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Assume we are given some constant $\alpha$ and a subset $\Omega\subset\mathbb{R}^n$ such that $\lambda(\Omega)<\infty$, where lambda denotes the Lebesgue measure. We consider the mapping $$ \Lambda:L^1(\Omega)\to\mathbb{R} $$ with $$ \Lambda(f)=\lambda(\{x\in\Omega: f(x)>\alpha\}). $$

I am trying to figure out a way to control the error sensitivity of this problem. If the function values are close to $\alpha$, then a small perturbation of the function could dramatically change the measure. Any suggestions as to what tools I could use would be greatly appreciated.

Thanks

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  • $\begingroup$ More precisely, you want to control the difference $$|\Lambda(f)-\Lambda(g)|, $$ right? $\endgroup$ – Giuseppe Negro Oct 29 '18 at 13:22
  • $\begingroup$ Yes, exactly :) $\endgroup$ – Nicolas Bourbaki Oct 29 '18 at 13:25
  • $\begingroup$ You could try to use Chebyshev's inequality en.wikipedia.org/wiki/… $\endgroup$ – daw Oct 29 '18 at 16:36
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The following standard trick might be useful. Let $$ \Lambda_\alpha(f):=|\{f>\alpha\}|.$$ Consider another function $g\in L^1$. If $f(x)+g(x)>\alpha$, then either $f(x)$ or $g(x)$ or both must be bigger than $\alpha/2$; $$ \{f+g>\alpha\}\subset \{f>\tfrac\alpha2\}\cup\{g>\tfrac\alpha2\}.$$ Therefore $$\tag{1} \Lambda_\alpha(f+g)\le \Lambda_{\alpha/2}(f)+\Lambda_{\alpha/2}(g).$$ Now, adding and subtracting $g$, we see that $$ \Lambda_{2\alpha}(f)\le \Lambda_{\alpha}(f-g) +\Lambda_\alpha(g), $$ which gives the bound $$ \tag{2} \Lambda_{2\alpha}(f)-\Lambda_\alpha (g)\le \Lambda_{\alpha}(f-g).$$

This is not very pretty, but maybe it helps. If $\alpha>0$, then you can combine it with Chebyshev's inequality, yielding $$ \max(\Lambda_{2\alpha}(f)-\Lambda_\alpha(g), \Lambda_{2\alpha}(g)-\Lambda_\alpha(f))\le \frac{1}{\alpha}\|f-g\|_1.$$

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  • $\begingroup$ Thank you my friend! $\endgroup$ – Nicolas Bourbaki Nov 8 '18 at 11:48

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