Past exam of the functional analysis I course of the University of Oxford / 2023 / 1 / d

By the mean value theorem, a continuously differentiable function with zero derivative on $[-1,1]$ is constant. Thus the kernel of the differentiation operator isIt is a closed set by the finite-dimensional subspace is closed theorem, applied in the normed space $(C^1([-1,1]),\Vert \cdot {\Vert }_{\infty })$. It is proper because the function $x\mapsto x$ has nonzero derivative.

For $n\ge 1$, take $u_n(x)=\sin (nx)$. Its supremum norm is at most one, whereas $u_n^{\prime }(0)=n$ and $|u_n^{\prime }(x)|\le n$. Hence

Choose the continuously differentiable functionsThe estimate $\sqrt{a^2+b^2}\le a+b$ for $a,b\ge 0$ givesThus $f_n\to |\cdot |$ by uniform convergence, and the triangle inequality gives $\Vert f_n-f_m{\Vert }_{\infty }\le n^{-1}+m^{-1}$. The sequence is a Cauchy sequence in $X$, but its unique possible supremum norm limit is $x\mapsto |x|$, which is not differentiable at zero. That limit is outside $X$, so

The kernel $W=⟨1⟩$ is complete: the map $a\mapsto a1$ is an isometry from $\mathbb{R}$ with its absolute-value norm onto $W$ with the supremum norm.

Define a linear map $S:Y\to X$ byThe fundamental theorem of calculus gives $Sg\in C^1([-1,1])$ and $TSg=g$. Also,Thus $S$ is a bounded right inverse, and $Y=C([-1,1])$ is a Banach space. Applying part (c),