# Existence of Lebesgue integrable function with extreme growth

Assume $$f(x) \in L_1([a,b])$$ and $$x_0\in[a,b]$$ is a point such that $$f(x)\xrightarrow[x\to x_0]\ +\infty$$.

Is there always exists a function $$g(x) \in L_1([a,b])$$ such that $$f(x)=o(g(x))$$ where $$x\to x_0$$?

In particular case, when $$f(x) \in L_{1+\varepsilon}([a,b])$$ for every $$\varepsilon>0$$ suitable function $$g(x)$$ exists. So, if counterexample exists, it's in $$L_1$$ and can't be in $$L_{1+\varepsilon}$$.

## 2 Answers

Yes there is .

WLOG $$f\geq 0$$. Let $$M_t=\{f\geq t\}$$ and $$\lambda(t)= \int_{M_t}fdx$$. Then

1) $$\lambda(t)>0$$.

2) $$\lambda(t)\downarrow 0$$ as $$t\to\infty$$.

3) $$x_0\in M_t^o$$, $$\forall t\in\mathbb{R}_+$$.

Choose $$\{t_n\}_n$$ by induction, which satisfies:

1) $$t_{n+1}>t_n$$.

2) $$\lambda(t_n)\leq 3^{-n}$$.

Define $$g:=+\infty \cdot I_{\{f=+\infty\}}+\sum_{n=1}^\infty 2^nI_{\{M_{t_n}-M_{t_{n+1}}\}}$$. Then

$$\int fgdx= \sum_n 2^n\int_{M_{t_n}-M_{t_{n+1}}} fdx\leq\sum_n 2^n3^{-n}\leq 3.$$

$$fg$$ is what we want.

• Only one question. What is $M_t^0$? – Sergey Kopylov Apr 17 at 19:45
• @SergeyKopylov The interior of $M_t$. You need this property to show $\lim_{x\to x_0}g=+\infty$, which I omit here. – XIAODA QU Apr 17 at 19:56
• it seems to me that $x_0$ doesn't ought to belong to $M_t$, but for every deleted neighbourhood $V'$ of $x_0$ exists $t$ such that $V'\in M_t$. Because we can assign $f$ in $x_0$ an arbitrary value – Sergey Kopylov Apr 17 at 20:10
• @SergeyKopylov just assume $f(x_0)=+\infty$. This will simplify the proof. it doesn' matter actually. – XIAODA QU Apr 17 at 20:14

Hmmm... We can use the open mappimg theorem to conclude there must be an unbounded function h such that $$h \cdot f \in L^1$$ but I don't know how to guarantee the unbounded part is near $$x_0$$