Schwarzian derivative as the derivative of the logarithm.

Question

There is one interesting fact in Wikipedia article on Schwarzian derivative, namely, if $F(z, w)=\log{\frac{f(z)-f(w)}{z-w}}$ then (sign error fixed) $6\frac{\partial^2 F(z, w)}{\partial z \partial w}|_{z=w} = (Sf)(w).$ I cannot see it. We have $\frac{\partial^2 F(z, w)}{\partial z \partial w} = \frac{f'(w)f'(z)}{(f(z)-f(w))^2}-\frac{1}{(z-w)^2}.$ When $|z-w|$ is small both of summands are large so none of them has no limit but their difference has. I brought them to a common denominator, but nothing happened. How to compute this limit?

Answer 1

This is “just” an exercise in developing Taylor series.

Consider a fixed point $w$ where $f(w) \ne \infty$ and $f'(w) \ne 0$. Then $$ f(z) = a_0 + a_1(z-w) + a_2(z-w)^2 + a_3 (z-w)^3 + \ldots $$ where $a_n = \frac{f^{(n)}(w)}{n!}$, and $\ldots$ indicates “higher-order terms of $(z-w)$” to shorten the notation.

Now $$ \frac{1}{f(z)-f(w)} = \frac{1}{a_1(z-w) + a_2(z-w)^2 + a_3 (z-w)^3 + \ldots} \\ = \frac{1}{a_1(z-w)(1 + \frac{a_2}{a_1}(z-w)+ \frac{a_3}{a_1}(z-w)^2 + \ldots)} \\ = \frac{1}{a_1(z-w)} \left( 1 - \frac{a_2}{a_1}(z-w) + \bigl( \frac{a_2^2}{a_1^2} - \frac{a_3}{a_1} \bigr)(z-w)^2 + \ldots \right) $$ and therefore $$ \frac{1}{(f(z)-f(w))^2} = \frac{1}{a_1^2(z-w)^2} \left( 1 - 2\frac{a_2}{a_1}(z-w) + \bigl( 3\frac{a_2^2}{a_1^2} - 2\frac{a_3}{a_1} \bigr)(z-w)^2 + \ldots \right) $$ On the other hand, $$ f'(w)f'(z) = a_1 (a_1 + 2a_2(z-w) + 3a_3(z-w)^2 + \ldots) \\ = a_1^2 (1 + 2\frac{a_2}{a_1}(z-w) + 3\frac{a_3}{a_1}(z-w)^2 + \ldots) $$ so that $$ \frac{f'(w)f'(z)}{(f(z)-f(w))^2} = \frac{1}{(z-w)^2} \left( 1 + \bigl( -\frac{a_2^2}{a_1^2} +\frac{a_3}{a_1} \bigr)(z-w)^2 + \ldots \right) = \frac{1}{(z-w)^2} + \bigl( -\frac{a_2^2}{a_1^2} +\frac{a_3}{a_1} \bigr) + \ldots $$ which gives the desired $$ \lim_{z \to w} \left( \frac{f'(w)f'(z)}{(f(z)-f(w))^2} - \frac{1}{(z-w)^2} \right) = \frac{a_3}{a_1} - \frac{a_2^2}{a_1^2} = \frac 16 \frac{f'''(w)}{f'(w) } - \frac 14 \left(\frac{f''(w)}{f'(w)}\right)^2 = \frac 16 (Sf)(w) $$

Finally, due to the identity theorem, this relation holds everywhere and not only at points where $f'(w) \ne 0$.

Answer 2

Let us take a simple example, namely let $\;f(z):=z^3.\;$ Now $\;F(z, w):=\log{\frac{f(z)-f(w)}{z-w}}$ and in our example is $\;\log(w^2+wz+z^2).\;$ Computing $\;\frac{\partial^2 F(z, w)}{\partial z \partial w}= -\frac{w^2+4wz+z^2}{(w^2+wz+z^2)^2}\;$ noticing the minus sign. Letting $\;w=z\;$ simplifies to $\;-2/{3z^2}\;$ and $6$ times that is $\;-4/z^2,\;$ which is exactly $\;(Sf)(z).\;$

In general, let $\;f(z):=a_0+a_1 z+a_2 z^2 + a_3 z^3+O(z)^4.\;$ Let $\;z=tx,\;w=ux.\;$ Then $\;F(z,w) = \log(a_1) + a_2(t+u)x/a_1 + O(x)^2,\;$ $\;\frac{\partial^2 F(z, w)}{\partial z \partial w} = (a_1a_3-a_2^2)/a_1^2 + O(x),\;$ $\;\frac{\partial^2 F(z, w)}{\partial z \partial w}|_{w=z} = \frac{a_1a_3-a_2^2}{a_1^2}+O(z),\;$ and $\;(Sf)(z) = 6\frac{a_1a_3-a_2^2}{a_1^2}+O(z),\;$ but this only shows the two quantities agree up to $O(z)$.

A real proof is $\;\frac{\partial^2 F(z, w)}{\partial z \partial w}=\frac{f'(w)f'(z)}{f(w)-f(z))^2}-\frac1{(w-z)^2}.\;$ Let $t:=w-z$ and expand the expression as a power series in $t$ resulting in $\;\frac{2f'(z)f'''(z)-3f'(z)^2}{12f'(z)^2}+O(t).\;$ In the limit as $t\to0$ this is $((Sf)(z))/6.$