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FALSE: existence of partial derivatives satisfying Cauchy–Riemann everywhere on an open set implies holomorphy

Statement

False claim: if all four first coordinate partial derivatives of $f = u + i v$ exist at every point of an open set and satisfy the Cauchy–Riemann equations there, then $f$ is holomorphic on that set.

Facts & Assumptions

Given: The function on $C$ $f (0) = 0, f (z) = \exp (- z^{- 4}) (z \neq 0) .$

[L1]

The complex exponential is entire with derivative itself (The complex exponential is entire and its complex derivative is itself).

[L2]

Complex differentiation is linear and satisfies the product rule; where $g (a) \neq 0$ the reciprocal $1 / g$ is complex differentiable at $a$ with $(1 / g)^{'} = - g^{'} / g^{2}$ (Linearity, product, reciprocal, and quotient rules for complex derivatives); and a composite of complex differentiable maps is complex differentiable (The chain rule for complex derivatives). Iterating the product rule makes $z \mapsto z^{4}$ complex differentiable, so the reciprocal rule makes $z \mapsto z^{- 4}$ complex differentiable wherever $z \neq 0$ . No general complex-exponent power rule is used.

[L3]

For every natural $m$ and real $a > 0$ , $x^{m} / \exp (a x) \to 0$ as $x \to + \infty$ (The exponential dominates every fixed nonnegative integer power at $+ \infty$ ).

[L4]

Complex differentiability implies the Cauchy–Riemann equations, and it also implies continuity at the point (Complex differentiability is equivalent to real total differentiability together with a complex-linear derivative, with $\partial_{\overset{ˉ}{z}} f = 0$ , or with the Cauchy–Riemann equations, Complex differentiability at a point implies continuity there).

Refutation

technique · direct counterexample

1.1

On $C ∖ {0}$ , the power, reciprocal, exponential, and chain rules show that $f$ is holomorphic. Consequently all four partials exist and satisfy the Cauchy–Riemann equations there by [L1], [L2], and [L4].

L1L2L4

1.2

For nonzero real $t$ , both $t^{4}$ and $(i t)^{4}$ equal $t^{4}$ , so $f (t) = f (i t) = e^{- 1 / t^{4}} \in R .$

givenalgebra

1.3

Put $x = 1 / ∣ t ∣$ . As $t \to 0$ , $x \to + \infty$ ; for $x \geq 1$ , $0 \leq x e^{- x^{4}} \leq x e^{- x}$ , and [L3] with $m = 1$ , $a = 1$ makes the last expression tend to $0$ . Hence $e^{- 1 / t^{4}} / ∣ t ∣ \to 0$ .

L3algebra

1.4

Along $z = t (1 + i)$ with nonzero real $t$ , one has $z^{4} = - 4 t^{4}$ , and therefore $f (t (1 + i)) = e^{1 / (4 t^{4})} ⟶ + \infty (t \to 0) .$ So $f$ is unbounded in every neighbourhood of $0$ and is not continuous there.

givenalgebra

2.1

Steps 1.2–1.3 give $u_{x} (0) = u_{y} (0) = v_{x} (0) = v_{y} (0) = 0$ . Thus all four partials exist at $0$ and satisfy both Cauchy–Riemann equations there. Together with step 1.1, the false claim's hypotheses hold throughout the open set $C$ .

step 1.1step 1.2step 1.3

3.1

By [L4], the discontinuity in step 1.4 rules out complex differentiability at $0$ . Hence $f$ satisfies Cauchy–Riemann everywhere but is not holomorphic on $C$ , refuting the claim.

step 2.1step 1.4L4

4.1

There is also a Wirtinger warning. Off $0$ , $f$ is holomorphic, so conjugating its Cauchy–Riemann equations gives $(\overset{ˉ}{f})_{z} = 0$ ; at $0$ , step 2.1 gives the same value. Thus $(\overset{ˉ}{f})_{z}$ is identically zero and continuous. Nevertheless, off $0$ the chain rule gives $f^{'} (z) = 4 z^{- 5} e^{- z^{- 4}}$ , whose modulus along $z = t (1 + i)$ tends to infinity, so the coordinate partials of $f$ and $\overset{ˉ}{f}$ are not continuous at $0$ .

step 1.1step 2.1step 1.4L2L4algebra∎

Depends on

Used by

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Sources

J. Lebl, Guide to Cultivating Complex Analysis, Exercise 2.2.4 (standard reference, not scraped)