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How would one prove the following statement?

$f(x)=\frac x{\sin x}$ is convex in $[0,\pi)$.

After having posted the question, I came upon two proofs which I have now posted in the answer below. I would like to know other proofs as well, even the brute force ones. Can we utilize the identity $\frac{\pi x}{\sin(\pi x)}=\Gamma(1+x)\Gamma(1-x)$ where $\Gamma$ is the gamma function? How about the fact that $f$ is the reciprocal of the sinc function with all the nice properties? Or can we represent $f$ by an integral $\int g(t,x)d\mu(t)$ where $g(t,x)$ is convex with respect to $x$ for every $t$ and $\mu(t)$ is a measure of $t$?

Hans
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  • @JackD'Aurizio: I assure you it is not. I did not know a single proof when I first posted this question. Over the course of posting the question --- with one version deleted --- I came upon two proofs which I subsequently posted here. The reason I put the answer up was I was asked to show my work, as evidenced by the long conversation above. By the time I have written up the question accordingly, I have already found two proofs which I posted subsequently. How would you suggest one do when posting something appearing deceivingly simple? – Hans Apr 23 '18 at 16:56
  • @JackD'Aurizio: If I post no result, I would be accused of not showing any work. When I do, I would be accused of showing off too much. Do you not think I am put between a rock and hard place? Also, when I do not post any work, I get useless general hints on how to take second derivatives and am instructed to look at the sign of it, as if I am a befuddled student coming upon the subject of derivatives on the first day. What would be your advice to resolve this dilemma? – Hans Apr 23 '18 at 17:01
  • @JackD'Aurizio: These two are very different approaches and one came after the other. Besides, I think it is a good question worth exploring for more solutions. I am still searching for more approaches. That was why I asked in the question body if it was possible to use other different representations of the function to prove the statement. I would love to have one through integration with a parameter. You are a brilliant expert in this area. Do you have a proof via the integration method? I would love to see it if you do. – Hans Apr 23 '18 at 17:09
  • @Hans: by residues/the Lagrange inversion theorem, all the coefficients of the Maclaurin series of $\frac{x}{\sin x}$ are non-negative. Then it is straightforward that $\frac{x}{\sin x}$ is convex on $(0,\pi)$. – Jack D'Aurizio Apr 23 '18 at 17:14
  • @JackD'Aurizio: Excellent! Would you please put up your brilliant proof? – Hans Apr 23 '18 at 17:18
  • @JackD'Aurizio: I tried to move to chat, but you did not follow. The fact is that these two proofs are very different approaches. If I answer someone else's question with two very different approaches, is it a rule that I have to put these two very different answers as one? Had I been hunting for upvotes, I would not have put this question up. I had many downvotes immediately after my posting. Only after I put up the answer to show my work, my vote went into the positive territory. I see now the long conversation between me and other people on this when I first posted has been deleted. – Hans Apr 23 '18 at 17:31
  • @JackD'Aurizio: With the deletion of the comments by others, now you can not see the struggle I went through to even put this question up. From the leftover very first post you can still see the objection to showing no work. By the way, I have edited the question to reflect our conversation on this subject. – Hans Apr 23 '18 at 17:45

3 Answers3

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Here is an elegant proof given by @deyore.

We show the convexity of $f$ via that of $\ln f(x)$. Take the second derivative of the latter. $$\frac{d^2}{dx^2}\ln f(x)=-\frac1{x^2}+\frac1{(\sin x)^2}>0,\ \forall x\in[0,\pi),$$ since $0<\sin x<x$ there.

Hans
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Here is a brute force proof via taking the second derivative. \begin{align} f''(x) &= \frac1{\sin^3x}\big(x(1+\cos^2 x)-2\sin x\cos x\big) \\ &= \frac1{\sin^3 x}\int_0^x \sin t(3\sin t-2t\cos t)\,\mathrm dt \\ &>0 \end{align} since $\tan t>\frac23t,\forall x\in\big[0,\frac\pi2\big)$ and $\cos t<0,\,\forall x\in\big(\frac\pi2,\pi\big)$.

Hans
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Let us say that a function $f(x)$ is horribly convex if it is analytic in a neighbourhood of the origin and the coefficients of its Maclaurin series are non-negative. Quite clearly, the space of horribly convex functions is closed with respect to the multiplication by $x^n$, differentiation and termwise-integration. By the Weierstrass product for the sine function $$ \sin(z) = z\prod_{n\geq 1}\left(1-\frac{z^2}{n^2 \pi^2}\right) $$ and by applying the logarithmic derivative $\frac{d}{dz}\log(\cdot)$ to both sides we get $$ z\cot(z) = 1+\sum_{n\geq 1}\frac{2z}{n^2 \pi^2 - z^2}=1-2\sum_{n\geq 1}\frac{\zeta(2n)}{\pi^{2n}}z^{2n} $$ already exploited by Euler in one of his proofs of $\zeta(2)=\frac{\pi^2}{6}$ (see my notes).
The fact that $1-z\cot z$ is a horribly convex function is now evident and it can be used for strenghtening the Huygens and Shafer-Fink inequalities, for instance. The trigonometric identity $$ \frac{1}{\sin(x)}=\cot(x/2)-\cot(x) $$ then implies that $\frac{z}{\sin z}$ is a horribly convex function, in particular a convex function over $(0,\pi)$, since the radius of convergence of the Maclaurin series clearly equals $\pi$ ($z=\pm \pi$ are the poles closest to the origin). In equivalent terms, convexity is a straightforward consequence of Herglotz' decomposition or of $\operatorname*{Res}_{z=k\pi}\frac{1}{\sin(z)}=(-1)^k$. Of course $$ \frac{d^2}{dx^2}\log\left(\frac{x}{\sin x}\right) = \frac{1}{\sin^2 x}-\frac{1}{x^2}>0$$ is just as effective.

Jack D'Aurizio
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  • Excellent, Jack, that this proof touches on other deep results. I will study this later. Is this related to the Lagrange inversion theorem you mentioned in the comment? – Hans Apr 23 '18 at 18:37
  • @Hans: yes, the coefficients of the Maclaurin series of $\frac{x}{\sin x}$ can be computed by Lagrange inversion, too, but Euler's route is simpler in my opinion. The crucial point stays the same, i.e. that the singularities of $\frac{1}{\sin x}$ are very regular (no pun intended). – Jack D'Aurizio Apr 23 '18 at 18:40