Termwise differentiation | Complex Analysis | Mathematics

Theorem 6.1 (Termwise differentiation). Suppose \(f(z) = \sum _{n=0}^{\infty } a_n (z - z_0)^n\) has radius of convergence \(R > 0\). Then \(f\) is analytic on \(\abs {z - z_0} < R\), and

\[ f'(z) = \sum _{n=1}^{\infty } n a_n (z - z_0)^{n-1}, \]

where the differentiated series also has radius of convergence \(R\).

Source: [1, ch. 5]

Proof. Translating, we may take \(z_0 = 0\).

Same radius. Since \(n^{1/n} \to 1\),

\[ \limsup _{n \to \infty } \abs {n a_n}^{1/n} = \limsup _{n \to \infty } \abs {a_n}^{1/n}, \]

so \(\sum n a_n z^{n-1}\) has radius \(R\) by Theorem 5.1. Similarly, \(\sum n(n-1) \abs {a_n} r^{n-2}\) converges for every \(0 \leq r < R\).

The derivative. Fix \(\abs {z} < R\), choose \(r\) with \(\abs {z} < r < R\), and let \(g(z) = \sum n a_n z^{n-1}\). For \(\abs {w} < r\) with \(w \neq z\),

\[ \frac {f(w) - f(z)}{w - z} - g(z) = \sum _{n=2}^{\infty } a_n \left ( \frac {w^n - z^n}{w - z} - n z^{n-1} \right ). \]

Expanding \(\frac {w^n - z^n}{w - z} = \sum _{j=0}^{n-1} w^j z^{n-1-j}\), the bracket equals \(\sum _{j=1}^{n-1} z^{n-1-j} (w^j - z^j)\). Since \(\abs {w^j - z^j} \leq j r^{j-1} \abs {w - z}\), the bracket is at most

\[ \abs {w - z} \, r^{n-2} \sum _{j=1}^{n-1} j = \abs {w - z} \, \frac {n(n-1)}{2} \, r^{n-2} \]

in modulus. Hence

\[ \absStar{\frac {f(w) - f(z)}{w - z} - g(z)} \leq \frac {\abs {w - z}}{2} \sum _{n=2}^{\infty } n(n-1) \abs {a_n} r^{n-2} \longrightarrow 0 \quad \text {as } w \to z. \]

So \(f'(z) = g(z)\) exists at every point of the disc. That is, \(f\) is analytic there.