4
$\begingroup$

How can I find a function $f$ such that $f \notin L^{p} (\mathbb{R})$ for all $p$ but you can find a constant $c>0$ for it with

$m(x \in \mathbb{R} \, s.t. |f(x)|>t) \leq \frac{c}{t}$ for $\forall t$

I tried $f(x)={1 \over x^2}$ since $f \notin L^{p} (\mathbb{R})$ for all $p$, but it looks like this is not the right answer. I guess we could use the Chevyshev's inequality on $f(x)$ , the measure theoretic one (see Section Measure-theoretic statement in https://en.wikipedia.org/wiki/Chebyshev%27s_inequality)

$\endgroup$
7
  • 2
    $\begingroup$ What is $p$? ${}{}{}$ $\endgroup$
    – copper.hat
    Oct 16, 2015 at 15:54
  • 1
    $\begingroup$ Do you mean $| f(x) | > t$ instead of $|f (x) > t|$? $\endgroup$
    – Hetebrij
    Oct 16, 2015 at 15:56
  • $\begingroup$ Perhaps you can tell us what you tried for $f.$ $\endgroup$
    – zhw.
    Oct 16, 2015 at 16:05
  • $\begingroup$ @copper.hat for all $p$ $\endgroup$
    – Susan_Math123
    Oct 16, 2015 at 16:19
  • 1
    $\begingroup$ You didn't try $f(x) = 1/x$? $\endgroup$
    – zhw.
    Oct 16, 2015 at 16:22

1 Answer 1

Reset to default
4
$\begingroup$

Let $f(x) = 1/x, x \ne 0.$ Then $f\not \in L^p(\mathbb {R}), 0 < p \le \infty,$ but $m(\{x: |f(x)| > t\}) = 2/t$ for all $t>0.$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .