# lower bound for absolute value of a hypergeometric function

I am working with a certain Gauss hypergeometric function, $_{2}F_{1}(a,a-b;2a;1-z)$, where $a, b \in {\mathbb R}$ with $a, a-b > 0$ and $0<b<1$.

It appears that $\left| _{2}F_{1}(a,a-b;2a;1-z) \right| \geq 1$ for all $|z| \leq 1$.

More generally, it seems that $$\left| _{2}F_{1}(a,b;c;1-z) \right| \geq 1$$ for all $|z| \leq 1$ when $a,b,c \in {\mathbb R}$ with $a,b>0$ and $c \geq \max( a,b)$.

For example, if $0 \leq z \leq 1$, then this is true since all the coefficients in the expansion of $_{2}F_{1}(a,b;c;z)$ are positive.

But I have been unable to prove the more general result for all $z$ with $|z| \leq 1$ myself or find a proof in the literature.

Would anyone have any ideas on this, or references I may have missed?

• For those who are afraid of the word "hypergeometric" as much as I am, the problem is to show that $\left|\int_0^\infty t^\alpha(t+1)^{-\beta}(t+z)^{-\gamma}\,\frac{dt}t\right|$ ($\alpha,\beta,\gamma>0, \beta+\gamma>\alpha$) attains its minimum in the unit disk at $z=1$. Some partial cases are easy but I still cannot see it in the full generality. The trick seems to be to change the contour in some smart way. – fedja May 12 '18 at 2:09
• Numerical computations indicate that (at least in the original version of the problem and assuming that ${}_2F_1(a,b;c;z)$ is understood as the main branch for $|z|>1$) there always exists a contour that one can use to prove the inequality. However, this contour is the union of two gradient descents from a saddle point and it is not quite clear (to me) how to show that it always has all the desired properties (the pictures are really nice though). – fedja May 13 '18 at 0:57
• I believe I have it (for the principal branch, i.e., for the extension to the plane with the cut $[1,+\infty)$. BTW, your claim that the case $z\in[-1,0)$ follows from the positivity of coefficients is a total nonsense: you cannot argue like that beyond the circle of convergence). The proof is rather strange (contour integration over an implicitly defined contour) but, I hope, correct though too long to fit in the comment box. I have no time for typing it now (honestly :-( ) but if you indicate that you are still interested, I'll try to post the argument in small chunks later. – fedja May 14 '18 at 1:35
• thanks, fedja. I would be very interested in your solution (when you have a chance). Thanks too for your time and effort! – user124217 May 14 '18 at 7:39
• You are certainly welcome. I've finished typing. Feel free to ask questions if something is unclear. – fedja May 14 '18 at 14:59

As I said, the problem is to show that the absolute value of $$z\mapsto\int_0^\infty t^\alpha(t+1)^{-\beta}(t+z)^{-\gamma}\frac {dt}t$$ attains its minimum in $\Omega=\{z:|z|\le 1, z\notin[-1,0]\}$ at $z=1$. Note that we are free to change the contour of integration to any curve that stays in the open angle bounded by the rays $\{-\tau w:\tau>0\}$ and $\{-\tau:\tau>0\}$ containing the positive semi-axis. So, I'll immediately change it to the ray $\{\tau\sqrt w:\tau>0\}$, which reduces the problem to the following. Let $z,w\in\mathbb C, \Re z,\Re w>0, zw\in\mathbb R_+$. Then $$\left|\int_0^\infty t^\alpha(t+w)^{-\beta}(t+z)^{-\gamma}\frac {dt}t\right|\ge \left|\int_0^\infty t^\alpha(t+|w|)^{-\beta}(t+|z|)^{-\gamma}\frac {dt}t\right|\,.$$ Note that this inequality is immediate if $\beta=\gamma$ and $z=\bar w$ (which corresponds to $c=2a$, $|z|=1$ in the original formulation) because the integrand on the left is then positive and obviously greater than the one on the right. However, in general the integrand on the left oscillates so we cannot take advantage of its being large in absolute value immediately.

Still it suggests the following idea: change the contour (we are now free to choose any curve $\Gamma$ connecting $0$ and $\infty$ in the right half-plane) so that the argument of the product $F(t)=t^\alpha(t+w)^{-\beta}(t+z)^{-\gamma}$ stays fixed (say, $\theta$) on $\Gamma$. If, in addition, we knew that $\Gamma$ crossed each circle centered at the origin just once, we would be able to write $$\left|\int_\Gamma f(t)\frac{dt}t\right|\ge \Re\left[\int_\Gamma e^{-i\theta}F(t)\frac{dt}t\right]=\int_\Gamma |F(t)|d(\log|t|) \\ \ge\int_\Gamma |t|^\alpha(|t|+|w|)^{-\beta}(|t|+|z|)^{-\gamma}d(\log|t|) =\int_0^\infty t^\alpha(t+|w|)^{-\beta}(t+|z|)^{-\gamma}\frac{dt}{t}\,,$$ making everything obvious again. However we face two problems here: we have to demonstrate the existence of such a curve and the circle crossing property. It turns out that the fixed argument condition determines the curve uniquely and we'll define it shortly. However, as far as proving the circle crossing property is concerned (numerical experiments indicate it should be true), I have lost my competition with the tree stump in my backyard: we both couldn't do it after two days of thinking but the stump was obviously much more concentrated on the problem since it wasn't getting distracted by anything (hey, it was so deep in thought that it even didn't move for 2 days!), so we'll have to circumvent it.

The next thing we need is that for $\Re z>0$, the function $$[-\pi/2,\pi/2]\ni \theta\mapsto |re^{i\theta}+z|^2=r^2+|z|^2+2sr\Im z+2\sqrt{1-s^2}r\Re z$$ is a non-negative strongly (in the sense of non-degenerate second derivative) concave function of $s=\sin\theta\in[-1,1]$. It follows that the function $$\theta\mapsto -\beta\log|w+re^{i\theta}|-\gamma\log|z+re^{i\theta}|$$ is a strongly convex function of $s$. Moreover, when going slightly away from the endpoints $\theta=\pm\pi/2$ it obviously decreases (both absolute value increase). Thus on each semicircle $S_r=\{re^{i\theta}:\theta\in[-\frac\pi 2,\frac\pi 2]\}\ni t$ the function $L(t)=\alpha\log|t|-\beta\log|w+t|-\gamma\log|z+t|$ has a unique strong minimum $t(r)$, which, thereby, depends on $r$ continuously, and when you move from $t(r)$ towards either endpoint of $S_r$, it increases monotonically. Clearly, $L(t(r))\to -\infty$ for $r\to 0$ and $r\to\infty$ (that's where we use the condition $\alpha<\beta+\gamma$), so there is at least one maximum $r_*$ of $r\mapsto L(t(r))$. Any such maximum corresponds to a non-degenerate saddle point of $L(t)$. However, the critical points of $L(t)$ are just the solutions of $$D(t)=\frac \alpha t-\frac\beta{t+w}-\frac\gamma{t+z}=0\,,$$ i.e., the roots of the quadratic polynomial $(\alpha-\beta-\gamma)t^2+[\text{some junk}]t+\alpha zw$. Since the leading coefficient is negative and the free term positive, there are two roots whose product is negative none of which lies on the imaginary axis (because $\Re D(t)<0$ for imaginary $t\ne 0$), so the right half-plane contains exactly one root, which means that $r_*$ and the associated saddle point $t_*=t(r_*)$ are unique.

Now we are ready to define $\Gamma$. It is just the union of two steepest descents for $L(t)$ originating at $t_*$ one to $\infty$ and one to $0$. Since the boundary of the right half-plane is repelling for the descending gradient flow of $L$ except at the origin (it is the same condition $\Re D(t)<0$ for imaginary $t\ne 0$; the vector field in question is just $-\bar D(t)$), we are guaranteed to stay in the right half-plane all the way. The descents cannot cross $S_{r_*}$ because the values of $L$ on them are smaller than $L(t_*)=\min_{S_{r_*}}L(t)$, and $L$ goes down at fixed rate on any compact subset of $\mathbb C_+\setminus\{0,t_*,\infty\}$, so the only thing the descents can do is to escape to $0$ inside the circle and to $\infty$ outside. Near $0$ and $\infty$ the flow is asymptotically like a straight line (the direction of the field is quite clear there), so we, indeed, get an admissible contour analytic everywhere except, perhaps, $t_*$. Finally, the argument of $F$ stays fixed because the gradient curves of a harmonic function $L=\Re\log F$ are exactly the level curves of its harmonic conjugate $\arg F=\Im\log F$.

As I said, if we had the circle crossing property, we would be done by now. Alas, we (or at least I) don't know how to prove that, so we'll have to consider the possibility of multiple crossings.

In the multiple crossing case the formula gets a bit more complicated: $$\Re\left[\int_\Gamma e^{-i\theta}F(t)\frac{dt}t\right]=\int_0^\infty \left[\sum_{t\in\Gamma\cap S_r}\varepsilon(t)|F(t)|\right]d(\log r)$$ where $\varepsilon(t)$ is $+1$ if $\Gamma$ (oriented from $0$ to $\infty$) crosses $S_r$ at $t$ in the outward direction and $-1$ if it does so in the inward direction.

What saves the day is that the outward and the inward crossings alternate on $S_r$ (note that it is by no means necessary that the adjacent crossings on $S_r$ are adjacent on $\Gamma$; still the alternation is always there because if we complement $\Gamma$ by a piece of the imaginary line and close the contour near infinity by a circular arc, we'll get a Jourdan curve bounding a domain and the signs at the crossings will correspond to going in or out of that domain when moving along the semi-circle and traversing the boundary of the domain). Now we have the following simple

Lemma: Assume that a function $f(s)$ has a unique minimum on $[-1,1]$ and is monotonically increasing if we go from that minimum to any endpoint. Then for any $-1\le s_0<s_1<\dots<s_{2m}\le 1$, we have $$\sum_{j=0}^{2m}(-1)^jf(s_j)\ge\min_j f(s_j)\ge\min_{s\in[-1,1]}f(s)$$ Applying this lemma to $f(\sin\theta)=|F(re^{i\theta})|, \theta\in[-\frac\pi 2,\frac\pi 2]$ finishes the story.

To prove the lemma just consider two cases. First, suppose that all points are on the same side of the minimum. By symmetry, WLOG, we may assume that they are to the right. Then $f(s_{2k})-f(s_{2k-1})\ge 0$ for $k=1,\dots,m$, so the sum is at least $f(s_0)$. Assume now that the minimum is between $s_j$ and $s_{j+1}$. Again, by symmetry, we may assume that $j=2j'-1$. Then $f(2k)-f(2k+1)\ge 0$ for $k=0,\dots,j'-1$ and $f(2k)-f(2k-1)\ge 0$ for $k=j'+1,\dots,m$, so the sum is at least $f(2j')=f(j+1)$.

The End!

Note that it proves that for $c\ge a,b\ge 0$, we have $|{}_2F_1(a,b;c,1-z)|\ge {}_2F_1(a,b;c,1-r)>0$ in $|z|<r$ for the principal branch. In particular, the principal brunch has no zeroes. The literature devoted to zeroes of ${}_2F_1$ is quite extensive, so such simple fact should certainly be known and its proof may shed some light on your question too (however I leave the really difficult task of searching the literature to you or somebody else).

Also, if you are the same person as Roger in Conjectured bound on Kummer's function (confluent hypergeometric function), I suggest that you merge the accounts. Multiple accounts are not a strict no-no but they are generally frowned upon and impede the communication. If not, just ignore this side remark :-)

• That is a really terrific answer, fedja. Thank you so much! Again, I appreciate very much the time and thought you spent on this. It will help me a lot. Regards, not-Roger :-) – user124217 May 15 '18 at 12:34
• I have a few questions about your w. What is w here? It appears to be related to z from your remark that $z=\bar{w}$ corresponds to $|z|=1$. Why do you take the ray formed from its square root? I.e., why the square root, in particular? How do you convert from the initial integral with the $(t+1)^{-\beta}$ term to the one with a $(t+w)^{-\beta}$ term? I thought change-of-variables, but could not reproduce your integrand with the $w$ in this way. – user124217 Jun 19 '18 at 19:41
• @user124217 $w$ is just the image of $1$ under the rotation. There is nothing going on at this stage, just shifting from one ray to another. And $z$ is the image of $z$, so it is a different $z$. Also I guess $z$ and $w$ got swapped in the first couple of sentences, so it should be $\sqrt z$ really. Sorry if that confused you. The idea is just to integrate over the bisector between $1$ and the original $z$. What do you get when making this change of variable? – fedja Jun 19 '18 at 21:29
• thanks. That makes it clearer to me. One other small thing. Should it be $\sqrt{\bar{z}}$ rather than $\sqrt{z}$ (here $z$ is the original $z$)? It seems so in order for your condition that $zw \in {\mathbb R}_{+}$ (where $z$ there is the different'' $z$) holds. That also seems necessary for your statement that $z=\bar{w}$ (different'' z) corresponds to the original $z$ being on the unit circle. Not trying to be picky, but just to make sure I have the details correct, and also because the case $c=2a$ is of particular interest to me. Thanks again for your help. – user124217 Jun 21 '18 at 22:16
• @user124217 I would say that you can rotate half-way from $1$ to $z$ without hitting singularities, so $\sqrt z$ looks correct to me in the ray definition with $w$ having the argument of $\sqrt{\bar z}$ and absolute value $1$ after rotation... I would hate to edit it all because it will bump the post to the front page, but, as I said, just switch to the bisector whatever it is in terms of the square roots. The best way to see it is to draw a picture. And be as picky as you need: the goal of the communication is to get the message across, so if something does not go through, ask more questions. – fedja Jun 21 '18 at 23:52