Complete this definition: “\(f: \RR^n\to \RR^m\) is real-differentiable a point \(p\in \RR^n\) iff there exists a linear transformation…”
Give an example of a function \(f:\RR^2\to \RR\) whose first-order partial derivatives exist everywhere but \(f\) is not differentiable at \((0, 0)\).
Give an example of a function \(f: \RR^2 \to \RR\) which is real-differentiable everywhere but nowhere complex-differentiable.
Solution
(a) A map \(f:\mathbb R^n\to\mathbb R^m\) is differentiable at \(p\) if there exists a linear map \(L:\mathbb R^n\to\mathbb R^m\) such that \[ \lim_{h\to0} \frac{\|f(p+h)-f(p)-Lh\|}{\|h\|}=0. \] The map \(L\), necessarily unique, is the derivative \(Df(p)\).
(b) Define \[ f(x,y)= \begin{cases} \dfrac{x^3}{x^2+y^2},&(x,y)\ne(0,0),\\ 0,&(x,y)=(0,0). \end{cases} \] Away from the origin it is smooth. At \((0,0)\), \[ f_x(0,0)=1, \qquad f_y(0,0)=0. \] If \(f\) were differentiable there, its derivative would be \(L(x,y)=x\). Along \(y=x\), \[ f(x,x)-L(x,x)=\frac{x}{2}-x=-\frac{x}{2}, \] whose absolute value divided by \(\sqrt2|x|\) does not tend to \(0\). Hence \(f\) is not differentiable at the origin although both first partial derivatives exist everywhere.
(c) Identify \(\mathbb R^2\) with \(\mathbb C\) and \(\mathbb R\) with the real axis in \(\mathbb C\). The function \[ f(x,y)=x=\Re z \] is real-linear, hence real-differentiable everywhere. Its real and imaginary parts are \(u(x,y)=x\) and \(v(x,y)=0\), so the Cauchy–Riemann equation \(u_x=v_y\) would require \(1=0\). Thus it is nowhere complex-differentiable.