This shows you the differences between two versions of the page.
Both sides previous revision Previous revision | |||
world:pinsker [2022/03/16 07:40] 127.0.0.1 external edit |
world:pinsker [2025/06/10 19:10] (current) rdouc [The proof] |
||
---|---|---|---|
Line 31: | Line 31: | ||
</WRAP> | </WRAP> | ||
- | Indeed, set $h(y)=\frac{2}{3}(2y+1)(-\log y+y-1)-(y-1)^2$ and check that $h(1)=h'(1)=0$ and $h''(y)=2(x-1)^2/(3y^2)>0$ for all $y \geq 0$. Therefore, $h$ is a convex function on $\rset^+$ and attains its minimum at $y=1$. Therefore, $h(y) \geq 0$ for all $y\geq 0$, which proves $(\star)$. Then, using $(\star)$ with $y=\frac{\pi(x)}{\sigma(x)}$, and taking $\bar f=f-\mu(f)$ where $\mu=2\pi/3 +\sigma/3$, | + | Indeed, set $h(y)=\frac{2}{3}(2y+1)(-\log y+y-1)-(y-1)^2$ and check that $h(1)=h'(1)=0$ and $h''(y)=2(y-1)^2/(3y^2)>0$ for all $y \geq 0$. Therefore, $h$ is a convex function on $\rset^+$ and attains its minimum at $y=1$. Therefore, $h(y) \geq 0$ for all $y\geq 0$, which proves $(\star)$. Then, using $(\star)$ with $y=\frac{\pi(x)}{\sigma(x)}$, and taking $\bar f=f-\mu(f)$ where $\mu=2\pi/3 +\sigma/3$, |
\begin{align*} | \begin{align*} |