Riesz representation for Hilbert spaces

If $f=0$, then $y=0$ represents $f$ because $⟨x,0⟩=0$ for every $x$, and $\mathrm{\parallel }f\mathrm{\parallel }=0=\mathrm{\parallel }y\mathrm{\parallel }$; the same $f$ has no other representing vector, since a vector $y^{\mathrm{\prime }}$ representing $f$ satisfies $⟨y^{\mathrm{\prime }},y^{\mathrm{\prime }}⟩=0$ and hence $y^{\mathrm{\prime }}=0$.

If $f\mathrm{\ne }0$, its kernel is a proper linear subspace of $H$: it is linear because $f(au+bv)=af(u)+bf(v)$, it is proper because $f$ takes a nonzero value, and it is closed because $x_n\to x$ and $f(x_n)=0$ give $\mathrm{\mid }f(x)\mathrm{\mid }\le \mathrm{\mid }f(x)-f(x_n)\mathrm{\mid }\le \mathrm{\parallel }f\mathrm{\parallel }\mathrm{\parallel }x-x_n\mathrm{\parallel }\to 0$.

Choose $z$ with $f(z)\mathrm{\ne }0$ and put $u=z\mathrm{/}f(z)$, so $f(u)=1$; decompose $u=m+n$ with $m\in \ker f$ and $n\in (\ker f{)}^{\perp }$, then $1=f(u)=f(m)+f(n)=f(n)$ shows $n\mathrm{\ne }0$ and $f(n)=1$.

For arbitrary $x$, the vector $v=x-f(x)u$ satisfies $f(v)=f(x)-f(x)\cdot 1=0$, hence $⟨v,n⟩=0$ and $⟨x,n⟩=f(x)⟨u,n⟩$; moreover $⟨u,n⟩=⟨m+n,n⟩=⟨m,n⟩+⟨n,n⟩=0+\mathrm{\parallel }n{\mathrm{\parallel }}^2$, so $f(x)=⟨x,n⟩\mathrm{/}\mathrm{\parallel }n{\mathrm{\parallel }}^2=⟨x,n\mathrm{/}\mathrm{\parallel }n{\mathrm{\parallel }}^2⟩$ with $y:=n\mathrm{/}\mathrm{\parallel }n{\mathrm{\parallel }}^2$.

Norm and uniqueness: Cauchy–Schwarz gives $\mathrm{\mid }f(x)\mathrm{\mid }=\mathrm{\mid }⟨x,y⟩\mathrm{\mid }\le \mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }$, so $\mathrm{\parallel }f\mathrm{\parallel }\le \mathrm{\parallel }y\mathrm{\parallel }$, while $f(y)=\mathrm{\parallel }y{\mathrm{\parallel }}^2$ gives $\mathrm{\parallel }f\mathrm{\parallel }\ge \mathrm{\parallel }y\mathrm{\parallel }$; hence $\mathrm{\parallel }f\mathrm{\parallel }=\mathrm{\parallel }y\mathrm{\parallel }$, and this contains the case $f=0$. If also $f(x)=⟨x,y^{\mathrm{\prime }}⟩$ for all $x$, then $⟨x,y-y^{\mathrm{\prime }}⟩=0$ for all $x$, and the choice $x=y-y^{\mathrm{\prime }}$ gives $\mathrm{\parallel }y-y^{\mathrm{\prime }}{\mathrm{\parallel }}^2=0$, so $y=y^{\mathrm{\prime }}$.

Conjugate linearity: if $f_i(x)=⟨x,y_i⟩$ for $i=1,2$, then for all $x$ and scalars $a,b$ one has $(a{f}_1+b{f}_2)(x)=a⟨x,y_1⟩+b⟨x,y_2⟩=⟨x,\bar{a}{y}_1+\bar{b}{y}_2⟩$ by conjugate-linearity in the second argument; uniqueness of the representing vector therefore gives the representing vector $\bar{a}{y}_1+\bar{b}{y}_2$, so the map $f\mapsto y$ is conjugate-linear, and by step 4.1 it is isometric.

Steps 1.1, 3.1 and 4.1 produce the unique representing vector together with the norm identity for every bounded $f$, and step 5.1 records its conjugate-linear isometric dependence; Countable Choice is used exactly through the orthogonal decomposition of step 2.1.