2
$\begingroup$

While trying to understand a proof, i have stumbled upon the following statement:

Let $f \in L^p(a,b)$ be a $p$-integrable function. Then the inequality

$$\liminf_{s \rightarrow t} \frac{1}{t-s} \int_s^t f(r) \mathbb{\lambda}(r) \leq f(t)$$

is true $L^p$ almost everywhere. Where $\mathbb{\lambda}$ denotes the standard Lebesgue-measure. Although the statement obviously holds for continuous functions, I have difficulties to verify it for $L^p$ ones.

I tried the following approach: Verify the statement for characteristic functions i.e. $ 1_A(x)=1$ if $x\in A$ and $0$ else. Then verify the statement for linear combinations of characteristic functions. Now proof the statement for functions $f= \lim f_n$ where $f_n$ is a linear combination of characteristic functions (Here i got stuck).

$\liminf_{s \rightarrow t} \frac{\int_s^t f(r) \mathbb{\lambda}(r)}{t-s} = \liminf_{s \rightarrow t} \lim_{n \rightarrow \infty} \frac{\int_s^t f_n(r) \mathbb{\lambda}(r)}{t-s} \leq^{?} \lim_{n \rightarrow \infty} \liminf_{s \rightarrow t} \frac{\int_s^t f_n(r) \mathbb{\lambda}(r)}{t-s} \leq \lim_{n \rightarrow \infty} f_n(t)=f(t) $

Does anybody have an idea how to justify the $lim$ changing positions, or any other suggestions on how to prove the statement?

Thanks for your help!

$\endgroup$
  • 1
    $\begingroup$ What does "$L^p$ almost everywhere" mean? Also, why not just use the Lebesgue differentiation theorem? $\endgroup$ – user147263 Jul 27 '14 at 21:51
  • $\begingroup$ The theorem does solve the problem (proof can be adapted with weak $L^p$ estimates to the $L^p$ case). I did not know the Theorem. Thank you! $\endgroup$ – asterisk Jul 28 '14 at 7:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.