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.

  • $\begingroup$ The open interval is only required when considering the differentiability of the function. This restriction isn't required in general for points where the function is not differentiable such as the boundary points. $\endgroup$ Apr 27, 2021 at 4:02
  • $\begingroup$ @CyclotomicField, I did not get your last sentence. $\endgroup$
    – RFZ
    Apr 27, 2021 at 4:17
  • $\begingroup$ Some functions are convex but not differentiable everywhere. For example $f(x)=\vert x \vert$ is convex even though it's not differentiable at $0$. $\sin (x)$ won't be differentiable at the points $0$ and $\pi / 2$ but including won't break convexity because the epigraph will still be a convex set. Proposition $5$ would be false if they didn't restrict it to differentiable functions. $\endgroup$ Apr 27, 2021 at 4:49
  • $\begingroup$ @CyclotomicField, let me ask you question please. I know that $\sin x$ is convex on $(0,\pi/2]$. How to show that it is convex on $[0,\pi/2]$? We just need to show that $\sin (\alpha t)\geq \alpha \sin t$ where $\alpha\in [0,1]$ and $t\in (0,\pi/2]$. $\endgroup$
    – RFZ
    Apr 29, 2021 at 17:04
  • $\begingroup$ This follows from the fact that $\sin x \leq x$. To see this geometrically note that the line through $0$ and any point on $\sin x$ with $x \in (0, \pi /2]$ is above the curve $\sin x$ on that region. $\endgroup$ Apr 29, 2021 at 17:11

1 Answer 1


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.

  • 1
    $\begingroup$ Hi! Thanks a lot for your help! I really appreciate it. Actually I was able to show that $\sin x\geq \frac{2x}{\pi}$ but I have some issues to show that it should be strictly inequality. BTW how did you come up with this reasoning using LMVT? That is really nice $\endgroup$
    – RFZ
    Apr 30, 2021 at 13:59
  • $\begingroup$ @ZFR: I'm glad I could help. The reasoning followed from trying to solve the equation $\sin t=\frac {2 t } \pi$ on $(0,\pi/2)$. $\endgroup$
    – Koro
    Apr 30, 2021 at 14:34
  • $\begingroup$ Oh I see. Also I guess we can write $g(x)$ on $[0,\pi/2]$ as $\sin x-\frac{2}{\pi}x$, right? It is not necessary to define it separately for $x=0$ and $x=\pi/2$. $\endgroup$
    – RFZ
    Apr 30, 2021 at 14:41
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    $\begingroup$ @ZFR: Right, that's not required. I did that in order to simplify things. You may use generalized LMVT on $(a,b)$: In appropriate situations, there exists some $c\in (a,b)$ such that $f'(c)=\frac{\lim_{x\to b^- }f(x) -\lim_{x\to a^+ } f(x)}{b-a }$. Now you don't have to define $g$ at all. $\endgroup$
    – Koro
    Apr 30, 2021 at 14:45
  • 1
    $\begingroup$ @ZFR: To prove this, define $g$ as I have done in my answer above. $\endgroup$
    – Koro
    Apr 30, 2021 at 14:58

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