An example of a function that is $L^p(X)$ but not $L^\infty(X)$ is $\log x^{-1}$, where $x\in X=[0,1]$ and $p\in[1,\infty)$. How do i show this?
A function that is $L^p$ but not $L^{\infty}$
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real-analysis
measure-theory
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0Thank you Davide, but i think you forgot to put the negative during the substitution so as to even make use of $-|f|\le-f$. Otherwise, great staff! – 2012-03-30
1 Answers
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First, $f$ is not essentially bounded, since if you fix $A$, then $f(x)\geq A$ if $x\leq e^{-A}$ (since $x\leq e^{-A}$ iff $1/x\geq e^A$ iff $f(x)\geq A$).
Now, we show that $f\in L^p$ for all $p\geq 1$ finite. We have for $\varepsilon>0$ fixed that $\int_{[\varepsilon,1]}|f(x)|^pdx=\int_{[1,1/\varepsilon]}|\ln t|^p\frac 1{t^2}dt,$ using the substitution $x=\frac 1t$. Now, you have to show that the integral $\int_1^{+\infty}\frac{(\ln t)^p}{t^2}dt$ is convergent, for example using the fact that $\lim_{t\to+\infty}\frac{(\ln t)^p}{\sqrt t}=0$.
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0Thank you Davide, but i think you forgot to put the negative during the substitution so as to even make use of $-|f|\le-f$. Otherwise, great staff! – 2012-03-30