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?
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$\begingroup$ 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. $\endgroup$– Adam HughesCommented Feb 19, 2014 at 5:25
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$\begingroup$ 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 WeaverCommented Feb 19, 2014 at 5:46
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$\begingroup$ 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. $\endgroup$– LingyunCommented Feb 19, 2014 at 5:48
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1$\begingroup$ @NikWeaver, a nitpick: aren't we just asking that each level set of $\tilde{f}$ is contained in a level set of $g$? $\endgroup$– VectornautCommented Feb 19, 2014 at 5:52
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$\begingroup$ @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. $\endgroup$– LingyunCommented Feb 19, 2014 at 6:09
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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 $\|\tilde{f} - f\|_2 \geq \frac{1}{2\sqrt{3}}$.
answered Feb 19, 2014 at 17:02