I am reading through Wadim Zudilin's lecture notes and at the beginning of page 76 (i.e. in last part of Lemma 5.9.) things are getting really confusing :
- How $g(y) = g(0) − 2^{3/2}y^2 + O(y^3)$ is true?
- How each equality in $f(z-z_0) = f(0) + 2^{3/2}(z-z_0)^2 + O((z-z_0)^3) = g(0) − 2^{3/2}(z-z_0)^2 + O((z-z_0)^3)$ are true?
- How the maximum of $|e^{f(z)}|$ being equal to $e^{f(z_0)}$ implies $\displaystyle\lim_{n\to\infty} r_n^{\frac{1}{n+1}}=e^{2f(z_0)}$? There is a rule that huge integral reduced to this euality but I don't know what theorem it is!
It is easier to start by item 2.
First, note that the equality at the beginning of page 76 is:
$$f(z-z_0) = f(z_0) + 2^{3/2}(z-z_0)^2 + O((z-z_0)^3) = g(0) + 2^{3/2}(z-z_0)^2 + O((z-z_0)^3)$$
Let us prove it directly. We know (page 75) that
$$f(z) = (1-z) \ln (1-z)+ 2z\ln z - (1+z)\ln (1+z)$$
and $z_0=\frac{1}{\sqrt{2}}$.
So, $f(z)$ is holomorphic in an open neighborhood of $z_0$. So, we have \begin{align*} f'(z)& = - \ln (1-z) - 1 + 2 \ln z +2 -\ln (1+z) -1 = \\ & = \ln \left(\frac{z^2}{1-z^2} \right)= \\ &= - \ln (z^{-2}-1) \tag{1} \end{align*} For $z= z_0=\frac{1}{\sqrt{2}}$, we have $$f'(z_0)= - \ln (2-1) = - \ln (1)=0 \tag{2}$$
Now let us calculate $f''(z)$. Since $$ f'(z)= - \ln (z^{-2}-1)$$ we have \begin{align*} f''(z)& = - \frac{1}{z^{-2}-1}(-2 z^{-3}) \\ & = \frac{2}{z(1-z^2)} \tag{3} \end{align*} For $z= z_0=\frac{1}{\sqrt{2}}$, we have $$f''(z_0)= 4 \sqrt{2} = 2 \cdot 2^{3/2} \tag{4}$$
From $(2)$ and $(4)$ and the fact that $f(z)$ is analytic in an open neighborhood of $z_0$, we have that
$$f(z-z_0) = f(z_0) + 2^{3/2}(z-z_0)^2 + O((z-z_0)^3)$$
Now, since $g(y)= \operatorname{Re} f(z_0+i y)$ and $f(z_0)$ is real, we have $g(0) = \operatorname{Re} f(z_0)= f(z_0)$, So we have $$f(z-z_0) = f(z_0) + 2^{3/2}(z-z_0)^2 + O((z-z_0)^3) = g(0) + 2^{3/2}(z-z_0)^2 + O((z-z_0)^3)$$
Now, let us prove item 1. Given any holomorphic function $f$, it is easy to see that:
$$ \frac{\partial \operatorname{Re} f }{\partial y} = - \operatorname{Im} f' \tag{5}$$ and $$ \frac{\partial \operatorname{Im} f }{\partial y} = \operatorname{Re} f' \tag{6}$$
So, let $G(x,y) = \operatorname{Re} f(x+iy)$. It is easy to see, using $(5)$, that $$ \frac{\partial G}{\partial y}(x,y) = - \operatorname{Im} f'(x+iy) \tag{7} $$ and considering $H(x,y) = \operatorname{Im} f'(x+iy)$, we have, using $(6)$, that $$ \frac{\partial H}{\partial y}(x,y) = \operatorname{Re} f''(x+iy) $$ which means $$ \frac{\partial^2 G}{\partial y^2}(x,y) = - \operatorname{Re} f''(x+iy) \tag{8}$$ Applying $(7)$ to $z_0$ (that means $x=\frac{1}{\sqrt{2}}$ and $y=0$), we have, by $(2)$,
$$ \frac{\partial G}{\partial y}\left (\frac{1}{\sqrt{2}},0 \right) = - \operatorname{Im} f'(z_0)= 0$$ Applying $(8)$ to $z_0$ (that means $x=\frac{1}{\sqrt{2}}$ and $y=0$), we have, by $(4)$, $$ \frac{\partial^2 G}{\partial y^2}\left (\frac{1}{\sqrt{2}},0 \right) = - \operatorname{Re} f''(z_0) = -2 \cdot 2^{3/2} $$ Since $g(y) = G \left (\frac{1}{\sqrt{2}},y \right)$, it follows immediately that $$ \frac{\operatorname{d} g}{\operatorname{d} y }(0) = \frac{\partial G}{\partial y}\left (\frac{1}{\sqrt{2}},0 \right) = 0$$ $$ \frac{\operatorname{d}^2 g}{\operatorname{d} y^2 }(0) = \frac{\partial^2 G}{\partial y^2}\left (\frac{1}{\sqrt{2}},0 \right) = -2 \cdot 2^{3/2}$$ So we have $g(y) = g(0) − 2^{3/2}y^2 + O(y^3)$.
Now, item 3, it is a consequence of $$r_n=\frac{2 \pi}{n i}\int_{z_0-i\infty}^{z_0+i\infty}e^{2(n+1)f(z)}\frac{1+z}{z^2(1 -z)}(1+O(n^{-1}))dz$$ and that fact that $|e^{f(z)}|$ has its maximum at $z=z_0$ and it is equal to $e^{f(z_0)}$.
In fact, we have $$ r_n^{\frac{1}{n+1}} = e^{2f(z_0)} \frac{(2 \pi)^{\frac{1}{n+1}}}{n^{\frac{1}{n+1}} i^{\frac{1}{n+1}}}\left (\int_{z_0-i\infty}^{z_0+i\infty}\left (\frac{e^{f(z)}}{e^{f(z_0)}}\right )^{2(n+1)}\frac{1+z}{z^2(1 -z)}(1+O(n^{-1}))dz \right)^{\frac{1}{n+1}} \tag{9}$$
It is easy to see that \begin{align*} \lim_{n \to \infty} & \left (\int_{z_0-i\infty}^{z_0+i\infty}\left (\frac{e^{f(z)}}{e^{f(z_0)}}\right )^{2(n+1)}\frac{1+z}{z^2(1 -z)}(1+O(n^{-1}))dz \right)^{\frac{1}{n+1}} = \\ &= \lim_{n \to \infty} \left (\int_{z_0-i\infty}^{z_0+i\infty}\left (\frac{e^{f(z)}}{e^{f(z_0)}}\right )^{2(n+1)}\frac{1+z}{z^2(1 -z)}dz \right)^{\frac{1}{n+1}} \tag{10} \end{align*}
Now, let $X=\{z_0+iy : y \in \Bbb R\}$ and let $\mu$ be the complex (finite) measure define on the Borel subsets of $X$ by $$\mu(E) = \int_E \frac{1+z}{z^2(1 -z)}dz$$.
Now, using that $|e^{f(z)}|$ has its maximum at $z=z_0$ and it is equal to $e^{f(z_0)}$, note that the function $u$, defined by $ u(z)= \left (\frac{e^{f(z)}}{e^{f(z_0)}}\right )^{2}$ is in $L^\infty(X,\mu)$ and $\|u\|_\infty=1$.
So, \begin{align*} \lim_{n \to \infty} \left (\int_{z_0-i\infty}^{z_0+i\infty}\left (\frac{e^{f(z)}}{e^{f(z_0)}}\right )^{2(n+1)}\frac{1+z}{z^2(1 -z)}dz \right)^{\frac{1}{n+1}} & = \lim_{n \to \infty} \|u\|_{n+1}= \\ &= \lim_{n \to \infty} \|u\|_n = \|u\|_\infty=1 \tag{11} \end{align*} (see Remark). So, from $(9)$, $(10)$ and $(11)$, we have that $\displaystyle\lim_{n\to\infty} r_n^{\frac{1}{n+1}}=e^{2f(z_0)}$.
Remark: For $(11)$ we used the following lemma:
Proof: First, suppose $\mu$ is a finite positive measure. If $\|f\|_\infty=0$, then the lemma is trivially true. Suppose $\|f\|_\infty>0$. Then, given any $\delta$ such that $0<\delta < \|f\|_\infty$, let $X_\delta:=\{x,|f(x)|\geqslant \lVert f\rVert_\infty-\delta\}$. We have $$\| f\|_p\geqslant \left(\int_{X_\delta}(\| f\|_\infty-\delta)^pd\mu\right)^{1/p}=(\| f\|_\infty-\delta)\mu(X_\delta)^{1/p},$$ So $$\liminf_{p\to \infty}\| f\|_p\geqslant\| f\|_\infty - \delta$$ Since this is true for any $\delta$, $0<\delta < \|f\|_\infty$, we have, $$\liminf_{p\to \infty}\| f\|_p\geqslant\| f\|_\infty $$
On the other hand, since $\mu$ is finite, $f \in L^1$, and since $|f(x)|\leqslant\| f\|_\infty$ for almost every $x$, we have for $p>1$, $$ \| f\|_p\leqslant\left(\int_X|f(x)|^{p-1}|f(x)|d\mu\right)^{1/p}\leqslant \| f\|_\infty^{\frac{p-1}p}\| f\|_1^{1/p}$$ So $$\limsup_{p\to \infty}\| f\|_p\leqslant\| f\|_\infty $$ So $$\lim_{p\to \infty}\| f\|_p = \| f\|_\infty $$
To complete the proof, note that the case where $\mu$ is a complex (finite) measure is a direct consequence of the case above.