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The question is

Let $f \in {L^\infty }[0,1]$ and assume that $f$ is not identically zero. Show that $\mathop {\lim }\limits_{p \to \infty } \frac{{\int_0^1 {|f{|^{p + 1}}dx} }}{{\int_0^1 {|f{|^p}dx} }}$ exists and compute the limit.

I can see $\mathop {\lim }\limits_{p \to \infty } {(\frac{{\int_0^1 {|f{|^{p + 1}}dx} }}{{\int_0^1 {|f{|^p}dx} }})^{\frac{1}{{p + 1}}}} = \mathop {\lim }\limits_{p \to \infty } \frac{{{{\left\| f \right\|}_{p + 1}}}}{{{{({{\left\| f \right\|}_p})}^{\frac{p}{{p + 1}}}}}} = \frac{{{{\left\| f \right\|}_\infty }}}{{{{({{\left\| f \right\|}_\infty })}^1}}} = 1$,

So $\mathop {\lim }\limits_{p \to \infty } \frac{{\int_0^1 {|f{|^{p + 1}}dx} }}{{\int_0^1 {|f{|^p}dx} }} = \mathop {\lim }\limits_{p \to \infty } {[{(1 + \frac{{\int_0^1 {|f{|^{p + 1}}dx} - \int_0^1 {|f{|^p}dx} }}{{\int_0^1 {|f{|^p}dx} }})^{\frac{1}{{p + 1}}}}]^{p + 1}} = {e^{\mathop {\lim }\limits_{p \to \infty } \frac{{\int_0^1 {|f{|^{p + 1}}dx} - \int_0^1 {|f{|^p}dx} }}{{\int_0^1 {|f{|^p}dx} }} \times (p + 1)}}$.

Here I got stuck, because ${\mathop {\lim }\limits_{p \to \infty } \frac{{\int_0^1 {|f{|^{p + 1}}dx} - \int_0^1 {|f{|^p}dx} }}{{\int_0^1 {|f{|^p}dx} }} \times (p + 1)}$ is of type "$0\times\infty$".

Thank you!

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  • $\begingroup$ The title is misleading. If $f$ is constantly $2$ on $[0,1]$, the limit in the title cannot be one. $\endgroup$ – Jack D'Aurizio Aug 20 '15 at 0:40

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