Vtyjkyuk

How to construct a function $fgeq0$ such that i) $fin L^p([0,1])forall p>0$ and ii) $esssup_I f=+infty$ for any interval $Isubset[0,1]$ ?

Thank you!

How about
$$
f(x)=|log x|?
$$

The above is a wrong answer, since it does not fulfill 3. So let's try again.

If $0<a<1$, then
$$
int_0^1|log |x-a||,dx=1 - (1 - a) log(1 - a)- a log a<2.
$$
Order the rationals in $[0,1]$ in a sequence $a_n$, let $f_n(x)=bigl|,log |x-a_n|,bigr|$ and
$$
f(x)=sum_n=1^infty2^-nf_n(x) .
$$
Clearly $f$ is positive. Moreover
$$
int_0^1f(x),dx=sum_n=1^infty2^-nint_0^1f_n(x),dxle2 sum_n=1^infty2^-n<infty,
$$
so that $f$ is finite almost everywhere and in $L^1$. We have to check that $fin L^p$ for $p>1$. Since
$$
|f|_plesum_n=1^infty2^-n|f_n|_p,
$$
it is enough to show that $|f_n|_ple C_p$ for some constant $C_p$ depending only on $p$. The function $|x-a|^1/(2p)bigl|,log |x-a|bigr|$ is continuous, and in particular bounded, as a function of $(x,a)$ on $[0,1]times[0,1]$. It follows that there exists a constant $C_p$ such that
$$
bigl|,log |x-a|^pbigr|lefracC_psqrt,quad0le xle1,quad0le ale1,
$$
and the claim follows.

Let
$$mathbbQcap [0,1]=leftx_nright_nin mathbbN $$
and
$$f_n(x)=-log|x-x_n| $$
so that $f_ngeq0$, $f_nin L^p(0,1)$ for all $pin (0,infty)$, and
$$|f_n|_p^p=int_0^1|f_n|^p=int_0^1|log|x-x_n||^pdx leqint_-1^1|log |x||^pdx=:C<infty$$
Let
$$f=sum_nin mathbbNfrac12^nf_n $$
then by Minkowski's inequality and Minkowski's inequality for $pin (0,1)$, we have
$$|f|_pleq maxleft2^frac1p-1;1rightsum_nin mathbbNfrac12^n|f_n|_pleq max left2^frac1p-1;1rightC^1/psum_nin mathbbNfrac12^n<infty $$
Hence $fin L^p$ for all $pin (0,infty)$, and for any interval $Isubset [0,1]$, there is $nin mathbbN$ such that $x_nin I$, and so $$operatornameess,supfgeq operatornameess,supf_n=+infty$$