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Feb 19, 2014 at 17:46 vote accept Lingyun
Feb 19, 2014 at 17:02 answer added Nik Weaver timeline score: 3
Feb 19, 2014 at 6:09 comment added Lingyun @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.
Feb 19, 2014 at 5:52 comment added Vectornaut @NikWeaver, a nitpick: aren't we just asking that each level set of $\tilde{f}$ is contained in a level set of $g$?
Feb 19, 2014 at 5:48 comment added Lingyun 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.
Feb 19, 2014 at 5:46 comment added Nik Weaver 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.
Feb 19, 2014 at 5:25 comment added Adam Hughes 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.
Feb 19, 2014 at 4:25 history asked Lingyun CC BY-SA 3.0