Definition 1. A convex function $f:(a,b)\to \mathbb{R}$ defined on an open interval $(a,b)\subset \mathbb{R}$ is convex if the
inequality $$f(\alpha_1x_1+\alpha_2x_2)\leq
\alpha_1f(x_1)+\alpha_2f(x_2)$$ holds for any points $x_1,x_2\in
(a,b)$ and any numbers $\alpha_1\geq 0,\ \alpha_2\geq 0$ such that
$\alpha_1+\alpha_2=1$. If this inequality is strict whenever $x_1\neq
x_2$ and $\alpha_1\alpha_2\neq 0$, the function is strictly convex
on $(a,b)$.Then he proves the following
Proposition 5. A necessary and sufficient condition for a function $f:(a,b)\to \mathbb{R}$ that is differentiable on the open interval
$(a,b)$ to be convex (downward) on that interval is that its
derivative $f'$ be nondecreasing on $(a,b)$. A strictly increasing
$f'$ corresponds to a strictly convex function.Corollary. A necessary and sufficient condition for a function $f:(a,b)\to \mathbb{R}$ having a second derivative on the open
interval $(a,b)$ to be convex (downward) on $(a,b)$ is that
$f''(x)\geq 0$ on that interval. The condition $f''(x)>0$ on $(a,b)$
is sufficient to guarantee that $f$ is strictly convex.Example 12. Let us study the convexity of $f(x)=\sin x$.
Since $f''(x)=-\sin x$, we have $f''(x)<0$ on the intervals $\pi\cdot
2k<x<\pi(2k+1)$ and $f''(x)>0$ on $\pi(2k-1)<x<\pi\cdot 2k$, where
$k\in \mathbb{Z}$. It follows from this, for example, that the arc of
the graph of $\sin x$ on the closed interval $0\leq x \leq
\frac{\pi}{2}$ lies above the chord it subtends everywhere except at
the endpoints; therefore $\sin x>\frac{2}{\pi}x$ for
$0<x<\frac{\pi}{2}$.
This is an excerpt from Zorich's book and I am a bit confused with the following moment: The author defines the convexity on the finite open intervals $(a,b)$. He shows that $f(x)=\sin x$ is strictly concave on $(0,\pi)$. The line which connects $(0,0)$ and $(\frac{\pi}{2},1)$ has an equation $y=\frac{2}{\pi}x$. Then he somehow includes the endpoints and deduce that $\sin x>\frac{2}{\pi}x$ on $(0,\frac{\pi}{2})$. Intuitively this is clear but I am a bit confused with the rigorous proof since he defines it for open intervals then he moves to closed interval $[0,\frac{\pi}{2}]$.
I'd be thankful if someone can show the more detailed and rigorous explanation of that.
Best Answer
You have shown that $\sin x$ is strictly concave downwards on $(0,\pi)$. Now for $x\in (0,\pi/2)$, note that:
$\begin{align}\sin x= &\sin (\frac 2\pi x (\frac \pi 2))\\=& \lim_{n\to \infty} \sin (\frac 1n (1-\frac 2\pi x) + \frac 2\pi x (\frac \pi 2)) \end {align}$
Noting that $\{2x/\pi,1-2x/\pi\}\subset (0,1)$ and by by strictly concave downward nature of $\sin $ on $(0,\pi)$, we have $\sin (\frac 1n (1-\frac 2\pi x) + \frac 2\pi x (\frac \pi 2)\gt (1-\frac 2\pi x) \sin (\frac 1n) + \frac 2\pi x \sin (\frac \pi 2)$
It follows that: $\lim _{n\to \infty} \sin (\frac 1n (1-\frac 2\pi x) + \frac 2\pi x (\frac \pi 2)\ge \lim_{n\to \infty}(1-\frac 2\pi x) \sin (\frac 1n) + \frac 2\pi x \sin (\frac \pi 2)=\frac 2\pi x$
So it follows that $\sin x\ge \frac 2 \pi x$ for $x\in (0,\pi/2)$.
Now it remains to rule out the equality. If $\sin t=\frac 2\pi t$ for some $t\in (0,\pi/2)$, then define $g:[0,\pi/2]\to \mathbb R$ as follows:
$g(x)=\begin{cases} 0; \text{ when }x=0\\ \sin x-\frac 2\pi x; \text{ when } x\in (0,\pi/2)\\0;\text{ when } x=\pi/2\end{cases}$
By LMVT on $g,\exists c_1\in (0,t)\land c_2\in (t,\pi/2)$ such that $g'(c_1)=\cos c_1=\frac 2\pi=g'(c_2)=\cos c_2$, which is a contradiction as $\cos $ is a decreasing function in $(0,\pi/2)$ in strict sense.