How bad an $L_1$ function can get in terms of having $\operatorname{esssup}$ as $\infty$ at many places?

Question

Let $f:[0,1]\to \mathbb{R}$ be a $L_1$ function (i.e., $\int_0^1 |f(x)|dx<\infty$) with $\operatorname{esssup} f=\infty$. Think of functions like $\frac1{\sqrt{x}}$ or $\frac{1}{\sqrt{x(1-x)}}$ for example (If you want to be pedantic, you can define $f(x)=0$ to be zero at the points where the functions blows up). For the first function we have singularity at $x=0$ and for the second one we have singularities at $x=0,1$.

In general I can always cook up some function $f$ that has singularity in a given set of finite points and still be in $L_1$. Something like $$f(x)=\frac{1}{\sqrt{\prod_k |x-a_k|}}$$ should work.

My question is do there exist a function $f\in L_1([0,1])$ such that this essential supremum is attained on a dense set? More precisely the converse is: For $f\in L_1([0,1])$ do there exist an open interval $I$ of $[0,1]$ such that $f_{I}$ has a finite essential supremum?

My guess is that it is not true. Given a function $f\in L_1$, I believe that the "points" where essential supremum is attained are isolated. But I can neither prove or disprove it.

Answer 1

I don't know if this completely answers your question but if $(r_n)$ is an ennumeration of rationals in $[0,1]$ then $f(x)=\sum_n \frac 1 {2^{n}} \frac 1 {\sqrt {|x-r_n|}}$ defines an integrable function which is (essentially) unbounded on every open interval.

Note that (by Tonelli's Theorem) $\int f(x)dx=2\sum \frac 1{2^{n}} (\sqrt {|1-r_n|} +\sqrt {r_n}) <\infty$ since $\sqrt {|1-r_n|} +\sqrt {r_n}$ is bounded. Hence $f(x)<\infty$ for almost all $x$.

Answer 2

Define

$$g(x)= \chi_{(0,1)}(x)\cdot x^{-1/2}.$$

Let $q_1,q_2,\dots $ be the rationals in $(0,1).$ Now define, for $x\in [0,1],$

$$f(x) =\sum_{n=1}^{\infty}\frac{g(x-q_n)}{2^n}.$$

Then $f:[0,1]\to [0,\infty]$ is measurable, and by the MCT

$$\int_0^1 f(x)\,dx = \sum_{n=1}^{\infty}\int_0^1\frac{g(x-r_n)}{2^n}\,dx = \sum_{n=1}^{\infty}\frac{1}{2^n} \int_{q_n}^1 (x-q_n)^{-1/2}\,dx.$$

The $n$th integral equals $2(1-q_n)^{1/2}\le 2.$ It follows that $\int_0^1 f<\infty.$ So $f\in L^1.$ But near and to the right of each $r_n$ we have $f(x) > (x-r_n)^{-1/2}/2^n.$ Thus the essential supremum of $f$ is $\infty$ on any subinterval of $[0,1]$ of positive length.