CG is the Krylov Galerkin method, and the $m$th iterate uniquely minimizes the $A$-norm of the error over $x_0+K_m(A,r_0)$

The CG recurrence gives $x_m-x_0$ as a linear combination of $p_0,\dots ,p_{m-1}$, so $x_m\in x_0+\mathrm{span}\{p_0,\dots ,p_{m-1}\}$. Also each residual satisfies $r_j=r_0-A(x_j-x_0)$, so $r_j\in K_{j+1}(A,r_0)\subseteq K_m(A,r_0)$ for $j<m$, and each $p_j=r_j+{\beta }_{j-1}{p}_{j-1}$ therefore also lies in $K_m(A,r_0)$. By [L2], the nonzero residuals $r_0,\dots ,r_{m-1}$ are mutually orthogonal, hence linearly independent; since they all lie in the $m$-generated space $K_m(A,r_0)$, they form a basis of that space, and so do $p_0,\dots ,p_{m-1}$ by the triangular relation $p_j\in r_j+\mathrm{span}\{r_0,\dots ,r_{j-1}\}$. Thus $x_m\in x_0+K_m(A,r_0)$.

By [L2], the residual $r_m$ is orthogonal to each earlier residual. Since step 1.1 shows that $p_0,\dots ,p_{m-1}$ form a basis of $K_m(A,r_0)$ and each $p_j$ lies in $\mathrm{span}\{r_0,\dots ,r_j\}$, the same orthogonality implies $r_m$ is orthogonal to $K_m(A,r_0)$. Therefore $x_m$ is a Krylov Galerkin iterate by [F1], and [L1] makes it unique.

Let $x\in x_0+K_m(A,r_0)$ and put $v:=x-x_m\in K_m(A,r_0)$. By [L1] and step 2.1, the error $x_{\ast }-x_m$ is $A$-orthogonal to $v$, so $\mathrm{\parallel }{x}_{\ast }-x{\mathrm{\parallel }}_A^2=\mathrm{\parallel }(x_{\ast }-x_m)-v{\mathrm{\parallel }}_A^2=\mathrm{\parallel }{x}_{\ast }-x_m{\mathrm{\parallel }}_A^2+\mathrm{\parallel }v{\mathrm{\parallel }}_A^2\mathrm{.}$ Hence $\mathrm{\parallel }{x}_{\ast }-x_m{\mathrm{\parallel }}_A\le \mathrm{\parallel }{x}_{\ast }-x{\mathrm{\parallel }}_A$, with equality only if $v=0$, that is, $x=x_m$. Using [L3], this is equivalent to saying that $x_m$ also uniquely minimizes the quadratic energy over the same affine Krylov space.