0

I have recently been trying to prove the following using only knowledge from chapters 1 through 6 of Rudin's "Principles of Mathematical Analysis". In other words, I cannot use the concept of measure.

$\lim_{p\rightarrow \infty}(\int_ {0}^{1}|f(x)|^p)^{1/p}=\max_{x\in[0,1]}|f(x)|$ if $f:[0,1]\rightarrow \mathbb{R}$ is continuous on its domain

where $| \cdot|$ is the absolute value function. However, I am unsure where to start and most proofs I have found online rely on concepts not covered in the mentioned chapters.

My attempt involved writing $\lim_{p\rightarrow \infty}(\int_ {0}^{1}|f(x)|^p)^{1/p}$ in the form $\lim_{p\rightarrow \infty}(\lim_{n \rightarrow \infty}(\frac{1}{n}\sum_{i=1}^{n}|f(\frac{i}{n})|^p))$. From there, I tried to use properties of Darboux lower and upper sums and integrals to reach an $\epsilon-\delta $ limit conclusion but I was not able to make much progress.

Could anyone point me in the right direction? Any hints, tips, or proof outlines would be greatly appreciated.

  • 1
    Hint: Prove two direction of inequalities: one direction is easy, and for the other direction, choose small interval s.t $|f(x)|\ge{\rm max}|f(x)|-\varepsilon$. – Kangyeon Moon Mar 10 '23 at 06:15
  • 1
    You don't need the “concept of measure”, only that $f(z) > M - \epsilon$ on some interval of positive length, which follows from the continuity. Perhaps this one is clearer? https://math.stackexchange.com/q/1766241/42969 – Martin R Mar 10 '23 at 06:26
  • See also https://math.stackexchange.com/q/3059678/42969 and the linked questions – all found with Approach0 – Martin R Mar 10 '23 at 06:30

0 Answers0