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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 $L_{1+\varepsilon}$.

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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.

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  • $\begingroup$ Only one question. What is $M_t^0$? $\endgroup$ – Sergey Kopylov Apr 17 at 19:45
  • $\begingroup$ @SergeyKopylov The interior of $M_t$. You need this property to show $\lim_{x\to x_0}g=+\infty$, which I omit here. $\endgroup$ – XIAODA QU Apr 17 at 19:56
  • $\begingroup$ 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 $\endgroup$ – Sergey Kopylov Apr 17 at 20:10
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    $\begingroup$ @SergeyKopylov just assume $f(x_0)=+\infty$. This will simplify the proof. it doesn' matter actually. $\endgroup$ – XIAODA QU Apr 17 at 20:14
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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 $x_0$

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