 Solution

If $F$ is empty, the zero linear functional satisfies the conclusion vacuously. Otherwise choose one $f_0\in F$, put $\delta =ϵ/2$, and form the convex setIt contains $B(0,\delta )$ because $f_0\in F$. The point $x=-f_0$ is outside $E$: membership would imply $0=f+u$ for some $f\in F$ with $\Vert u\Vert <\delta$, contradicting the assumed absence of points of $F$ in $B(0,ϵ)$.

Apply part (i). There is a continuous linear functional $T$ with $T(-f_0)\ge 1$ and $T(f+u-f_0)\le 1$ for every $f\in F$, $\Vert u\Vert <\delta$. Thus $T\ne 0$ andFor fixed $f$, take the supremum of $Tu$ over this open ball; it is $\delta \Vert T\Vert$. Consequently $Tf\le -\delta \Vert T\Vert$ for every $f\in F$. The normalizationis a continuous linear functional satisfying $Sf\ge 1$ for every $f\in F$. No compactness, closedness or attainment of the supremum on the open ball is needed.