3.4 A Cauchy-Riemann Converse

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The Cuachy-Riemann equations tell us how the partial derivatives of $u$ and $v$ interact when $f=u+iv$ is holomorphic. A natural question is to ask whether these equations can guarantee that $f$ is holomorphic. The following example shows that the answer will not be an unqualified "yes".

Example

Define $f:\mathbb{C}\to \mathbb{C}$ by $$f(z)=\{\begin{array}{ll}1& xy=0\\ 0& xy\ne 0\end{array}$$ for all $z\in \mathbb{C}$. All of the partial derivatives $$({\partial }_{1}u)(0,0)({\partial }_{2}u)(0,0)({\partial }_{1}v)(0,0)({\partial }_{2}v)(0,0)$$ exist and are equal to 0. However $f$ is not continuous at 0, because for instance $$\underset{t\to 0}{lim}f(t+it)=\underset{t\to 0}{lim}0=0\ne 1=f(1)$$ so it cannot be differentiable there.

We conclude from the above example that a function $f$ need not be differentiable at a point $z=x+iy$ just because the partial derivatives $$({\partial }_{1}u)(0,0)({\partial }_{2}u)(0,0)({\partial }_{1}v)(0,0)({\partial }_{2}v)(0,0)$$ exist and satisfy the Cauchy-Riemann equations. What additional assumptions will guarantee $f$ is differentiable there? In our example, there is no open ball centered at zero inside which all the partial derivatives exist. A reasonable additional assumption would be to add the requirement that all the partial derivatives exist inside $\mathsf{B}(z,r)$ for some $r>0$. As the following example shows, this is not quite enough.

Example

Define $f$ by $$f(z)=\{\begin{array}{ll}(\bar{z}{)}^2/z& z\ne 0\\ 0& z=0\end{array}$$ for all $z\in \mathbb{C}$. First of all, note that $f$ is not differentiable at $z=0$ because $$\frac{f(z+0)-f(0)}{z}={\left({\displaystyle \frac{x-iy}{x+iy}}\right)}^2$$ has different limits as $z\to 0$ along the real and imaginary axes.

For this function $$u(x,y)={\displaystyle \frac{x^3-3x{y}^2}{x^2+y^2}}v(x,y)=\frac{y^3-3x^{2}y}{x^2+y^2}$$ whenever $(x,y)\ne 0$ and we calcalculate that the Cauchy-Riemann equations hold at $(0,0)$. However, for example, the partial derivative ${\partial }_{1}u$ is not continuous at $(0,0)$. We have $$({\partial }_{1}u)(x,y)=\{\begin{array}{ll}{\displaystyle \frac{x^4-3y^4+6x^2{y}^2}{(x^2+y^2{)}^2}}& (x,y)\ne (0,0)\\ 0& (x,y)=(0,0)\end{array}$$ but $\underset{t\to 0}{lim}({\partial }_{1}u)(t,t)=4$.

If, in addition to existing inside $\mathsf{B}(z,r)$ for some $r>0$ the partial derivatives are all continuous on $\mathsf{B}(z,r)$ then we can conclude from the Cauchy-Riemann equations that $f$ is holomorphic at $z$.

Theorem

Fix $z\in \mathbb{C}$ and $r>0$. Suppose $u,v:\mathsf{B}(z,r)\to \mathbb{R}$ satisfy all of the following properties.

1. ${\partial }_{1}u,{\partial }_{2}u,{\partial }_{1}v,{\partial }_{2}v$ all exist on all of $\mathsf{B}(z,r)$
2. ${\partial }_{1}u,{\partial }_{2}u,{\partial }_{1}v,{\partial }_{2}v$ are all continuous on all of $\mathsf{B}(z,r)$
3. ${\partial }_{1}u={\partial }_{2}v$ and ${\partial }_{2}u=-{\partial }_{1}v$ on all of $\mathsf{B}(z,r)$

Then $f(z)=u(x,y)+iv(x,y)$ is differentiable at $z$.

Proof:

We already know that if $f^{\prime }(z)$ exists then its derivative must be $({\partial }_{1}u)(x,y)+i({\partial }_{1}v)(x,y)$. Fix $ϵ>0$. We start by looking at $u(x+c,y+d)$ for $c^2+d^2<r^2$. Since ${\partial }_{2}u$ exists at $(x+c,y)$ we can estimate that $$|u(x+c,y+d)-u(x+c,y)-d({\partial }_{2}u)(x+c,y)|<ϵ|d|$$ whenever $|d|$ is small enough. Similarly, since ${\partial }_{1}u$ exists at $(x,y)$ we can estimate that $$|u(x+c,y)-u(x,y)-c({\partial }_{1}u)(x,y)|<ϵ|c|$$ whenever $|c|$ is small enough. Lastly, continuity of ${\partial }_{2}u$ at $(x,y)$ guarantees $$|({\partial }_{2}u)(x+c,y)-({\partial }_{2}u)(x,y)|<ϵ$$ whenever $|c|$ is small enough. These three inequalities combined give $$|u(x+c,y+d)-u(x,y)-c({\partial }_{1}u)(x,y)-d({\partial }_{2}u)(x,y)|<(|c|+2|d|)ϵ$$ whenever $|c|$ and $|d|$ are small enough. The same argument applied to $v$ gives $$|v(x+c,y+d)-v(x,y)-c({\partial }_{1}v)(x,y)-d({\partial }_{2}v)(x,y)|<(|c|+2|d|)ϵ$$ so that $$\begin{array}{rl}& |\frac{u(x+c,y+d)+iv(x+c,y+d)-u(x,y)-iv(x,y)}{c+id}\\ & -(({\partial }_{1}u)(x,y)+i({\partial }_{1}v)(x,y))|\\ \le & \frac{|u(x+c,y+d)-u(x,y)-c({\partial }_{1}u)(x,y)+d({\partial }_{1}v)(x,y)|}{\sqrt{c^2+d^2}}\\ & +\frac{|v(x+c,y+d)-v(x,y)-d({\partial }_{1}u)(x,y)-c({\partial }_{1}v)(x,y)|}{\sqrt{c^2+d^2}}\\ \le & 4\frac{|c|+|d|}{\sqrt{c^2+d^2}}ϵ\end{array}$$ if $|c|$ and $|d|$ are small enough, by the Cauchy-Riemann equations and the above estimates. Finally, note that $$\frac{|c|+|d|}{\sqrt{c^2+d^2}}\le \frac{2max\{|c|,|d|\}}{\sqrt{2}max\{|c|,|d|\}}=\sqrt{2}$$ giving the desired conclusion. $◻$