# 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 $\int_0^1\lvert f+h_\epsilon\rvert^2=+\infty$.