Conditions for the Converse of the Cauchy-Riemann Equations to Hold
Theorem
Suppose a complex function $f: A \to \mathbb{C}$ defined on a complex domain $A \subseteq \mathbb{C}$ can be expressed, for real-valued functions $u,v$, as $$ f(z) = f(x+iy) = u(x,y) + iv(x,y) $$ and $u,v$ have continuous first-order partial derivatives with respect to $x,y$ while also satisfying the system of differential equations $$ \begin{cases} u_{x} (x,y) = v_{y} (x,y) \\ u_{y} (x,y) = -v_{x} (x,y) \end{cases} $$ then $f$ is analytic on $A$.
Explanation
The problem with analysis is that statements are always this long-winded, making them unpleasant to read. To summarize briefly, for the converse of the Cauchy-Riemann equations to hold, the partial derivatives must be continuous. Naturally, most of the functions we deal with easily satisfy this condition.
Proof
Part 1. The form of $u_{x}, u_{y}$
Without loss of generality, for two real numbers $\alpha , \beta > 0$, let $h := \alpha + i \beta$; then for $f$ we obtain $$ f(z+h) - f(z) = [ u(x+\alpha,y+\beta) - u(x,y) ] + i [ v(x+\alpha,y+\beta) - v(x,y) ] $$ Likewise, without loss of generality, considering only $u$, by the mean value theorem there exist $0<\theta<1$ and $0<\phi<1$ satisfying $$ \begin{align*} & u(x+\alpha,y+\beta) - u(x,y) \\ =& u(x+\alpha,y+\beta) - u(x,y+\beta) + u(x,y+\beta) - u(x,y) \\ =& [u(x+\alpha,y+\beta) - u(x,y+\beta)] + [u(x,y+\beta) - u(x,y)] \\ =& \alpha u_{x} (x+\theta \alpha,y+\beta) + \beta u_{y} (x,y+\phi \beta) \end{align*} $$
Mean value theorem: If a function $f(x)$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists at least one $c$ in $(a,b)$ satisfying $\displaystyle f '(c)={{f(b)-f(a)}\over{b-a}}$.
Here, the reason we may set $\theta \in (0,1)$ is that when the mean value theorem is applied to $u \left( x , y + \beta \right)$ on $\left[ x , x + \alpha \right]$, as in $$ \begin{align*} {{ \partial u } \over { \partial x }} \left( c , y + \beta \right) =& {{ u(x+\alpha,y+\beta) - u(x,y+\beta) } \over { \left( x + \alpha \right) - x }} \\ \implies \alpha u_{x} \left( c , y + \beta \right) =& u(x+\alpha,y+\beta) - u(x,y+\beta) \end{align*} $$ the number $c \in \left( x , x + \alpha \right)$, lying between $x$ and $\left( x + \alpha \right)$, can be written as $x + \theta \alpha$.
Part 2. The form of $f ' (z)$
Now, for some $\varepsilon_{1}$ and $\varepsilon_{2}$, set $$ \begin{align*} u_{x} (x+\theta \alpha,y+\beta) =& u_{x} + \varepsilon_{1} \\ u_{y} (x,y+\phi \beta) =& u_{y} + \varepsilon_{2} \end{align*} $$ Since we assumed that $u_{x}$ and $u_{y}$ are continuous, as $(\alpha,\beta) \to 0$ we will have $\varepsilon_{1}, \varepsilon_{2} \to 0$. (This is why continuity is needed.) Therefore $$ u(x+\alpha,y+\beta) - u(x,y) = \alpha u_{x} + \beta u_{y} + \alpha \varepsilon_{1} + \beta \varepsilon_{2} $$ and, in the same way, for $v$ we can likewise set, for some $\eta_{1}$ and $\eta_{2}$, $$ v(x+\alpha,y+\beta) - v(x,y) = \alpha v_{x} + \beta v_{y} + \alpha \eta_{1} + \beta \eta_{2} $$ Returning to $f(z+h) - f(z)$, since $u,v$ satisfy the Cauchy-Riemann equations, $$ \begin{align*} & f(z+h) - f(z) \\ =& [ u(x+\alpha,y+\beta) - u(x,y) ] + i [ v(x+\alpha,y+\beta) - v(x,y) ] \\ =& [\alpha u_{x} + \beta u_{y} + \alpha \varepsilon_{1} + \beta \varepsilon_{2}] + i [\alpha v_{x} + \beta v_{y} + \alpha \eta_{1} + \beta \eta_{2}] \\ =& h(u_{x} + i v_{x}) + \alpha \xi_{1} + \beta \xi_{2} \end{align*} $$ where $\xi_{1} := \varepsilon_{1} + \eta_{1}$ and $\xi_{2} := \varepsilon_{2} + \eta_{2}$. Now, since $$ f ' (z) = \lim_{h \to 0} {{f(z+h) - f(z)} \over h} = \lim_{h \to 0} \left( u_{x} + i v_{x} + {{\alpha \xi_{1} + \beta \xi_{2}} \over h} \right) $$ it suffices to show that $\lim_{h \to 0} {{\alpha \xi_{1} + \beta \xi_{2}} \over h} = 0$ to complete the proof.
Part 3. $\displaystyle \lim_{h \to 0} {{\alpha \xi_{1} + \beta \xi_{2}} \over h} = 0$
From the inequality $$ \left| {{\alpha \xi_{1} + \beta \xi_{2}} \over {h}} \right| \le { {\max (|\alpha|,|\beta|)} \over {\sqrt{\alpha^2 + \beta^2}} } |\xi_{1} + \xi_{2}| \le |\xi_{1} + \xi_{2}| \le |\xi_{1}| + |\xi_{2}| $$ since $$ \lim_{h \to 0} \xi_{1} = 0 \\ \lim_{h \to 0} \xi_{2} = 0 $$ the following holds. $$ \lim_{h \to 0} {{\alpha \xi_{1} + \beta \xi_{2}} \over h} = 0 $$
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Renewal
- August 19, 2023, 류대식, substantially reinforced the content related to the mean value theorem