Lower bound $\int_0^1 \frac{|f'(x)|^2}{f} \,\mathrm{d} x$ by $\int_0^1 |f-1|^2\, \mathrm{d} x$

Question

Assume that $f$ is a probability density on $x \in (0,1)$, I want to obtain a bound of the following form (if it is possible at all): $$ \int_0^1 \frac{|f'|^2}{f} \,\mathrm{d} x \geq C\,\int_0^1 |f-1|^2\, \mathrm{d} x$$ for some universal constant $C > 0$. My strategy is as follows: by the log-Sobolev inequality, we have $$\int_0^1 \frac{|f'|^2}{f} \,\mathrm{d} x \geq C\,\int_0^1 f\,\ln f \,\mathrm{d} x.$$ Thus, I am expecting that a $L^p$-type (with $p=2$) Csiszar–Kullback-Pinsker inequality of the form $$\int_0^1 f\,\ln f \,\mathrm{d} x \geq C\,\int_0^1 |f-1|^2\, \mathrm{d} x \quad (*)$$ can hold. However, I searched very hard in the literature and the only relevant material I can found is this $L^p$-type Csiszar–Kullback-Pinsker inequality, which is not enough to deduce $(*)$ as the infimum of $\psi''(s) = 1/s$ over $s \in (0,\infty)$ is not strictly greater than zero (here $\psi(s) = s\,\ln(s) - s + 1$). Thus, it seems that unless we have certain (pointwise) boundedness assumptions imposed on $f$, there is no hope for the validity of $(*)$. I get stuck at this point I would appreciate any ideas which can help me out.

Answer 1

First of all such a Csiszar–Kullback-Pinsker inequality or whatever cannot possibly be true since $x^2$ explodes faster than $x\log x$ so you can make a local adjustment so that the right-hand side is infinite while the left-hand side is not. Yet, your original inequality is true and here is the proof:

By the Cauchy-Schwarz inequality we have $$\int_0^1 \frac{|f'(x)|^2}{f(x)}dx \int_0^1 f(x)dx \ge \left(\int_0^1 |f'(x)|dx\right)^2.$$

Since $\int_0^1 f(x)dx = 1$, we just have to prove that the right-hands side of this is at least $C\int_0^1 |f(x)-1|^2dx$. I will assume that since we are talking about $f'(x)$, the function $f$ is at least continuous. Since we also have $\int_0^1 f(x)dx = 1$, by the intermediate value theorem there must exist $x_0\in (0, 1)$ such that $f(x_0) = 1$. Applying the fundamental theorem of calculus, we can see that for all $x\in (0, 1)$ we have

$$|f(x) -1| = |f(x) - f(x_0)| \le \int_0^1 |f'(x)|dx.$$

Squaring this and integrating over $(0, 1)$ we get

$$\int_0^1 |f(x)-1|^2dx \le \left(\int_0^1 |f'(x)|dx\right)^2,$$

that is the required inequality holds with $C = 1$. I'm pretty sure $C$ can be optimized with this method, but I will leave it to someone else.