f}$ must be constant, within the unit cube, on every plane of the form $y = c$. If $\tilde{f}$ were not constant on any such plane then for every $c'$ sufficiently close to $c$ there would be points on the $y = c$ and $y = c'$ planes at which $\tilde{f}$ took the same value, contradicting the condition given above.

We can put a lower bound on $\|f - \tilde{f}\|_2$ by integrating over the unit cube and recalling that on each constant $y$ plane we have $\tilde{f}(x,c,z)$ constant and $f(x,c,z) = x$. An easy calculation shows that the smallest possible value of $\int_{[0,1]^3} |\tilde{f}(x,y,z) - f(x,y,z)|^2$ is $\frac{1}{12}$. So $\|\tilde{f} - f\|_2 \geq \frac{1}{2\sqrt{3}}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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