Divergence for original function implies large value eventually for some simple function approximation

Question

Let $g\in L^1[0,1]$ with $\|g\|_{L^1[0,1]}\leq1$. We are given a sequence of continuous, linear functionals $f_n$ such that $f_n(g)\to\infty$ as $n\to\infty$.

Now given some $L>0$, I'm wondering whether it is possible to construct a simple function $g_L$ such that $\liminf_{n\to\infty}f_n(g_L)\geq L$?

For the particular application I have in mind, it actually suffices to find some normalized indicator function or simple function $g^*$ with $\|g^*\|_{L^1}\leq1$ such that $\liminf_{n\to\infty}f_n(g^*)>0$. So the idea is, you give me a sequence of continuous linear functionals $f_n:L^1[0,1]\to\mathbb R$ and a $g\in L^1[0,1]$ such that $\lim_{n\to\infty}f_n(g)=\infty$; and I have to find a simple function $g^*$ such that $\liminf_{n\to\infty}f_n(g^*)>0$.

This seems kind of obvious but I'm still in doubt whether something like this holds in general: for example, what happens if the divergent limiting behaviour is in some sense completely caused by an infinite spike (of finite $L^1$ size) of $g$? Is such a behaviour possible?

Any help is much appreciated. If necessary, I'm willing to make additional assumptions on $g$ or on the approximating simple function.

Answer 1

Answer for the earlier version of the question:

There are easy counter-examples even with constant functions $g_n$ and $g$. Let $g=1$ and $g_m=1-\frac 1m$. Let $f_n(x)=0$ for $0 \leq x \leq 1-\frac 1 n$ and $f_n(x)=n^{2}(x-1+\frac 1 n)$ for $1-\frac 1 n <x\leq 1$. Note that for any $m$, $f_n(g_m)=0$ for $n$ sufficiently large.

Answer 2

It turns out that my conjecture is wrong.

First of all, the existence of a simple function $g^*$ satisfying the desired property implies the existence of an indicator function $\mathbf1_A$ having the same property, for some $A\in\mathcal B[0,1]$ of positive measure.

Since $L^1[0,1]^*$, the dual of $L^1[0,1]$, is isomorphic to $L^\infty[0,1]$, the problem is equivalent to the following: we are given a sequence $(\phi_n)_{n\in \mathbb N}\subset L^\infty[0,1]$ and $g\in L^1[0,1]$ such that $\int_0^1\phi_n(x)g(x)\ \mathrm dx\to\infty$ as $n\to\infty$, and we want to find an indicator function $\mathbf1_A$ such that $\liminf_{n\to\infty}\int_0^1\phi_n(x)\mathbf1_A(x)\ \mathrm dx=\liminf_{n\to\infty}\int_A\phi_n(x)\ \mathrm dx>0$.

Now note that $g(x)=\frac1{2\sqrt x}$ and $\phi_n=n^{3/4}\mathbf1_{[0,\frac1n]}$ satisfy the given conditions. Write $\lambda$ for the Lebesgue measure on $[0,1]$. Then for any $A\in \mathcal B[0,1]$ it holds that $\int_A\phi_n(x)\ \mathrm dx=n^{3/4}\lambda([0,\frac1n]\cap A)\leq n^{-1/4}\to0$ as $n\to\infty$. Hence we cannot find an indicator function or a simple function $g^*$ such that $\liminf_{n\to\infty}\int_0^1\phi_ng^*>0$.

To conclude, we really need to assume more on $f_n$ to draw such conclusions. E.g. if $f_n$ is the $L^1$ norm of iterates of a compact, self-adjoint operator, i.e. there exists a compact, self-adjoint operator $K:L^1[0,1]\to L^1[0,1]$ such that $f_n(g)=\int_{[0,1]}K^ng$, then by the Krein-Rutman theorem such conclusions are possible.