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If $g(w)$ and $f(z)$ are holomorphic functions, show that $g(f(z))$ is also holomorphic.

From the assumptions we have that for some $A,B$,

$$\lim_{h\rightarrow 0}\frac{g(f(z)+h)-g(f(z))}{h}=A,$$

$$\lim_{h\rightarrow 0}\frac{f(z+h)-f(z)}{h}=B.$$

We want to prove that there exists $C$ with

$$\lim_{h\rightarrow 0}\frac{g(f(z+h))-g(f(z))}{h}=C.$$

The second equation means, by definition of the limit, that for any $\varepsilon > 0$, there exists $\delta$ such that $0<|h|<\delta$ implies $$\left|\frac{f(z+h)-f(z)}{h}-B\right|<\varepsilon.$$

I don't know how to get to $\dfrac{g(f(z+h))-g(f(z))}{h}$ from here.

PJ Miller
  • 8,193

1 Answers1

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I like the denominator-free version. So let $z_0$ arbitrary in the domain of $f$ and $w_0 = f(z_0)$.

That $g$ is complex differentiable in $w_0$ means there exists a function $\Delta_g$, which is continuous in $w_0$, such that for all $w$ in the domain of $g$ you have

$$g(w) = g(w_0) + \Delta_g(w)\cdot (w-w_0).$$

Similarly, the complex differentiability of $f$ in $z_0$ means there is a function $\Delta_f$, continuous in $z_0$, such that

$$f(z) = f(z_0) + \Delta_f(z)\cdot (z-z_0).$$

Thus we find

$$\begin{align} g(f(z)) &= g(f(z_0)) + \Delta_g(f(z))\cdot \bigl(f(z) - f(z_0)\bigr)\\ &= g(f(z_0)) + \Delta_g(f(z))\cdot\Delta_f(z)\cdot (z-z_0). \end{align}$$

Since $\Delta_g$ is continuous in $w_0 = f(z_0)$ and $f$ is continuous in $z_0$, the function $\Delta_g \circ f$ is continuous in $z_0$. $\Delta_f$ is also continuous in $z_0$, hence

$$\Delta_{g\circ f} \colon z \mapsto \Delta_g(f(z))\cdot \Delta_f(z)$$

is continuous in $z_0$ and

$$(g\circ f)(z) = (g\circ f)(z_0) + \Delta_{g\circ f}(z)\cdot (z-z_0),$$

and that means $g\circ f$ is complex differentiable in $z_0$.

Since $z_0$ was arbitrary, $g\circ f$ is holomorphic for $g$ and $f$ holomorphic.

Daniel Fischer
  • 206,697