Identify $X$ with its canonical image in $X^{**}$ and put
$$
d=d(\Phi,X)>0,
\qquad
c=\frac d{d+1}.
$$
For $x\in S_X$ and $\lambda\in\mathbb R$, if $|\lambda|\leq1/(d+1)$ then
$$
\lVert x-\lambda\Phi\rVert
\geq1-|\lambda|\lVert\Phi\rVert
\geq\frac d{d+1}=c.
$$
If $|\lambda|\geq1/(d+1)$, then
$$
\lVert x-\lambda\Phi\rVert
\geq d(\lambda\Phi,X)=|\lambda|d\geq c.
$$
Therefore $d(x,\operatorname{span}\{\Phi\})\geq c$.

The restriction of $x\in X^{**}$ to $\ker\Phi\subseteq X^*$ has norm
$$
\sup_{g\in B_{\ker\Phi}}|g(x)|
=d(x,(\ker\Phi)^\perp)
=d(x,\operatorname{span}\{\Phi\})
\geq c,
$$
where the first equality is the <Hahn-Banach distance formula> and $(\ker\Phi)^\perp=\operatorname{span}\{\Phi\}$. Scaling from $S_X$ gives
$$
c\lVert x\rVert\leq\sup_{g\in B_{\ker\Phi}}|g(x)|
$$
for every $x\in X$. Hence $\ker\Phi$ is $c$-norming for $X$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 106 / 5 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-106/5/b/solution