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In this question $f$ is a Lebesgue measurable function on $[0,1]$ with the property that $\|f\|_{2}=1,\|f\|_{1}=1/2$. I am trying to prove that $$ \frac{(1-\lambda)^{2}}{4}\leq m\left\{x\in [0,1]: |f(x)| \geq \frac{\lambda}{2} \right\}\>, $$ for all $0\leq \lambda \leq1$.

Here, $m$ denotes the Lebesgue measure on $[0,1]$. I think I should use the Chebyshev's inequality to prove this; also I need to split the interval in order to get use $\|f\|_{2}=1,\|f\|_{1}=1/2$, but so far I have no idea about what to do next. This is an old qual problem.

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Hint: With $b = 1/2$,

$$\newcommand{\rd}{\,\mathrm d}\newcommand{\I}[1]{\mathbf 1_{(#1)}} b = \int |f| \rd x = \int |f| \,\I{|f| \geq \lambda b} \rd x + \int |f| \,\I{|f| < \lambda b} \rd x \>. $$

Now use Cauchy–Schwarz and monotonicity in appropriate ways.

Follow-up exercise: Explain the more general result you've actually just proved replacing $1/2$ with $b \geq 0$ and Lebesgue measure with an arbitrary probability measure.

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  • $\begingroup$ Thanks a lot, I will try now. $\endgroup$ – user53800 Jan 2 '13 at 0:08
  • $\begingroup$ Can you give me some more hint? $\endgroup$ – user53800 Jan 2 '13 at 0:25
  • $\begingroup$ @user: $|f| \I{|f| < \lambda b} < \lambda b$. $\endgroup$ – cardinal Jan 2 '13 at 0:30
  • $\begingroup$ Thanks so much, I got it. It's an elegant proof.Is that right for the probability measure$P(f>b\lambda)\geq b^{2}(1-\lambda)^{2}$? $\endgroup$ – user53800 Jan 2 '13 at 2:44
  • $\begingroup$ @user: You surely meant $\mathbb P(|f| \geq b \lambda)$, but yes, that is correct, under the constraints $\mathbb E |f| \geq b$ and $\mathbb E |f|^2 = 1$. :) $\endgroup$ – cardinal Jan 2 '13 at 2:51

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