# A function that is $L^p$ but not $L^{\infty}$

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?

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possible duplicate of $L^p$-spaces in real analysis – Davide Giraudo Mar 30 '12 at 11:06
How do i show this? First thing, you look at the definition of $L^\infty$. Second, you prove your function is not $L^\infty$. Then we'll see... – Did Mar 30 '12 at 11:09
@Davide: I suggest that we close (and delete) the other one (which already has four close votes). – t.b. Mar 30 '12 at 11:23
@t.b. Sorry, in fact I belived that joe was about to delete the post. I voted to close the other one and deleted the corresponding answer. – Davide Giraudo Mar 30 '12 at 11:25
Thank 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! – jm sol Mar 30 '12 at 20:19

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 $\lim_{t\to+\infty}\frac{(\ln t)^p}{\sqrt t}=0$.
Thank 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! – jm sol Mar 30 '12 at 20:17