The sample mean uniquely minimizes squared error on a nonempty leaf

Enumerate $I$ in increasing sample-index order as $i_1,\dots ,i_N$ and define $T(c)=(c,\dots ,c)$ and $b=(y_{i_1},\dots ,y_{i_N})$. For all scalars $a,c,d\in \mathbb{R}$, $T(ac+d)=aT(c)+T(d)$ coordinatewise, so $T$ is linear.

Identify $\mathbb{R}$ with ${\mathbb{R}}^1$. By [F2] for $n=1$ and $n=N$, the standard coordinate inner products make $\mathbb{R}$ and ${\mathbb{R}}^N$ finite-dimensional inner-product spaces. Hence $\parallel Tc-b{\parallel }^2={\sum }_{i\in I}(y_i-c{)}^2$.

Define $S:{\mathbb{R}}^N\to \mathbb{R}$ by $S(v_1,\dots ,v_N)={\sum }_{k=1}^N{v}_k$. This map is linear, and for every $c\in \mathbb{R}$ and $v\in {\mathbb{R}}^N$, $⟨T(c),v{⟩}_{{\mathbb{R}}^N}=c{\sum }_{k=1}^N{v}_k=⟨c,S(v){⟩}_{\mathbb{R}}$. Thus $S$ is an adjoint of $T$ by [F3]. If another map also satisfied this identity, its value at each $v$ would differ from $S(v)$ by a scalar pairing to zero with every $c\in \mathbb{R}$, hence would equal $S(v)$ by positive definiteness. Therefore $T^{\ast }=S$.

Now [F4] applies to the linear map $T$ between these finite-dimensional inner-product spaces, with its adjoint established in step 3.1. The normal equation $T^{\ast }(Tc-b)=0$ is $Nc-{\sum }_{i\in I}{y}_i=0$, whose unique solution is $c={\bar{y}}_I$ because $N\ge 1$. By [F4], this solution is the unique minimizer.

Expanding around ${\bar{y}}_I$ gives ${\sum }_{i\in I}(y_i-c{)}^2={\sum }_{i\in I}(y_i-{\bar{y}}_I{)}^2+N(c-{\bar{y}}_I{)}^2$, since ${\sum }_{i\in I}(y_i-{\bar{y}}_I)=0$. Therefore the minimum value is ${\sum }_{i\in I}(y_i-{\bar{y}}_I{)}^2$.