If $f$ is bounded away from $0$, are the functions that are close to $f$ in $L^2$ bounded away from zero as well?

Question

Consider a probability space $(\Omega,\mathcal F,\mathbb P)$, and let $X$ be a $\mathbb R^d$-valued random variable defined on that space with distribution $\rho$ (i.e. $\rho(A) = \mathbb P(X\in A)$ for all $A\in\mathcal B(\mathbb R^d)$). Furthermore, assume that $X\in \mathcal X$ with probability $1$, where $\mathcal X$ is a compact subset of $\mathbb R^d$.

Now let $f^*:\mathcal X\to\mathbb R$ be a real-valued and continuous function such that $$\mathbb P\left(|f^*(X)|>\delta\right) = 1 $$ For some constant $\delta > 0$. In words, this means that $f^*$ is such that $f^*(X)$ is almost surely a margin of $\delta$ away from zero.

My question is the following : assuming that I can find a continuous function $\hat f:\mathcal X\to\mathbb R$ such that $\|\hat f - f^*\|_{L^2} \le \varepsilon$ for a small $\varepsilon>0$ (here, $\|\cdot\|_{L^2}$ refers to the $L^2(\rho)$ norm), is it possible to get a lower bound of the form $$\mathbb P\left(|\hat f (X)|>\alpha(\delta,\varepsilon)\right) \ge 1-\beta(\delta,\varepsilon)$$ For some positive $\alpha$ and non-negative $\beta$ which will depend on $\delta$ and $\varepsilon$ ? The intuition is that, if $\hat f$ and $f^*$ are "very close", then with high-probability $\hat f$ should be bounded away from $0$, for some (perhaps smaller) margin $\alpha$. I do not know however how to formalize that intuition, and the fact that the approximation is in the $L^2$ sense rather than $L^\infty$ makes the problem even more challenging.

Alternatively, could there be some non-trivial assumptions I could make on $\rho$ to guarantee that $\mathbb P\left(|\hat f (X)|>\alpha(\delta,\varepsilon)\right) = 1 $ ?

Answer 1

Obviously we will need to have that $\varepsilon < \delta$ (otherwise if $f^* = \delta$ is constant then $\|f^*\|_{L^2} \le \varepsilon$).

Then we write for $\alpha$ small, $\mathbb{P}(|\hat{f}(X)| < \alpha) \le \mathbb{P}(|\hat{f}(X) - f^*(X)| > \delta - \alpha)$; where we use the fact that $|f^*(X)| > \delta$ almost surely.

By Markov's inequality, this yields the bound $$\mathbb{P}(|\hat{f}(X)| < \alpha) \le \frac{\|\hat{f} - f^*\|_{L^2}^2}{(\delta - \alpha)^2} \le \frac{\varepsilon^2}{(\delta - \alpha)^2}$$

Then to get something explicit for $\alpha, \beta$, e.g. since $\delta > \varepsilon$, we can find some $0 < a < 1$ such that $\varepsilon^a < \delta$. Take $\alpha = \delta - \varepsilon^a$ which yields a bound of order $\varepsilon^{2 - 2a}$ so that $\beta = 1 - \varepsilon^{2-2a}$.

Answer 2

Let $A=[|\hat f|\le\alpha]\subset\Bbb R^d,$ for some $\alpha\in[0,\delta].$

$\rho$-almost everywhere on $A,$ we have both $|\hat f|\le\alpha$ and $|f^*|>\delta$ hence $|\hat f-f^*|>\delta-\alpha,$ so that $$\varepsilon^2\ge\int_A|\hat f-f^*|^2d\rho\ge(\delta-\alpha)^2\rho(A).$$ Therefore, $$\mathbb P\left(|\hat f (X)|>\alpha\right)\ge 1-\frac{\varepsilon^2}{(\delta-\alpha)^2}.$$