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Let $f$ be a bounded measurable function on $[0,1]$ with lebesgue measure $m$, then $\lim_{p \rightarrow \infty}\|f\|_{p} = \|f\|_{\infty}$.

There are several solutions to more general versions of this on SE already, but I'm wondering if the assumptions that $f$ is bounded and measurable on $[0,1]$ admits a simpler solution?

Here is one such solution: Limit of $L^p$ norm

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  • $\begingroup$ Adding a link to the more general versions, and one or two additional tags, would improve your question. $\endgroup$ – 6005 Aug 9 '16 at 20:14
  • $\begingroup$ A tricky solution may be to use a sort of radial rearrangement: there is a bounded and weakly decreasing function $g$ on $[0,1]$ such that for any $p$ we have $\|f\|_p=\|g\|_p$, but the claim is kind of trivial for bounded and weakly decreasing functions. $\endgroup$ – Jack D'Aurizio Aug 9 '16 at 20:33
  • $\begingroup$ ... especially since we may assume $g(0)=1$ without loss of generality. $\endgroup$ – Jack D'Aurizio Aug 9 '16 at 21:09
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The assumption that $f$ is bounded and measurable on a set $E$ with $m(E) = 1$ does simplify things somewhat. In this case, $\|f\|_p$ is monotonically increasing as a function of $p$.

To see this, suppose that $1 < p < q < \infty$. Let $r$ be the number such that $1/p = 1/q + 1/r$. Then $1 < r < \infty$, and by this generalization of Hölder's inequality we have $$\|f\|_p = \|f \chi_E\|_p \leq \|f\|_q \|\chi_E\|_r = \|f\|_q m(E)^{1/r} = \|f\|_q$$ Moreover, $$\|f\|_p^p = \int_E |f|^p \leq \|f\|_{\infty}^p m(E) = \|f\|_{\infty}^p$$ so $$\|f\|_p \leq \|f\|_{\infty}$$ As $p \mapsto \|f\|_p$ is increasing and bounded, the limit $\lim_{p \to \infty}\|f\|_p$ exists and does not exceed $\|f\|_{\infty}$. It remains to show that the limit is in fact $\|f\|_{\infty}$. If $\|f\|_{\infty} = 0$, there is nothing to prove. Otherwise, fix any $\epsilon$ satisfying $0 < \epsilon < \|f\|_{\infty}.$ Then $|f(x)| > \|f\|_{\infty} - \epsilon > 0$ on a set $M$ of positive measure. Consequently, $$\int_E |f|^p \geq \int_M |f|^p \geq (\|f\|_{\infty} - \epsilon)^p m(M)$$ and therefore $$\|f\|_p \geq (\|f\|_{\infty} - \epsilon)m(M)^{1/p}$$ Now $m(M) \leq m(E) = 1$, so $m(M)^{1/p} \uparrow 1$ as $p \to \infty$, and therefore we have $$\lim_{p \to \infty}\|f\|_p \geq \|f\|_{\infty} - \epsilon$$ As this holds for arbitrarily small positive $\epsilon$, the result follows.

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  • $\begingroup$ By the way, essentially the same argument works if we assume $m(E) \leq 1$ instead of $m(E) = 1$. On the other hand, if $m(E) > 1$, then $p \mapsto \|f\|_p$ is no longer necessarily monotonic, so the argument is somewhat more complicated (but the result is still true). $\endgroup$ – Bungo Aug 9 '16 at 20:45

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