Norm-attaining functionals on a Hilbert space

Indeed, if $x\mathrm{\notin }M$, then

$r=\frac{\mathrm{\mid }f(x)\mathrm{\mid }}{2\mathrm{\parallel }f\mathrm{\parallel }}>0,$

and $\mathrm{\parallel }h-x\mathrm{\parallel }<r$ implies

$\mathrm{\mid }f(h)\mathrm{\mid }\ge \mathrm{\mid }f(x)\mathrm{\mid }-\mathrm{\mid }f(h-x)\mathrm{\mid }>\mathrm{\mid }f(x)\mathrm{\mid }-\mathrm{\parallel }f\mathrm{\parallel }r=\frac{\mathrm{\mid }f(x)\mathrm{\mid }}{2}>0.$

Thus the open ball $B(x,r)$ misses $M$, so the complement of $M$ is open. By [F5], $M$ is therefore a Banach space with the restricted norm. [F1, F2, F5]

For every $u\in M$,

$1=\mathrm{\mid }f(v-u)\mathrm{\mid }\le \mathrm{\parallel }f\mathrm{\parallel }\mathrm{\parallel }v-u\mathrm{\parallel },$

and hence $d\ge 1\mathrm{/}\mathrm{\parallel }f\mathrm{\parallel }>0$. [step 1.2, F2, F3]

$A_n=\{u\in M:\mathrm{\parallel }v-u\mathrm{\parallel }<d+\frac{1}{n+1}\}\mathrm{.}$

Each $A_n$ is nonempty by [F3]. Apply ${\mathrm{A}\mathrm{C}}_{\omega }$ once to this family and choose $m_n\in A_n$ for every $n$. Thus, with $r_n=\mathrm{\parallel }v-m_n\mathrm{\parallel }$,

$d\le r_n<d+\frac{1}{n+1}\mathrm{.}$

This is the sole use of choice in the proof. [F3, F6]

$\mathrm{\parallel }{a}_n-a_k{\mathrm{\parallel }}^2+\mathrm{\parallel }{a}_n+a_k{\mathrm{\parallel }}^2=2\mathrm{\parallel }{a}_n{\mathrm{\parallel }}^2+2\mathrm{\parallel }{a}_k{\mathrm{\parallel }}^2\mathrm{.}$

Since $(m_n+m_k)\mathrm{/}2\in M$, the definition of $d$ gives $\mathrm{\parallel }(a_n+a_k)\mathrm{/}2\mathrm{\parallel }\ge d$. Therefore

\|m_n-m_k\|^2\le2r_n^2+2r_k^2-4d^2.\tag{1}

This is the estimate used below. [step 2.1, step 3.1, F1]

Given $\epsilon >0$, set

$\eta =\min \{1,\frac{{\epsilon }^2}{8d+4}\}>0.$

By [F4], choose $N$ so that $1\mathrm{/}(N+1)<\eta$. For $n,k\ge N$, step 3.1 and the eventual monotonicity of reciprocals give $r_n,r_k<d+\eta$. Using $\eta \le 1$ in (1),

$\mathrm{\parallel }{m}_n-m_k{\mathrm{\parallel }}^2<4(d+\eta {)}^2-4d^2=8d\eta +4{\eta }^2\le (8d+4)\eta \le {\epsilon }^2\mathrm{.}$

Both sides before squaring are nonnegative, so $\mathrm{\parallel }{m}_n-m_k\mathrm{\parallel }<\epsilon$. [step 3.1, step 4.1, F4]

The triangle inequality gives

$d\le \mathrm{\parallel }v-m\mathrm{\parallel }\le \mathrm{\parallel }v-m_n\mathrm{\parallel }+\mathrm{\parallel }{m}_n-m\mathrm{\parallel }\mathrm{.}$

Given $\epsilon >0$, step 3.1, [F4], and convergence let us make the two terms on the right smaller than $d+\epsilon \mathrm{/}2$ and $\epsilon \mathrm{/}2$, respectively. Thus $\mathrm{\parallel }v-m\mathrm{\parallel }<d+\epsilon$ for every $\epsilon >0$, while $d$ is a lower bound, so $\mathrm{\parallel }v-m\mathrm{\parallel }=d$. [step 1.2, step 2.1, step 3.1, step 5.1, F1, F4, F5]

For $u\in M$ and $t\in \mathbb{R}$, minimality of $m$ and $m+tu\in M$ give

\|z-tu\|^2-\|z\|^2 =t^2\|u\|^2-2t\operatorname{Re}\langle z,u\rangle\ge0.\tag{2}

If $\mathrm{Re}⟨z,u⟩\mathrm{\ne }0$, then $u\mathrm{\ne }0$ and taking $t=\mathrm{Re}⟨z,u⟩\mathrm{/}\mathrm{\parallel }u{\mathrm{\parallel }}^2$ makes the right side of (2) negative. Hence $\mathrm{Re}⟨z,u⟩=0$. Over $\mathbb{C}$, apply the same conclusion to $iu\in M$; the linear-first convention gives $⟨z,iu⟩=-i⟨z,u⟩$, whose real part is $\mathrm{Im}⟨z,u⟩$. Thus in either scalar field $⟨z,u⟩=0$ for every $u\in M$. [step 2.1, step 6.1, F1]

Indeed, the vector $u=x-f(x)z$ lies in $M$. Step 7.1 and conjugate symmetry give $⟨u,z⟩=0$, so

$⟨x,z⟩=f(x)\mathrm{\parallel }z{\mathrm{\parallel }}^2\mathrm{.}$

Consequently, with $y=z\mathrm{/}\mathrm{\parallel }z{\mathrm{\parallel }}^2$,

$f(x)=⟨x,y⟩(x\in H)\mathrm{.}$

If another vector $y^{\mathrm{\prime }}$ represented $f$, then $⟨x,y-y^{\mathrm{\prime }}⟩=0$ for all $x$; choosing $x=y-y^{\mathrm{\prime }}$ and using positive definiteness gives $y=y^{\mathrm{\prime }}$. [step 7.1, F1]

By [F7], $\mathrm{\mid }f(x)\mathrm{\mid }\le \mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }$, so $\mathrm{\parallel }f\mathrm{\parallel }\le \mathrm{\parallel }y\mathrm{\parallel }$. Conversely the unit vector $x_0=y\mathrm{/}\mathrm{\parallel }y\mathrm{\parallel }$ satisfies

$f(x_0)=⟨\frac{y}{\mathrm{\parallel }y\mathrm{\parallel }},y⟩=\mathrm{\parallel }y\mathrm{\parallel },$

where the last number is positive real. Thus $\mathrm{\parallel }f\mathrm{\parallel }=\mathrm{\parallel }y\mathrm{\parallel }=\mathrm{\mid }f(x_0)\mathrm{\mid }$, and $f$ attains its norm at $x_0$. Together with the zero case in step 1.1, this proves every clause, including both scalar fields and the zero Hilbert space. [step 1.1, step 8.1, F1, F2, F7] ∎