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Real total derivatives transport through the complex coordinate isometry

Statement

Let $U \subseteq C$ be open in $d_{C}$ , let $z \in U$ , and let $f : U \to C$ . Define $f_{R} : = Φ \circ f \circ Φ^{- 1} : Φ [U] \to R^{2}$ . The set $Φ [U]$ is open by The complex coordinate isometry transports topology and affine segments. A map $L : C \to C$ is real-linear when it preserves addition and multiplication by embedded real scalars. Then $f_{R}$ is totally differentiable at $Φ (z)$ exactly when there is a real-linear $L$ such that, with $r (h) : = f (z + h) - f (z) - L (h)$ ,

$\frac{∣ r (h) ∣}{∣ h ∣} ⟶ 0$

as $h \to 0$ , $h \neq 0$ , and $z + h \in U$ . In that case

$D f_{R} (Φ (z)) = Φ \circ L \circ Φ^{- 1} .$

The map $L$ need not be complex-linear; this is a real total-derivative statement.

Facts & Assumptions

Given: The open set $U$ , the point $z \in U$ , the map $f$ , and the coordinate map $Φ$ .

[F1]

The complex coordinate isometry transports topology and affine segments and The complex coordinate map preserves real scalar multiplication: $Φ$ is a bijective real-linear isometry, and $Φ [U]$ is open.

[F2]

A linear map $L : R^{m} \to R^{n}$ in Euclidean coordinates: A linear map between Euclidean coordinate spaces preserves real linear combinations.

[F3]

The total (Fréchet) derivative $D f (a)$ as the linear first-order approximation with $o (∥ h ∥_{2})$ remainder: Total differentiability is a linear first-order approximation whose Euclidean remainder divided by the input norm tends to zero.

Proof

1.1givenF1algebrachoose

The inverse $Φ^{- 1}$ is real-linear: for arbitrary $u, v \in R^{2}$ , choose $ζ, ω \in C$ with $Φ (ζ) = u$ and $Φ (ω) = v$ , possible by [F1]. Additivity of $Φ$ gives $Φ^{- 1} (u + v) = Φ^{- 1} (Φ (ζ + ω)) = ζ + ω = Φ^{- 1} u + Φ^{- 1} v$ ; for every $λ \in R$ , its real homogeneity gives $Φ^{- 1} (λ u) = Φ^{- 1} (Φ (λ ζ)) = λ ζ = λ Φ^{- 1} u$ .

2.1givenF1F2step 1.1algebra

The correspondence $L \mapsto Φ \circ L \circ Φ^{- 1}$ sends each real-linear $L$ to a real-linear map $A$ by [F1], [F2], and step 1.1. Conversely, for each real-linear $A$ , the map $L = Φ^{- 1} \circ A \circ Φ$ is real-linear by the same facts, and the two constructions are inverse because $Φ^{- 1} \circ Φ$ and $Φ \circ Φ^{- 1}$ are identity maps. Thus this is a bijection between the real-linear maps on the two coordinate spaces.

3.1givenF1step 2.1algebra

For corresponding $L, A$ from step 2.1, put $k = Φ (h)$ . Additivity gives $Φ (z + h) = Φ (z) + k$ . If $r (h) = f (z + h) - f (z) - L (h)$ and $ρ (k) = f_{R} (Φ (z) + k) - f_{R} (Φ (z)) - A (k)$ , then the definition of $f_{R}$ and $A (k) = Φ (L (h))$ give $ρ (k) = Φ (r (h))$ . The isometry [F1] therefore gives $∥ ρ (k) ∥_{2} / ∥ k ∥_{2} = ∣ r (h) ∣ / ∣ h ∣$ for $h \neq 0$ ; the bijective isometry also sends $h \to 0$ exactly to $k \to 0$ and sends the condition $z + h \in U$ exactly to $Φ (z) + k \in Φ [U]$ . Hence the two remainder limits are equivalent.

4.1givenF3step 2.1step 3.1∎

By [F3], total differentiability of $f_{R}$ at $Φ (z)$ means that some real-linear $A$ has Euclidean remainder ratio tending to zero. Steps 2.1 and 3.1 identify such maps $A$ bijectively with real-linear $L$ and identify their remainder ratios, so this condition is equivalent to existence of the stated $L$ with the complex remainder ratio tending to zero. For each corresponding pair, the Euclidean derivative is $A = Φ \circ L \circ Φ^{- 1}$ .

Depends on

Used by

Dependency tree · two levels

22 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. Lebl, Guide to Cultivating Complex Analysis, Sections 2.1.2 and 2.2.2 (standard reference, not scraped)