Suppose $\mathcal{F}=\left\{f\in L^2([a,b]): 0<\underline{c}\leq f\leq\overline{c} \right\}$. Consider the following transformation $$\tilde{\mathcal{F}} := \left\{\frac{f}{\int f d\mu}: f\in \mathcal{F}\right\}$$ Want to show the claim that $\mathcal{F}$ and $\tilde{\mathcal{F}}$ have the $\epsilon$-metric entropy (log of covering/packing number under $L_2$ norm) of same order.
To put this in a more concrete context, suppose that $\mathcal{F}$ is a Sobolev ellipsoid, i.e $$\mathcal{F} = \mathcal{E}_k(A) = \left\{f\in L^2([a,b]): f = \sum_{j=0}^\infty \theta_j\phi_j(X), \sum_{j=0}^\infty \theta_j^2j^{2k}< A\right\}$$ Suppose $k>1$ and $\phi_j$'s are uniformly bounded so that $\mathcal{E}_k(A)$ is uniformly bounded, say by a constant $\rho$. We can see that a transformation of $\mathcal{E}_k(A)$, say $$ \tilde{\mathcal{E}}_k(A) := \left\{\frac{f + \rho + 1}{\int f d\mu + \rho + 1}: f\in \mathcal{E}_k(A) \right\}$$ is a subset of $\mathcal{E}_k(A')$ for some $A'$. In some well-published papers (eg. Yang and Barron 1999, p. 1591-), the authors claim that for $A$ large enough, $\tilde{\mathcal{E}}_k(A)$ and and $\mathcal{E}_k(A)$ have the same order of $L^2$ metric entropy. They stated in the paper "it is easy to see". While I can "see" it intuitively, I have yet to be able to come up with a rigorous proof. In Yang and Barron's paper, they also give some other function classes that they claim such property holds.
Note 1: for the Sobolev ellipsoid, I think the convex property may play a role.
Note 2: One direction is easy. Consider any $f,g \in\mathcal{F}$, and let $\tilde{f} = f/\int f$ and $\tilde{g} = g/\int g$. We can show that for some constant $C$, $$ \|\tilde{f}-\tilde{g}\|_2 \leq C \|f - g\|_2 $$
The question is however if there's a reverse inequality of the form $\|f - g\|_2\leq C'\|\tilde{f}-\tilde{g}\|_2 $ for some constant $C'$.
This is not true generally. Take $\underline{c}=1$, $\overline{c}=2$, $a=1$, $b=2$. For $n\in\mathbb{N}$, consider the class of functions: $$ \mathcal{F}_n:=\left\{x\mapsto 1+\frac{k}{n}, k\in\{0,\dots,n\}\right\} $$ For any non-identical $f,g\in \mathcal{F}_n$, we have $||f-g||_2\geq 1/n$. So, for $\epsilon<1/n$, the covering number is $n+1$. However, $\tilde{\mathcal{F}}=\{x\mapsto 1\}$, which has covering number $1$.