lajos/The W function -- a continuous approach

In this note we prove analogous statements (cf. the “discrete version”) about the unimodality and the location of the maximum of the expression \(\frac{m^{p-1} }{(p-m)! (m-1)!}\), but now the variable \(m\) will be allowed to be a “continuous” one. The final result will be the same, but the proofs are different and more general.

Similarly to the above mentioned page, \(p\ge 2\) is an arbitrary integer throughout this page also. Earlier we have had \[ F_p(m):=\frac{m^{p-1} }{(p-m)! (m-1)!}=\frac{m^p}{p!}\binom{p}{m}=\frac{m^{p} }{m! (p-m)!} \] with \(m\in [2,p]\cap\mathbb{N}\), but now let us extend naturally the definition of \(F_p(m)\) to any \(m\in [2,p]\), \(m\in\mathbb{R}\) as
\[ \widetilde{F_p}(m):=\frac{m^{p} }{\Gamma(1+m)\Gamma(1+p-m) }, \] where \(\Gamma\) denotes the gamma function.

We are again interested in the quantity \[ \max_{2\le m\le p, m\in\mathbb{R}} \widetilde{F_p}(m), \] if \(p\) is large.

It will be easier if we extend a little bit and normalize the domain of definition of \(\widetilde{F_p}\): instead of the real interval \([2,p]\), we are going to work on the real interval \([0,1]\), and define for any \(2\le p\in\mathbb{N}\) and \(x\in [0,1]\) \[ A_p(x):=\frac{(p x)^p}{\Gamma (1+p x)\Gamma (1+p-p x)}. \] Clearly, if \(m\in [2,p]\) and \(x=\frac{m}{p}\), then \(\widetilde{F_p}(m)=A_p(x)\).

As another useful notation, for any \(2\le p\in\mathbb{N}\) and \(x\in [0,1]\) we set

\[ B_p(x):=\frac{p x}{\sqrt[p]{\Gamma (1+p x)\Gamma (1+p-p x)}}. \]

In this document, we are going to use the adjective “unimodal” in the following (more restrictive) sense: we say here that a function \(f\) is unimodal on \([0,1]\), if there is a unique \(x^*\in[0,1]\) such that \(f\) is strictly increasing on \([0,x^*]\), has a strict global maximum at \(x^*\), then decreasing on \([x^*,1]\).

Main claim. For each fixed \(2\le p\in\mathbb{N}\), the function \(A_p(\cdot)\) is unimodal on \([0,1]\), and if \(m_A(p)\) denotes the abscissa of its maximum, then \[ \lim_{p\to\infty}m_A(p)=L:=\frac{1}{1+W\left(\frac{1}{e}\right)}\approx 0.78218829428. \]

In the rest of this document, the above claim is verified in a series of (interesting auxiliary) steps. We are going to use some properties of the gamma function and the digamma function.

\(\bullet\) Since the \(\Gamma\) function is analytic (for example) in the complex open right half-plane, and positive on \([1,+\infty)\), both \(A_p(\cdot)\) and \(B_p(\cdot)\) are differentiable on \([0,1]\).

\(1^{\mathbf{st}}\) claim. For any fixed \(p\ge 2\), \(A_p\) is unimodal on \([0,1]\).

Proof. The derivative of \(A_p(\cdot)\) evaluated at \(x\in [0,1]\) is \[ p^{p+1} x^{p-1}\frac{x \Gamma(1+p x)\ \Gamma^\prime(1+p (1-x))-x \Gamma^\prime(1+p x)\ \Gamma(1+p (1-x))+\Gamma(1+p x)\ \Gamma(1+p (1-x))}{\Gamma^2(1+p x)\ \Gamma^2(1+p (1-x)) }= \]

\[ \frac{p^{p+1} x^{p-1}}{\Gamma(1+p x)\ \Gamma(1+p (1-x)) }\left[1+x \left(\frac{\Gamma^\prime(1+p(1- x))}{\Gamma(1+p (1-x))}-\frac{\Gamma^\prime(1+p x)}{\Gamma(1+p x)}\right)\right]. \]

The essential part of the proof is to analyze the sign of the \([...]\) factor above. First let us use the following integral representation of the digamma function (traditionally denoted by \(\psi\), and by PolyGamma[z] in Mathematica) \[ \frac{\Gamma^\prime(z)}{\Gamma(z)}=\int_0^1 \frac{1-t^{z-1}}{1-t}dt -\gamma \] valid in the open right half-plane, where \(\gamma\approx 0.577216\) is the Euler-Mascheroni constant.

Then

\[ 1+x \left(\frac{\Gamma^\prime(1+p(1- x))}{\Gamma(1+p (1-x))}-\frac{\Gamma^\prime(1+p x)}{\Gamma(1+p x)}\right)=1+x\int_0^1 \frac{t^{px}-t^{p(1-x)}}{1-t}dt=\int_0^1 I_p(t,x)dt=:{\mathcal{I}}_p(x), \] with \[I_p(t,x):=1+x\frac{t^{px}-t^{p(1-x)}}{1-t}\] for \(0<t<1\), \(0\le x\le 1\) and \(p\ge 2\).

** Fix \(p\ge 2\) in the rest of the proof, and for any \(0\le x\le 1\), let us extend the definition of \(I_p(t,x)\) for \(t=0\) and \(t=1\) as well: we set

\[I_p(0,x):=1\quad \mathrm{if }\ x\in [0,1),\quad I_p(0,1):=0,\quad \mathrm{and}\quad I_p(1,x):=1-p x(2x-1)\quad \mathrm{if}\ x\in [0,1].\]

We remark that \(I_p(t,0)=1\) (for all \(t\in (0,1)\)) and \(I_p(t,1)=1+\frac{t^p-1}{1-t}\) (for all \(t\in (0,1)\)). It is clear that the function \(I_p\) is continuous on the open unit square \((0,1)^2\), but it cannot be continuously extended to the closed unit square \([0,1]^2\): there is a problem near the corner point \((t,x)=(0,1)\).

Nevertheless, with the above extension it is seen (for example, also via l’Hospital’s rule) that \[ \lim_{t\to 0^+}I_p(t,x)=I_p(0,x)\quad \mathrm{for\ all}\ x\in \{0\}\cup (0,1)\cup\{1\}, \]

\[ \lim_{t\to 1^-}I_p(t,x)=I_p(1,x)\quad \mathrm{for\ all}\ x\in \{0\}\cup (0,1)\cup\{1\}, \]

\[ \lim_{x\to 0^+}I_p(t,x)=I_p(t,0)\quad \mathrm{for\ all}\ t\in \{0\}\cup (0,1)\cup\{1\}, \] but \[ \lim_{x\to 1^-}I_p(t,x)=I_p(t,1)\quad \mathrm{for\ all}\ t\in (0,1)\cup\{1\}. \]

From these we see that the function \(I_p\) is continuous, hence bounded, for example, on the compact rectangle \((t,x)\in [1/2,1]\times [0,1]\). On the other hand, for any \((t,x)\in (0,1/2]\times [0,1]\) we have \[ |I_p(t,x)|\le 1+\frac{x}{1-t}\left(t^{px}+t^{p(1-x)}\right)\le 1+\frac{1}{1/2}(1+1)=5. \] Finally, for any \(x\in [0,1]\) we see that \(|I_p(0,x)|\le 1\). We can thus conclude that the function \(|I_p|\) is bounded on the closed unit square \([0,1]^2\) by some positive constant \({\mathcal{C}}_p\).

Let us choose now an arbitrary sequence \(x_n\in [0,1]\) (\(n\in\mathbb{N}\)) converging to some \(x\in [0,1]\), and define the corresponding sequence of functions \(g_n(t):=I_p(t,x_n)\) (for \(0\le t \le 1\)) and \(g(t):=I_p(t,x)\) (for \(0\le t \le 1\)). Then \(g_n(\cdot)\) is continuous on \([0,1]\), and \(g_n(t)\to g(t)\) for \(t\in (0,1]\) as \(n\to +\infty\), moreover, \(|g_n(t)|\le {\mathcal{C}}_p\) for \(t\in [0,1]\) and \(n\in\mathbb{N}\), so according to the Dominated Convergence Theorem, \(\int_0^1 g_n(t)dt\to \int_0^1 g(t)dt\), in other words, the function

\[ [0,1]\ni x\mapsto {\mathcal{I}}_p(x) \]

is continuous.

** We show now that \({\mathcal{I}}_p(x)\ge 1\) if \(x\in\left[0,\frac{1}{2}\right]\):

\[ {\mathcal{I}}_p(x)=\int_0^1 1+\frac{x t^{p x}\left(1-t^{p (1-2 x)}\right) }{1-t}dt \ge \int_0^1 1+\frac{x t^{p x}\cdot 0 }{1-t}dt=1. \]

** Then we prove that \({\mathcal{I}}_p(\cdot)\) is strictly decreasing on \(\left[\frac{1}{2},1\right]\). Indeed, for any fixed \(0 < t < 1\), we compute

\[ \partial_x I_p(t,x) = \frac{t^{-p x} \left(t^{2 p x}+p x \ln (t) t^{2 p x}+p x t^p \ln (t)-t^p\right)}{1-t}. \]

Here \(t^{2 p x}+p x \ln (t) t^{2 p x}+p x t^p \ln (t)-t^p<0\) is equivalent to

\[ t^{p(2 x-1)}\left(1 +p x \ln(t)\right)< \left(1-p x \ln (t)\right), \]

which is easily verified, since the left hand-side is \(<1\), while the right hand-side is \(>1\) for any \(0<t<1\), \(1/2\le x \le 1\) and \(p\ge 2\). Thus for any \(1/2 \le x_1 < x_2 \le 1\) and for all \(0<t<1\), \(I_p(t,x_1)>I_p(t,x_2)\), so the continuous (in \(t\)) integrand is strictly decreased on the whole interval \(t\in (0,1)\), hence the integral is also strictly decreased: \({\mathcal{I}}_p(x_1)>{\mathcal{I}}_p(x_2)\).

** We easily establish that \({\mathcal{I}}_p(1/2)>0>{\mathcal{I}}_p(1)\):

\[ {\mathcal{I}}_p(1/2)=\int_0^1 1 dt =1, \] and \[ {\mathcal{I}}_p(1)=\int_0^1 1+ \frac{t^p-1}{1-t} dt =-\sum_{k=2}^p\frac{1}{k}. \]

** The above imply that \({\mathcal{I}}_p(\cdot)\) has a unique zero on \([0,1]\). Let us denote this by \(m_A(p)\). From the above, we also have that \(m_A(p)\in\left(\frac{1}{2},1\right)\) and that \({\mathcal{I}}_p(\cdot)\) is positive on \([0,m_A(p))\) and negative on \((m_A(p),1]\).

The proof of this claim is finished by recalling that

\[ A_p^\prime(x)=\frac{p^{p+1} x^{p-1}}{\Gamma(1+p x)\ \Gamma(1+p (1-x)) }{\mathcal{I}}_p(x), \]

and taking into account that the factor in front of \({\mathcal{I}}_p(x)\) is positive for all \(x\in(0,1)\).

\(\bullet\) \(A_p(x)>0\) if \(0<x\le 1\), and \(A_p(0)=0\). (The same holds for \(B_p\).) Moreover, \(A_p=B_p^p\), so if \(p\) is fixed, then the monotonicity properties of \(A_p(\cdot)\) and \(B_p(\cdot)\) are the same: the unimodality of \(A_p\) implies the unimodality of \(B_p\), and if \(m_A(p)\in\left(\frac{1}{2},1\right)\) denotes the abscissa of the maximum of \(A_p\) (see in the previous proof), then \(B_p\) also attains its maximum there.

\(\bullet\) Next we establish that the graphs of the functions \(B_p(\cdot)\) converge to a limit curve.

\(2^{\mathbf{nd}}\) claim. For any fixed \(0<x<1\), we have \[ \lim_{p\to +\infty} B_p(x)= e \left(\frac{1-x}{x}\right)^{x-1}, \] moreover, \(\lim_{p\to +\infty} B_p(0)=0=\lim_{x\to 0^+} e \left(\frac{1-x}{x}\right)^{x-1}\) and \(\lim_{p\to +\infty} B_p(1)=e=\lim_{x\to 1^-} e \left(\frac{1-x}{x}\right)^{x-1}\).

Proof. When \(0<x<1\), we apply Stirling’s formula to get

\[ \lim_{p\to +\infty} B_p(x)= \lim_{p\to +\infty} \frac{p x}{\sqrt[p]{(px/e)^{px}\sqrt{2\pi px}}\sqrt[p]{(p(1-x)/e)^{p(1-x)}\sqrt{2\pi p(1-x)}}}\sqrt[p]{\frac{(px/e)^{px}\sqrt{2\pi px}}{{\Gamma (1+p x)}}}\sqrt[p]{\frac{(p(1-x)/e)^{p(1-x)}\sqrt{2\pi p(1-x)}}{{\Gamma (1+p(1-x))}}}= \]

\[ \lim_{p\to +\infty} \frac{p x}{{(px/e)^{x}\cdot\sqrt[2p]{2\pi px}}\cdot {(p(1-x)/e)^{1-x}\cdot\sqrt[2p]{2\pi p(1-x)}}}=\frac{x}{(x/e)^{x}\cdot ((1-x)/e)^{1-x}} \lim_{p\to +\infty} \frac{p }{{p^{x}}\cdot {p^{1-x}}}= e \left(\frac{1-x}{x}\right)^{x-1}. \]

The endpoint cases (\(x=0\) and \(x=1\)) are treated similarly and easily.

\(3^{\mathbf{rd}}\) claim. The function \(x\mapsto e \left(\frac{1-x}{x}\right)^{x-1}\) is also unimodal on \(x\in [0,1]\).

Proof. Its derivative evaluated at \(x\in (0,1)\) is \[ e \left(\frac{1-x}{x}\right)^{x-1}\widetilde{\lambda}(x) , \] where, for \(0 < x < 1\), \[\widetilde{\lambda}(x):=\frac{1}{x}+\ln \left(\frac{1-x}{x}\right).\] We see that \[ \lim_{ 0^+}\widetilde{\lambda}=+\infty,\quad \lim_{ 1^-}\widetilde{\lambda}=-\infty \] and \(\widetilde{\lambda}\) is differentiable on \((0,1)\).

Since \[ \widetilde{\lambda}^\prime(x)=-\frac{1}{(1-x) x^2}<0, \] \(\widetilde{\lambda}\) is strictly decreasing on \((0,1)\). These imply that there is a unique zero, say \(L\), of \(\widetilde{\lambda}\) in \((0,1)\), and \(\widetilde{\lambda}\Big|_{(0,L)}>0\) and \(\widetilde{\lambda}\Big|_{(L,1)}<0\). (Moreover, since \(\widetilde{\lambda}(1/2)=2>0\), we also have that \(1/2<L<1\).)

\(\bullet\) If, as in the preceding proof, \(L\in(1/2,1)\) denotes the abscissa of the unique maximum of the function \(x\mapsto e \left(\frac{1-x}{x}\right)^{x-1}\), then we know that \[\widetilde{\lambda}(L)=\frac{1}{L}+\ln\left(\frac{1-L}{L}\right)=0,\] from which we express \(L\) as \[ L=\frac{1}{1+W\left(\frac{1}{e}\right)}. \]

\(\bullet\) Finally, we show that the abscissa of the maximum of \(B_p(\cdot)\), \(m_A(p)\) converges to \(L\) as \(p\to +\infty\). We formulate this result as a general result on unimodal functions: in this context, pointwise convergence of the functions implies convergence of the sequence of the zeros of the derivatives.

\(4^{\mathbf{th}}\) claim. Suppose that the unimodal functions \(\varphi_n:[0,1]\to\mathbb{R}\) converge to a unimodal function \(\varphi:[0,1]\to\mathbb{R}\) pointwise on \([0,1]\). Let \(m_{\varphi}(n)\in [0,1]\) denote the abscissa of the unique maximum of \(\varphi_n\), and \(m_{\varphi}\in [0,1]\) denote the abscissa of the unique maximum of \(\varphi\). Then \(m_{\varphi}(n) \to m_{\varphi}\) as \(n\to +\infty\).

Proof. Suppose, to the contrary, that there is a subsequence, say \(n_k\), such that \(m_\varphi(n_k)\) converges to some \([0,1]\ni \widetilde{m}_{\varphi}\ne m_{\varphi}\) as \(k\to +\infty\). We can suppose that \(m_{\varphi}<\widetilde{m}_{\varphi}\), the other case would be analogous. If we set \(m^*_{\varphi}:=\frac{m_{\varphi}+\widetilde{m}_{\varphi}}{2}\), then \(m_{\varphi} < m^*_{\varphi} < \widetilde{m}_{\varphi}\), and it follows from the definition of unimodality that \(\varphi(m_{\varphi})>\varphi(m^*_{\varphi})\).

Now we let \(\varepsilon:=\frac{\varphi(m_{\varphi})-\varphi(m^*_{\varphi})}{2}>0\). Due to pointwise convergence, there is an index \(k_1=k_1(\epsilon)\) such that for all \(k\ge k_1\) we have \(|\varphi_{n_k}(m_\varphi)-\varphi(m_\varphi)|<\epsilon\), implying \[\varphi_{n_k}(m_\varphi)>\frac{\varphi(m_\varphi)+\varphi(m^*_\varphi)}{2}=\varphi(m^*_\varphi)+\epsilon.\quad\quad\quad (*)\]

On the other hand, since \(m_\varphi(n_k)\to \widetilde{m}_{\varphi}\) as \(k\to +\infty\), there is an index \(k_2\) such that for all \(k\ge k_2\) we have \[ m_\varphi (n_k)\ge m^*_\varphi.\quad\quad\quad (**) \]

We set \(k_3:=\max(k_1,k_2)\). Then the unimodality of \(\varphi_{n_k}\) with ( ** ), further ( * ) imply that for any \(k\ge k_3\) \[ \varphi_{n_k}(m^*_\varphi) > \varphi_{n_k}(m_\varphi)>\varphi(m^*_\varphi)+\epsilon, \] contradicting to the fact that, due to pointwise convergence, \[ \varphi_{n_k}(m^*_\varphi)\to \varphi(m^*_\varphi)\quad\quad (k\to +\infty) \] should also be true.

Remark. A simple consequence of the above is the following: for any fixed \(0<x<1\), \(A_p(x)\) is asymptotically \[\frac{1}{2 \pi \sqrt{x(1-x)}}\frac{\left(e \left(\frac{x}{1-x}\right)^{1-x}\right)^p}{p}\] (as \(p\to+\infty\)). Now by substituting \(x=L=\frac{1}{1+W\left(\frac{1}{e}\right)}\), and using the identity \(\frac{1}{W\left(\frac{1}{e}\right)}=e^{W\left(\frac{1}{e}\right)+1}\) twice (in both directions), we recover the asymptotic formula for the maximum from the bottom of the “discrete page”:

\[ \frac{1+W\left(\frac{1}{e}\right)}{2 \pi \sqrt{W\left(\frac{1}{e}\right)}}\frac{\left(\frac{1}{W\left(\frac{1}{e}\right)}\right)^p}{p} \approx 0.385588\cdot \frac{3.59112^p}{p}. \]