2

Suppose $m(E)<\infty$ and $f\in\mathcal{L}^{\infty}(E)$. The goal of this problem is to show \begin{align*} \lim_{p\to \infty}\|f\|_p=\|f\|_{\infty}. \end{align*} First, prove that \begin{align*} \lim_{p\to\infty}\|f\|_p\leq \|f\|_{\infty}. \end{align*}Proof: Assume $m(E)<\infty$ and $f\in\mathcal{L}^{\infty}(E)$. By definition, \begin{align*} \|f\|_p=\left(\int_E|f|^pdm\right)^{1/p}\leq \left(\int_E\|f\|_{\infty}^p\right)^{1/p}=\|f\|_{\infty}\cdot m(E)^{1/p}<\infty. \end{align*} Letting $p\to \infty$, we have that $m(E)^{1/p}\to 1$. Therefore, \begin{align*} \lim_{p\to\infty}\|f\|_p\leq \|f\|_{\infty}. \end{align*}

I'm not sure if I am missing anything here or need to justify being able to bring the limit inside and applying to $m(E)$.

Next, I need to prove that \begin{align*} \lim_{p\to\infty}\|f\|_p\geq \|f\|_{\infty}-\epsilon \end{align*} for any $\epsilon>0$. Hint: Look at the set \begin{align*} F=\{x\in E:|f|>\|f\|_{\infty}-\epsilon\}. \end{align*}

I'm stuck on this part and would appreciate any help, thanks.

user
  • 173

1 Answers1

1

You have that

$||f||_p^p=\int_F|f|^p+\int_{F^c}|f|^p>\int_F|f|^p >$

$>\int_F (||f||_\infty -\epsilon)^p=(||f||_\infty -\epsilon)^pm(F) $

Then

$||f||_p> (||f||_\infty -\epsilon)(m(F))^\frac{1}{p}$

So if you fixed $\epsilon$ (and $ F=F_{\epsilon}$ ) you have that for $p\to \infty$

$lim_p||f||_p>||f||_\infty -\epsilon$

for every $\epsilon>0$ so

$lim_p ||f||_p>||f||_\infty$

Federico Fallucca
  • 8,593
  • 1
  • 9
  • 20