$\left\{\,f\in L^1[0,1]\,\big\vert\,\int_0^1\lvert f\rvert^2>1\right\}$ is open Problem
(Here $\lVert f\rVert$ means the $L^1$-norm $\int_0^1\lvert f\rvert$)
Suppose $f\in L^1[0,1]$ such that $\int_0^1\lvert f\rvert^2>1$. I need to work out an explicit $\epsilon>0$ such that such that $\int_0^1\lvert g\rvert^2>1$ for each $g\in L^1[0,1]$ whenever $\lVert g-f\rVert\le\epsilon$.
Comment
One can find $\epsilon$ non-constructively as follows: for otherwise we can choose $\{g_n\}\subseteq L^1[0,1]$ such that $\lVert g_n-f\rVert\to0$ and $\int_0^1\lvert g_n\rvert^2\le1$, then $\lvert g_n\rvert^2\to\lvert f\rvert^2$ in measure, and by Fatou's lemma, $\int_0^1\lvert f\rvert^2\ge1$, contradiction!
However, I want to explicate an $\epsilon>0$, i.e. the perturbation radius, for each $f$. To start with, let's consider $f=c>1$ and fix $\epsilon>0$. We try to minimize $\int_0^1\lvert g\rvert^2$ under the assumption that $\lVert g-f\rVert=\eta\le\epsilon$. Let $h=f-g$, and it's natural to assume that $h>0$, since $\lvert g\rvert\ge\big\lvert\lvert f\rvert-\lvert h\rvert\big\rvert$. Then
$$I=\int_0^1\lvert g\rvert^2=c^2-2c\eta+\int_0^1\lvert h\rvert^2\ge c^2-2c\eta+\left(\int_0^1\lvert h\rvert\right)^2=(c-\eta)^2$$
If $\epsilon$ is sufficiently small, then $I_\mathrm{min}=(c-\epsilon)^2$ when taking $h=\epsilon$.
If $f$ is arbitrary, maybe we can approximate $f$ by simple functions then obtain some good estimations. I have no clear idea on how to proceed next, and I hope we can obtain a good, if not best, $\epsilon$ for each $f$.
Any idea? Thanks!
Caveat
One should note that the result is not that trivial, by seeing that on contrary, $\left\{\,f\in L^1[0,1]\,\big\vert\,\int_0^1\lvert f\rvert^2<1\,\right\}$ is however nowhere dense, and therefore $L^2[0,1]$ is meager in $L^1[0,1]$. It seems astonishing, but it should be clear after you recall the following facts:


*

*$h_\epsilon(x)=\epsilon/\sqrt x$ is small in $L^1$-norm but not square-integrable;

*The collection of continuous functions $C^0[0,1]$ with $L^1$-norm can be embedded densely in $L^1[0,1]$;

*For each $f\in C^0[0,1]$, we have $\int_0^1\lvert f+h_\epsilon\rvert^2=+\infty$.

 A: Answer: $\epsilon = \int_0^1 (|f|-M)^+$ where $M>0$ is a constant such that $\int_0^1 \min(|f|,M)^2=1$. Here $a^+=\max(a,0)$ and $\epsilon$ is uniquely determined although $M$ might not be. 
It's not difficult to reduce the problem to the case $f\ge 0$ (since we are essentially working with $|f|$), so I am going to assume $f\ge 0$. We can recast the question as: 

Minimize the functional $\phi(g)=\|f-g\|_1$ on the closed unit ball of $L^2$. 

The functional $\phi$ is bounded from below, norm-continuous, and convex. Therefore it is weakly lower semicontinuous on $L^2$, which implies it attains its infimum the closed unit ball. (You can argue this point directly, without invoking all this stuff from the calculus of variations: take an infimizing sequence, choose a weakly convergent subsequence, apply Mazur's lemma to get a strongly convergent sequence than is also infimizing). 
Let $g$ be a minimizer. Note that $g\ge 0$ (otherwise $|g|$ beats $g$) and $g\le f$ (otherwise $\min(f,g)$ beats $g$). So, $\|f-g\|=\int_0^1 f - \int_0^1 g$. In other words, $g$  maximizes $\int_0^1 g$ subject to two conditions $\|g\|_2\le 1$ and $g\le f$. 
Now it becomes apparent that $g$ must be essentially constant  on the set $\{x:g(x)<f(x)\}$. Indeed, otherwise there is $\epsilon>0$ such that $g$ is not essentially constant on the set $E=\{x:g(x)<f(x)-\epsilon\}$. Then we can find two constants $\alpha<\beta$ and two subsets  $A,B$ of $E$  of equal positive measure such that $g\le \alpha$ on $A$ and $g\ge \beta$ on  $B$. Consider the perturbation
$$
g_t = g+  t(\chi_A -   \chi_B),\quad t>0
$$
which for small $t$ preserves the condition $g\le f$, preserves $\int_0^1 g$, and strictly decreases $\|g\|_2$. Adding another small multiple of $\chi_A$ creates a function $h$ with $\int_0^1 h>\int_0^1 g$.
Since the minimizing function $g$ is equal to some constant $M$ a.e. on the set $\{f<g\}$, it can be written as $g=\min(f,M)$. The conclusion follows.
