Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 2 / 3G (Analysis II)

Differentiability here is understood in the sense of the Frechet derivative: there is a linear map $L:{\mathbb{R}}^n\to {\mathbb{R}}^m$ such thatThe map $L=Df(x)$ is the Frechet derivative. In finite dimensions a linear map is bounded, so $\Vert Lh\Vert \le C\Vert h\Vert$ for some $C$. The defining estimate therefore impliesThis is precisely continuity at $x$: differentiability implies continuity.

First suppose each $g_j$ is differentiable at $x_j$. Write $g_j(x_j+t)=g_j(x_j)+v_{j}t+r_j(t)$, where $v_j\in {\mathbb{R}}^m$ and $r_j(t)=o(|t|)$. Define the linear map $Lh={\sum }_j{v}_j{h}_j$. Given $\epsilon >0$, choose a common sufficiently small neighbourhood so that $\Vert r_j(h_j)\Vert \le \epsilon |h_j|$ for all $j$, with $r_j(0)=0$. ThenThis proves differentiability, withConversely, suppose $f$ is differentiable with Frechet derivative $L$. Along the $j$th coordinate direction $e_j$, the other summands cancel exactly:Thus $g_j$ is differentiable at $x_j$, with $g_j^{\prime }(x_j)=L{e}_j$. The equivalence works because each summand varies in only one coordinate; it is stronger than merely having partial derivatives.

The function $g(t)=|t{|}^{3/2}$ is differentiable everywhere: for $t\ne 0$, its derivative is $(3/2)\mathrm{sgn}(t)\sqrt{|t|}$, while at zero the difference quotient is $\mathrm{sgn}(t)\sqrt{|t|}\to 0$.

The function $k(t)=|t{|}^{1/2}$ is differentiable for $t\ne 0$, with derivative $\mathrm{sgn}(t)/(2\sqrt{|t|})$. At zero its difference quotient is $\mathrm{sgn}(t)/\sqrt{|t|}$, which has no finite limit. Applying the coordinatewise-sum argument givesAt those points,where the first coefficient is zero at $x=0$. At any point with $y=0$, the restriction to the vertical coordinate line already fails to be differentiable.