Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 8 / 2

Let $C=\Vert T\Vert$. We construct the extension by a deterministic recursion over the given dense list, using the one-dimensional dominated extension of a real linear functional. Neither Zorn's lemma nor any selection of a family of extensions is needed.

Suppose $f:W\to \mathbb{R}$ is already a linear functional satisfying $|f(w)|\le C\Vert w\Vert$, and $v\notin W$. An extension to $W+\mathbb{R}v$ must have the formDefineFor any $w,z\in W$, the bound on $f$ and the triangle inequality giveThus every quantity in the supremum defining $L_v$ is at most every quantity in the infimum defining $U_v$. Taking $w=0$ and $z=0$ gives finite bounds, specificallyThe completeness of the real numbers defines these endpoints uniquely. Set $a=L_v$; this is an explicit rule, not an arbitrary choice from a nonempty interval.

For $t>0$, the upper-endpoint inequality with $w/t$ yields $f(w)+ta\le C\Vert w+tv\Vert$. For $t<0$, write $t=-s$ with $s>0$ and use the lower-endpoint inequality with $w/s$ to obtain the same bound. For $t=0$ it is the original bound. Applying the upper bound also to the negative of the vector givesThe representation $w+tv$ is unique because $v\notin W$, so $f_v$ is a well-defined bounded linear functional extending $f$.

Start with $W_0=U$ and $f_0=T$. At step $n$, if $y_n\in W_{n-1}$ leave the functional unchanged; otherwise use exactly the supremum-endpoint rule with $v=y_n$. Put $W_n=U+\mathrm{span}\{y_1,\dots ,y_n\}$. This defines compatible functionals $f_n$ on $W_n$ with the common bound $C$. Their union $f_{\infty }$ on $W_{\infty }={\bigcup }_n{W}_n$ is linear and bounded by $C$. The subspace $W_{\infty }$ is dense in $V$ because it contains every $y_n$.

To make the final passage to the closure equally free of the axiom of choice, for $v\in V$ let $j_n(v)$ be the least positive integer with $\Vert y_{j_n(v)}-v\Vert <2^{-n}$. Density ensures that this integer exists, and leastness specifies it uniquely. DefineThe limit exists becauseAny other approximating sequence in $W_{\infty }$ gives the same limit, by the same norm bound. Applying this independence to sums and scalar multiples of approximating sequences proves linearity of $\tilde{T}$. It also shows $\tilde{T}=f_{\infty }$ on $W_{\infty }$ and $|\tilde{T}(v)|\le C\Vert v\Vert$. Restriction to $U$ therefore givesThe upper bound follows from construction and the lower bound from restriction to $U$. This also covers $C=0$, when all extensions are zero. No completeness of $V$ or closedness of $U$ was assumed. The argument establishes the choice-free Hahn-Banach extension in a separable space by uniquely specified operations at every stage.

A point $x$ of a set $S$ is an extreme point if $x=(1-t)y+tz$, with $y,z\in S$ and $0<t<1$, forces $y=z=x$. For a convex set it is enough to test midpoint decompositions: a nontrivial segment containing $x$ in its interior contains a smaller segment with midpoint $x$. The nonempty convex set ${\mathbb{R}}^2$ has no extreme points, since every $x$ is the midpoint of $x+e_1$ and $x-e_1$.

For the real l-infinity sequence space, a vector in the closed unit ball is extreme precisely when every coordinate is $1$ or $-1$. If $|x_j|<1$ for some $j$, choose $0<\delta \le 1-|x_j|$. The distinct vectors $x\pm \delta e_j$ remain in the unit ball and have midpoint $x$, so $x$ is not extreme. Conversely suppose all coordinates have modulus one and $x=(y+z)/2$, with $\Vert y{\Vert }_{\infty },\Vert z{\Vert }_{\infty }\le 1$. In each real interval $[-1,1]$, an endpoint can be a midpoint only if both entries equal that endpoint. Thus $y_j=z_j=x_j$ for every $j$ and $y=z=x$. Therefore

For the real absolutely summable sequence space, any point with $\Vert x{\Vert }_1<1$ is nonextreme: choose a nonzero coordinate perturbation of norm less than $1-\Vert x{\Vert }_1$. If $\Vert x{\Vert }_1=1$ and two coordinates $x_j,x_k$ are nonzero, take $0<\delta <\min (|x_j|,|x_k|)$ and putBoth perturbations preserve the signs of those two coordinates. One adds $\delta$ to one absolute value and subtracts it from the other, so $\Vert x+h{\Vert }_1=\Vert x-h{\Vert }_1=1$. Hence such an $x$ is not extreme. The only remaining candidates are $\pm e_j$. If $e_j=(y+z)/2$ with $\Vert y{\Vert }_1,\Vert z{\Vert }_1\le 1$, its $j$th coordinate forces $y_j=z_j=1$. That consumes the entire norm allowance, forcing all other coordinates to vanish. The argument for $-e_j$ is identical. Thus

For the real l2 sequence space, interior points of the unit ball again permit sufficiently small opposite perturbations and are not extreme. If $\Vert x{\Vert }_2=1$ and $x=(y+z)/2$, write $y=x+h$, $z=x-h$. The parallelogram law givesThe left side is at most one, so $h=0$ and $y=z=x$. ConsequentlyThese are the extreme points of real sequence-space unit balls: signs in ${\ell }^{\infty }$, coordinate spikes in ${\ell }^1$, and the whole sphere in ${\ell }^2$. The restriction to real scalars matters for the first description; the complex supremum ball has arbitrary unit-modulus coordinates instead.