Existence of Lebesgue integrable function with extreme growth

Question

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 $L_{1+\varepsilon}$.

Answer 1

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.

Answer 2

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 $x_0$