Include brief justification for your answers to parts (b) and (c).
Complete the definition: \(f:\mathbb{R}^n\to\mathbb{R}^m\) is (real) differentiable at a point \(a\in\mathbb{R}^n\) if there is a linear transformation \(\ldots\)
Give an example of a function \(g:\mathbb{R}^2\to\mathbb{R}\) whose first order partial derivatives exist everywhere, such that \(g\) is not differentiable at \((0,0)\).
Give an example of a function \(h:\mathbb{R}^2\to\mathbb{R}^2\) which is differentiable everywhere, but not complex differentiable anywhere.
Solution
The map \(f\) is differentiable at \(a\) if there is a linear map \(L:\mathbb R^n\to\mathbb R^m\) such that \[ \lim_{x\to a}\frac{\|f(x)-f(a)-L(x-a)\|}{\|x-a\|}=0. \] The map \(L\), necessarily unique, is \(Df(a)\).
Define \[ g(x,y)= \begin{cases} \dfrac{x^3}{x^2+y^2},&(x,y)\ne(0,0),\\ 0,&(x,y)=(0,0). \end{cases} \] Away from the origin it is smooth. At the origin both partial derivatives exist: \(g_x(0,0)=1\) and \(g_y(0,0)=0\). If \(g\) were differentiable there, its derivative would be \(L(x,y)=x\). But along \(y=x\), \[ g(x,x)-L(x,x)=\frac{x}{2}-x=-\frac{x}{2}, \] whose ratio to \(\sqrt{2}|x|\) does not tend to \(0\). Thus \(g\) is not differentiable at the origin.
Take \[ h(x,y)=(x,-y), \] which corresponds to the complex map \(z\mapsto\bar z\). It is real-linear, hence differentiable everywhere. Its real components \(u=x\), \(v=-y\) satisfy \(u_x=1\) and \(v_y=-1\), so the Cauchy–Riemann equations fail at every point. Thus it is nowhere complex differentiable.