0
$\begingroup$

Let $(X,\Sigma,\mu)$ be a measure space. Let $f$ be a measurable function and $t > 0, t\in \mathbb{R}.$.

Denote: $$C_f(t) = \mu \{x \in \Omega : |f(x)| \geq t \}.$$

In the first part of this question (from a past paper), I showed that $$C_{f+g}(t) \leq C_f(t/2) + C_g(t/2).$$ I'm not sure if this relates to the next part or not.

(b) For any integers $n,m \geq 1$ prove that $$C_f(t) \leq \frac{1}{t^n + t^m} \int (|f|^n + |f|^m)d\mu.$$

In all honesty, I'm not even sure where to start with this. It's only four marks which suggests I'm missing something simple.

$\endgroup$

1 Answer 1

1
$\begingroup$

\begin{align} \int(|f|^n+|f|^m)\ \mathsf d\mu &\geqslant \int_{\{|f|\geqslant t\}}(|f|^n+|f|^m)\ \mathsf d\mu\\ &\geqslant \int_{\{|f|\geqslant t\}} (t^n + t^m)\ \mathsf d\mu\\ &=C_f(t)(t^n+t^m). \end{align}

$\endgroup$
2
  • $\begingroup$ Cheers. Any way you could share your thought process here? I stared at this for good amount of time without spotting it. $\endgroup$ May 2, 2016 at 11:23
  • 1
    $\begingroup$ $x\mapsto x^n$ is increasing so if $|f(x)|\geqslant t$ then $|f(x)|^n\geqslant t$. There's not a whole lot to explain really. $\endgroup$
    – Math1000
    May 2, 2016 at 11:27

You must log in to answer this question.

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