Let $\mathcal{F}$ be a set of uniformly bounded measurable functions on interval $[0,1]$ with respect to the Lebesgue measure. Let $\tilde{\mathcal{F}} = \{e^f/\int e^f d\mu: f\in\mathcal{F}\}$ be the transformed density class. Let $\|\cdot\|_p$ be the $L_p$ norm for $1\leq p\leq\infty$. For any $f\in\mathcal{F}$, denote $\tilde{f} = e^f/\int e^f$. Show that for any $f_1$ and $f_2$ in $\mathcal{F}$, $$\| \tilde{f}_1 - \tilde{f}_2 \|_p\leq c_1 \|f_1 -f_2 \|_p $$ and $$\|\log(\tilde{f}_1) - \log(\tilde{f}_2) \|_p\leq c_2\|\tilde{f}_1 - \tilde{f}_2\|_p$$ for some constants $c_1$ and $c_2$ that only depends on $\sup_{f\in\mathcal{F}}\|f\|_\infty$.
In the paper the author says its "easy" to see these two inequalities but I've been struggling to show this explicitly for a while now. Any suggestions would be greatly appreciated.
Edit: I think the uniform boundedness and the fact that $\exp(\cdot)$ and $\ln(\cdot)$ are Lipschitz should give the results. Note that $\exp(x)$ on a bounded interval is always Lipschitz; for $\ln(x)$ to be Lipschitz, need $x$ to be on $[a,\infty)$ for $a>0$.
Indeed, the lipschitz continuity comes to our rescue. I gave some rough estimates. Feel free to refine them.
Let us define $C:= \sup_{f\in \mathcal{F}} \Vert f \Vert_\infty$. Then we compute
$$ \Vert \tilde{f} - \tilde{g}\Vert_p = \frac{1}{\int e^f \int e^g} \Vert (\int e^g) e^f - (\int e^f) e^g \Vert_p \leq \frac{1}{\int e^f \int e^g} \left[ (\int e^g ) \Vert e^f - e^g \Vert_p + \vert \int (e^f - e^g ) \vert \cdot \Vert e^g \Vert_p \right]. $$ Hence, we get $$ \Vert \tilde{f} - \tilde{g}\Vert_p \leq e^{3C} \left[ \Vert e^f - e^g \Vert_p + \Vert e^f - e^g \Vert_1 \right] \leq 2 e^{3C} \Vert e^f - e^g \Vert_p. $$ By the mean value theorem we get $\vert e^{f(x)} - e^{g(x)} \vert \leq e^C \vert f(x) - g(x) \vert$. Therefore, we get $$ \Vert \tilde{f} - \tilde{g}\Vert_p \leq 2 e^{4C} \Vert f -g \Vert_p. $$
We have $$ \frac{e^{f(x)}}{\int e^f} \geq \frac{e^{-C}}{e^C} = e^{-2C}. $$ Thus, again by the mean value theorem, we get $\vert \log(\tilde{f}(x)) - \log(\tilde{g}(x)) \vert \leq e^{2C} \vert \tilde{f}(x) - \tilde{g}(x) \vert$ and therefore we get $$ \Vert \log(\tilde{f}) - \log(\tilde{g}) \Vert_p \leq e^{2C} \Vert \tilde{f} - \tilde{g} \Vert_p. $$