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On a convex open set the difference quotient is an average of the derivative along the segment

Statement

Let $V \subseteq C$ be open and convex (A convex subset of $R^{m}$ contains every line segment between two of its points) and let $f : V \to C$ be holomorphic. Then for all $z, w \in V$

$f (w) - f (z) = (w - z) \int_{0}^{1} f^{'} (z + t (w - z)) d t,$

the integral being the componentwise integral of a continuous $R^{2}$ -valued function of $t$ (The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral). In particular, for $w \neq z$ ,

$\frac{f (w) - f (z)}{w - z} = \int_{0}^{1} f^{'} (z + t (w - z)) d t,$

while for $w = z$ the displayed integral equals $f^{'} (z)$ and both sides of the first identity are $0$ .

Facts & Assumptions

Given: An open convex $V \subseteq C$ , a holomorphic $f : V \to C$ and points $z, w \in V$ ; write $Φ (a + b i) = (a, b)$ , and read convexity as that of $Φ [V]$ in A convex subset of $R^{m}$ contains every line segment between two of its points. The bijective isometry $Φ$ preserves affine segments by The complex coordinate isometry transports topology and affine segments, so $(1 - t) z + t w \in V$ for $0 \leq t \leq 1$ .

[L1]

If $F$ is a primitive of a continuous $f$ on an open set containing the trace of a rectifiable contour $γ : [a, b] \to C$ and $F^{'} = f$ is continuous, then $\int_{γ} f (z) d z = F (γ (b)) - F (γ (a))$ (The line integral of a continuous function admitting a primitive is that primitive's endpoint increment along every rectifiable path, A primitive of a complex function on an open set).

[L2]

For a piecewise- $C^{1}$ contour $γ$ and $f$ continuous on its trace, $\int_{γ} f (z) d z = \sum_{j} \int_{t_{j}}^{t_{j + 1}} f (γ (t)) γ_{j}^{'} (t) d t$ (For piecewise-C1 contours the Riemann–Stieltjes integral agrees with the parametric complex integral and the published real line integrals).

[L3]

A holomorphic $f = u + i v$ on an open subset of $C$ has $(u, v)$ of class $C^{k}$ for every natural $k$ , hence smooth (Holomorphic functions are real analytic and smooth in their two real coordinates), and every holomorphic function has complex derivatives of every natural order (All higher complex derivatives exist and satisfy Cauchy's integral formula on an interior circle).

[L4]

A subset $U \subseteq R^{m}$ is convex when $(1 - t) x + t y \in U$ for all $x, y \in U$ and $t \in [0, 1]$ (A convex subset of $R^{m}$ contains every line segment between two of its points).

[L5]

Integrals of $R^{m}$ -valued functions are taken componentwise and are real-linear (The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral); the real integral is linear in the integrand (Integrable functions on $[a, b]$ form a set closed under sums and scalar multiples, and $\int_{a}^{b} (λ f + μ g) = λ \int_{a}^{b} f + μ \int_{a}^{b} g$ ).

[L6]

A function complex differentiable at a point is continuous there (Complex differentiability at a point implies continuity there).

[L7]

A continuous path differentiable with a continuous derivative on each piece of a partition is rectifiable (A continuous piecewise- $C^{1}$ path is rectifiable and its length is the sum of the speed integrals over its pieces).

Proof

technique · direct

1.1givenL1L3L6

By [L3] the derivative $f^{'}$ is again holomorphic on $V$ , hence continuous there by [L6], so $f$ is a primitive of the continuous $f^{'}$ with continuous $F^{'}$ in the sense of [L1].

1.2givenL4L7

The map $ℓ (t) = z + t (w - z)$ on $[0, 1]$ has values in $V$ by [L4], since $V$ is convex, and is differentiable with the constant continuous derivative $w - z$ , so it is a piecewise- $C^{1}$ , hence rectifiable, contour with trace in $V$ ([L7]); its endpoints are $ℓ (0) = z$ and $ℓ (1) = w$ . In real coordinates the path is $Φ (ℓ (t)) = Φ (z) + t Φ (w - z)$ , so this real derivative and the complex difference quotient both give the same constant velocity $w - z$ .

2.1step 1.1step 1.2L1

By [L1] applied to $ℓ$ , $\int_{ℓ} f^{'} (ζ) d ζ = f (ℓ (1)) - f (ℓ (0)) = f (w) - f (z)$ .

2.2step 1.1step 1.2L2L5algebra

By [L2] applied to $ℓ$ , whose derivative is the constant $w - z$ , $\int_{ℓ} f^{'} (ζ) d ζ = \int_{0}^{1} f^{'} (z + t (w - z)) (w - z) d t$ , and multiplication by $w - z$ is the fixed real-linear coordinate map $(x, y) \mapsto (Re (w - z) x - Im (w - z) y, Im (w - z) x + Re (w - z) y)$ ; applying [L5] to its two real components therefore pulls this constant out of the integral, so this equals $(w - z) \int_{0}^{1} f^{'} (z + t (w - z)) d t$ .

3.1step 2.1step 2.2L5algebra∎

Steps 2.1 and 2.2 give $f (w) - f (z) = (w - z) \int_{0}^{1} f^{'} (z + t (w - z)) d t$ ; dividing by $w - z$ when $w \neq z$ gives the difference-quotient form, and when $w = z$ the integrand is the constant $f^{'} (z)$ , whose integral over $[0, 1]$ is $f^{'} (z)$ by [L5], while both sides of the first identity are $0$ .

Depends on

Dependency tree · 2 levels

Sources

M. Weber, Complex Analysis (Indiana University), Ch. 4 §4.1