Let \(f : \mathbb { R } ^ { 2 } \to \mathbb { R } ^ { 2 }\)
Define in terms of linear transformations, what it means for \(f\) to be differentiable at a point \((a, b) \in \RR^2\)
State a version of the inverse function theorem in this setting.
Identifying \(\CC\) with \(\RR^2\) in the usual way, give, with proof, a necessary and sufficient condition for a function satisfying the definition of real differentiability in Part a) to be complex differentiable at the point \(a + i b\)
Solution
Put \[ p=(a,b)\in\RR^2, \qquad z_0=a+ib\in\CC. \] When convenient, write \[ f=(u,v) \] and identify \((x,y)\in\RR^2\) with \(x+iy\in\CC\).
1. Part (a): \(f\) is real differentiable at \(p\) exactly when there is areal-linear map\[ L:\RR^2\longrightarrow\RR^2 \]such that\[ \lim_{h\to0} \frac{\norm{f(p+h)-f(p)-Lh}}{\norm h} = 0. \]In that case \(L\) is unique and is denoted \(Df(p)\).
The displayed condition is the definition of Fréchet differentiability in finite-dimensional real vector spaces.
For uniqueness, suppose both \(L\) and \(M\) satisfy the condition. For every fixed nonzero \(w\in\RR^2\) and every nonzero real \(t\), \[ \frac{\norm{(L-M)(tw)}}{\abs t\,\norm w} = \frac{\norm{(L-M)w}}{\norm w}. \] On the other hand, \[ (L-M)(tw) = \bigl(f(p+tw)-f(p)-M(tw)\bigr) - \bigl(f(p+tw)-f(p)-L(tw)\bigr), \] so the defining limits for \(L\) and \(M\) force the left-hand side to tend to \(0\) as \(t\to0\). Thus \((L-M)w=0\) for every nonzero \(w\), and the same is trivial for \(w=0\). Hence \(L=M\).
2. Part (b): a version of the inverse function theorem is the following.If \(f\) is \(C^1\) on a neighborhood of \(p\) and\[ \det Df(p)\neq0, \]then there are open neighborhoods\[ p\in U\subseteq\RR^2, \qquad f(p)\in V\subseteq\RR^2 \]such that\[ f|_U:U\longrightarrow V \]is a \(C^1\) diffeomorphism. Moreover,\[ D(f|_U^{-1})(f(x)) = \bigl(Df(x)\bigr)^{-1} \]for \(x\in U\) after \(U\) is chosen small enough that \(Df(x)\) remainsinvertible.
This is the inverse function theorem specialized to maps \(\RR^2\to\RR^2\).
3. If the real-differentiable map \(f\) is complex differentiable at\(z_0\), then \(Df(p)\) is complex-linear.
Let \[ \lambda=f_{\CC}'(z_0). \] Complex differentiability means \[ f(z_0+h)-f(z_0) = \lambda h+r(h), \qquad \frac{\abs{r(h)}}{\abs h}\longrightarrow0. \] Under the identification \(\CC\cong\RR^2\), multiplication by \(\lambda\) is a real-linear map. The displayed estimate therefore also exhibits multiplication by \(\lambda\) as the real derivative of \(f\) at \(p\). By the uniqueness in step 1, \[ Df(p)(h)=\lambda h. \] Hence \(Df(p)\) is complex-linear.
4. A real-linear map \(L:\CC\to\CC\) is complex-linear if and only if\[ L(ih)=iL(h) \]for every \(h\in\CC\).
Complex-linearity immediately implies the displayed identity.
Conversely, suppose the identity holds. For \[ \alpha+i\beta\in\CC \] and \(h\in\CC\), real-linearity gives \[ \begin{aligned} L((\alpha+i\beta)h) &= \alpha L(h)+\beta L(ih)\\ &= \alpha L(h)+i\beta L(h)\\ &= (\alpha+i\beta)L(h). \end{aligned} \] Thus \(L\) is complex-linear.
5. If \(f=(u,v)\) is real differentiable at \(p\), then \(Df(p)\) iscomplex-linear if and only if the Cauchy–Riemann equations\[ u_x(p)=v_y(p), \qquad u_y(p)=-v_x(p) \]hold.
Real differentiability gives \[ Df(p) = \begin{pmatrix} u_x(p)&u_y(p)\\ v_x(p)&v_y(p) \end{pmatrix}. \] Let \[ J= \begin{pmatrix} 0&-1\\ 1&0 \end{pmatrix}, \] the real matrix of multiplication by \(i\). By step 4, \(Df(p)\) is complex-linear exactly when \[ Df(p)J=JDf(p). \] Writing out this matrix equation gives \[ u_x(p)=v_y(p), \qquad u_y(p)=-v_x(p). \]
6. Conversely, if the Cauchy–Riemann equations in step 5 hold, then\(f\) is complex differentiable at \(z_0\).
By step 5, the real derivative \(Df(p)\) is complex-linear. Hence there is some \(\lambda\in\CC\) such that \[ Df(p)(h)=\lambda h. \] Real differentiability gives \[ f(z_0+h)-f(z_0) = \lambda h+r(h), \qquad \frac{\abs{r(h)}}{\abs h}\longrightarrow0. \] For \(h\neq0\), \[ \frac{f(z_0+h)-f(z_0)}h = \lambda+\frac{r(h)}h. \] Since \[ \abs{\frac{r(h)}h} = \frac{\abs{r(h)}}{\abs h} \longrightarrow0, \] the complex difference quotient tends to \(\lambda\). Thus \(f\) is complex differentiable at \(z_0\).
7. Part (c): for a real-differentiable \(f=(u,v)\),\[ \boxed{ f\text{ is complex differentiable at }a+ib \iff \begin{cases} u_x(a,b)=v_y(a,b),\\ u_y(a,b)=-v_x(a,b). \end{cases} } \]Equivalently, \(Df(a,b)\) is complex-linear.
5. Sufficiency follows from steps<1>5–<1>6.
8. Q.E.D.
Steps 1, 2, and 7 answer parts (a), (b), and (c), respectively.