Past exam of the mathematics course of the University of Cambridge / 2018 / ib / Paper 1 / 11F (Analysis II)

The map $f:U\to {\mathbb{R}}^n$ is differentiable at $a$ when there is a linear map $Df(a)$ such thatIt is $C^1$ when it is differentiable everywhere and $a\mapsto Df(a)$ is continuous. The definition immediately gives $\Vert f(a+h)-f(a)\Vert \le (\Vert Df(a)\Vert +o(1))\Vert h\Vert \to 0$, so differentiability implies continuity.

The inverse function theorem says that if $f$ is $C^1$ near $a$ and $Df(a)$ is invertible, then $f$ restricts to a diffeomorphism between neighbourhoods of $a$ and $f(a)$.

For distinct $x,y\in U$, convexity keeps the segment $x+t(y-x)$ in $U$. Continuity of $Df$ makesThe fundamental theorem of calculus along a line segment givesso $\Vert f(y)-f(x)\Vert \ge (1-q)\Vert y-x\Vert >0$. Thus $f$ is injective.

The bound $\Vert Df(a)-I\Vert <1$ makes $Df(a)$ invertible by the Neumann series. The inverse function theorem therefore makes $f$ locally open at every point, so $f(U)$ is open.

Surjectivity need not hold: the identity map on a proper convex open set has proper image. Even for $U={\mathbb{R}}^n$ it can fail. In one dimension, $f(x)=\arctan x$ satisfies $|f^{\prime }(x)-1|=x^2/(1+x^2)<1$, but its image is $(-\pi /2,\pi /2)$.

HereThis matrix acts as multiplication by the complex number $(2x-1)+2iy$, so its operator norm is $\sqrt{(2x-1{)}^2+4y^2}$. The largest set is therefore