# (Sharp) inequality for Beta function

I am trying to prove the following inequality concerning the Beta Function: $$\alpha x^\alpha B(\alpha, x\alpha) \geq 1 \quad \forall 0 < \alpha \leq 1, \ x > 0,$$ where as usual $$B(a,b) = \int_0^1 t^{a-1}(1-t)^{b-1}dt$$.

In fact, I only need this inequality when $$x$$ is large enough, but it empirically seems to be true for all $$x$$.

The main reason why I'm confident that the result is true is that it is very easy to plot, and I've experimentally checked it for reasonable values of $$x$$ (say between 0 and $$10^{10}$$). For example, for $$x=100$$, the plot is: Varying $$x$$, it seems that the inequality is rather sharp, namely I was not able to find a point where that product is larger than around $$1.5$$ (but I do not need any such reverse inequality).

I know very little about Beta functions, therefore I apologize in advance if such a result is already known in the literature. I've tried looking around, but I always ended on inequalities trying to link $$B(a,b)$$ with $$\frac{1}{ab}$$, which is quite different from what I am looking for, and also only holds true when both $$a$$ and $$b$$ are smaller than 1, which is not my setting.

I have tried the following to prove it, but without success: the inequality is well-known to be an equality when $$\alpha = 1$$, and the limit for $$\alpha \to 0$$ should be equal to 1, too. Therefore, it would be enough to prove that there exists at most one $$0 < \alpha < 1$$ where the derivative of the expression to be bounded vanishes. This derivative can be written explicitly in terms of the digamma function $$\psi$$ as: $$x^\alpha B(\alpha, x\alpha) \Big(\alpha \psi(\alpha) - (x+1)\alpha\psi((x+1)\alpha) + x\alpha \psi(x\alpha) + 1 + \alpha \log x \Big).$$ Dividing by $$x^\alpha B(\alpha, x\alpha) \alpha$$, this becomes $$-f(\alpha) + \frac{1}{\alpha} + \log x,$$ where $$f(\alpha) = -\psi(\alpha) + (x+1)\psi((x+1)\alpha) - x \psi(x\alpha)$$ is, as proven by Alzer and Berg, Theorem 4.1, a completely monotonic function. Unfortunately, the difference of two completely monotonic functions (such as $$f(\alpha)$$ and $$\frac{1}{\alpha} + C$$) can vanish in arbitrarily many points, therefore this does not allow to conclude.

Many thanks in advance for any hint on how to get such a bound!

[EDIT]: As pointed out in the comments, the link to the paper of Alzer and Berg pointed to the wrong version, I have corrected the link.

• There is no Theorem 4.1 in the quoted paper. There is a Theorem 4 there, but it does not talk about the digamma function. Can you please clarify? – GH from MO Dec 29 '18 at 21:58
• @GHfromMO Thanks for pointing out the wrong link, the one I inserted did not send to the most updated version. I have now corrected it! – Ester Mariucci Dec 29 '18 at 22:14

You can get away with the usual distribution function mumbo-jumbo. The general lemma is as follows:

Let $$\mu,\nu$$ be non-negative measures and $$f,g$$ be non-negative functions such that there exists $$s_0>0$$ with the property that $$\mu\{f>s\}\ge \nu\{g>s\}$$ for $$s\le s_0$$ and the reverse inequality holds for $$s\ge s_0$$. Suppose also that $$\int f^q\,d\mu=\int g^q\,d\nu<+\infty$$ for some $$q>0$$. Then, as long as the integrals in question are finite, we have $$\int f^p\,d\mu\ge \int g^p\,d\nu$$ for $$0 and the reverse inequality holds for $$p\ge q$$.

The proof of the lemma is rather straightforward. Let $$p\le q$$ (that is the case you are really interested in) $$\int f^p\,d\mu-\int g^p\,d\nu=p\int_0^\infty s^p[\mu\{f>s\}-\nu\{g>s\}]\frac{ds}s \\ =p\int_0^\infty [s^p-s_0^{p-q}s^q][\mu\{f>s\}-\nu\{g>s\}]\frac{ds}s\ge 0\,.$$

Now we use it with $$f(t)=t(1-t)^x$$, $$d\mu=\frac{dt}{t(1-t)}$$ on $$(0,1)$$, $$g(t)=t$$, $$d\nu=\frac{dt}{t}$$ on $$(0,\frac1x)$$. Since the maximum of $$t(1-t)^x$$ is attained at $$t=\frac{1}{x+1}$$, we see that the function $$s\mapsto \mu\{f>s\}$$ drops to $$0$$ before the function $$s\mapsto \nu\{g>s\}$$. Also, the first function has larger in absolute value negative derivative than the second one for each value of $$s$$ where it is still positive. To see it, notice that the set where $$f>s$$ is an interval $$(u,v)=(u(s),v(s))$$ that shrinks as $$s$$ increases and the left end $$u$$ of this interval satisfies $$du\left(\frac 1u-\frac x{1-u}\right)=\frac{ds}s\,,$$ so trivially $$\frac{du}{u(1-u)}\ge \frac{du}u>\frac {ds}s$$ The right end moving to the left can only increase the decay speed. Finally, for $$q=1$$, the integrals are equal (which also shows that the graphs of the distribution functions must indeed intersect), so for $$0 (which plays the role of $$\alpha$$), we have the desired inequality.

• That's a wonderful and clever proof, many thanks! I'm marking this as accepted (although @egs's one also proves the desired inequality) also because it came first. – Ester Mariucci Jan 1 at 19:59

One can also use Jensen's inequality. Let (for $$\sigma>0$$) $$G_\sigma$$ denote a random variable with $$\Gamma(1,\sigma)$$-distribution, i.e. having Lebesgue density $$f_\sigma(t)=\frac{t^{\sigma-1}}{\Gamma(\sigma)} e^{-t}\;1_{(0,\infty)}(t)\;,$$ then $$\mathbb{E}(G_\sigma)=\sigma$$. Since $$\alpha\in (0,1)$$ the functions $$t\mapsto t^\alpha$$ resp. $$t\mapsto t^{1-\alpha}$$ on $$\mathbb{R}_+$$ are concave. By Jensen's inequality $$\frac{\Gamma(\alpha+\alpha x)}{\Gamma(\alpha x)}=\mathbb{E}(G_{x\alpha}^\alpha)\leq \left(\mathbb{E}(G_{x\alpha})\right)^\alpha=(x\alpha)^{\alpha}$$

and $$\frac{1}{\Gamma(\alpha)}=\mathbb{E} G_\alpha^{1-\alpha}\leq\left(\mathbb{E}(G_{\alpha})\right)^{1-\alpha}=\frac{1}{\alpha^{\alpha-1}}$$ Using that gives $$B(\alpha,x \alpha)=\frac{\Gamma(\alpha)\,\Gamma(x\alpha)}{\Gamma(\alpha +x\alpha)}\geq \frac{\Gamma(\alpha)}{\alpha^\alpha x^\alpha}\geq \frac{\Gamma(\alpha)}{\alpha\,\Gamma(\alpha)\,x^\alpha}=\frac{1}{\alpha x^\alpha},$$ as desired.

• That's a very quick way to prove it, thanks! This proof is of particular interest to me because I was also aiming to prove \Gamma(\alpha) \geq \alpha^{\alpha+1}, which indeed is essentially equivalent to the inequality on Beta by using Stirling on both \Gamma(x\alpha) and \Gamma((x+1)\alpha). – Ester Mariucci Jan 1 at 20:03

This is an attempt to strengthen your claim.

If $$x$$ is large then $$B(x,y)\sim \Gamma(y)x^{-y}$$ and hence $$B(\alpha x,\alpha)\sim \Gamma(\alpha)(\alpha x)^{-\alpha};$$ where $$\Gamma(z)$$ is the Euler Gamma function.

On the other hand, for small $$\alpha$$, we have the expansion $$\Gamma(1+\alpha)=1+\alpha\Gamma'(1)+\mathcal{O}(\alpha^2).$$ Since $$\alpha\Gamma(\alpha)=\Gamma(1+\alpha)$$, it follows that $$\Gamma(\alpha)\sim \frac1{\alpha}-\gamma+\mathcal{O}(\alpha)$$ where $$\gamma$$ is the Euler constant.

We may now combine the above two estimates to obtain $$\alpha x^{\alpha}B(\alpha x,\alpha)\sim \alpha x^{\alpha}\left(\frac1{\alpha}-\gamma\right)(\alpha x)^{-\alpha}=\left(\frac1{\alpha}-\gamma\right)\alpha^{1-\alpha}\geq1$$ provided $$\alpha$$ is small enough. For example, $$0<\alpha<\frac12$$ works.