An unbounded function with integrability for all p>0

Question

How to construct a function $f\geq0$ such that i) $f\in L^p([0,1])\forall p>0$ and ii) $esssup_I f=+\infty$ for any interval $I\subset[0,1]$ ?

Thank you!

Answer 1

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 $$ f(x)=\sum_{n=1}^\infty2^{-n}f_n(x)\ . $$ Clearly $f$ is positive. Moreover $$ \int_0^1f(x)\,dx=\sum_{n=1}^\infty2^{-n}\int_0^1f_n(x)\,dx\le2 \sum_{n=1}^\infty2^{-n}<\infty, $$ so that $f$ is finite almost everywhere and in $L^1$. We have to check that $f\in L^p$ for $p>1$. Since $$ \|f\|_p\le\sum_{n=1}^\infty2^{-n}\|f_n\|_p, $$ it is enough to show that $\|f_n\|_p\le C_p$ 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 $$ \bigl|\,\log |x-a|^p\bigr|\le\frac{C_p}{\sqrt{|x-a|}},\quad0\le x\le1,\quad0\le a\le1, $$ and the claim follows.

Answer 2

Let $$\mathbb{Q}\cap [0,1]=\left\{x_n\right\}_{n\in \mathbb{N}} $$ and $$f_n(x)=-\log|x-x_n| $$ so that $f_n\geq0$, $f_n\in L^p(0,1)$ for all $p\in (0,\infty)$, and $$\|f_n\|_p^p=\int_0^1|f_n|^p=\int_{0}^{1}|\log|x-x_n||^pdx \leq\int_{-1}^{1}|\log |x||^pdx=:C<\infty$$ Let $$f=\sum_{n\in \mathbb{N}}\frac{1}{2^n}f_n $$ then by Minkowski's inequality and Minkowski's inequality for $p\in (0,1)$, we have $$\|f\|_p\leq \max\left\{2^{\frac{1}{p}-1};1\right\}\sum_{n\in \mathbb{N}}\frac{1}{2^n}\|f_n\|_p\leq \max \left\{2^{\frac{1}{p}-1};1\right\}C^{1/p}\sum_{n\in \mathbb{N}}\frac{1}{2^n}<\infty $$ Hence $f\in L^p$ for all $p\in (0,\infty)$, and for any interval $I\subset [0,1]$, there is $n\in \mathbb{N}$ such that $x_n\in I$, and so $$\operatorname{ess\,sup}f\geq \operatorname{ess\,sup}f_n=+\infty$$