Question: how to get the the identity
$$\left\|y_n-y_m\right\|^2=2\left\|x-y_n\right\|^2+2\left\|x-y_m\right\|^2-4\left\|x-\frac{y_n+y_m}{2}\right\|^2$$
The background of the question is from the book Theoretical Foundations of Functional Data Analysis, with an Introduction to Linear Operatorsby Randall L. Eubank and Tailen Hsing:
The following result is fundamental in optimization theory.
Let $\mathbb{M}$ be a closed convex set in a Hilbert space$\mathbb{H}$. For every $x \in \mathbb{H},\|x-y\|$ has a uniqueminimizer $\hat{x}$ in $\mathbb{M}$ that satisfies
$$ \langle x-\hat{x}, y-\hat{x}\rangle \leq 0 $$
for all $y \in \mathbb{M}$.
Proof: Our proof follows along the lines of that in Luenberger (1969).First we observe that by the definition of infimum, we can obtain asequence $y_n \in$$\mathbb{M}$ such that
$$ \lim _{n \rightarrow \infty}\left\|x-y_n\right\|^2=\inf _{y \in> \mathbb{M}}\|x-y\|^2 $$
The first step is to show that this sequence is Cauchy therebyinsuring that it converges to some element in $\mathbb{M}$. For thispurpose, we use the identity
$$> \left\|y_n-y_m\right\|^2=2\left\|x-y_n\right\|^2+2\left\|x-y_m\right\|^2-4\left\|x-\frac{y_n+y_m}{2}\right\|^2> $$
From convexity, $\left(y_n+y_m\right) / 2 \in \mathbb{M}$ and, hence
$$ \left\|y_n-y_m\right\|^2 \leq> 2\left\|x-y_m\right\|^2+2\left\|x-y_m\right\|^2-4 \inf _{y \in> \mathbb{M}}\|x-y\|^2 $$
Thus, $\left\{y_n\right\}$ is Cauchy and must have a limit $\hat{x}$that attains the infimum. If there were another element $\tilde{x}$ in$\mathbb{M}$ that also attained the infimum, we could create thealternating series having $y_n=\hat{x}$ for $n$ even and$y_n=\tilde{x}$ for $n$ odd. It is now trivially true that $\lim> \left\|x-y_n\right\|^2=\inf _{y \in \mathbb{M}}\|x-y\|^2$ and ourprevious argument can be used to see that the sequence is Cauchy. So,it must have a limit and that can only be true if $\hat{x}=\tilde{x}$.Suppose now that there is a $y \in \mathbb{M}$ for which $\langle> x-\hat{x}, y-\hat{x}\rangle>0$. Then, if we let $x(a)=a y+(1-a)> \hat{x} \in \mathbb{M}$ for $a \in(0,1)$, the derivative of$\|x-x(a)\|^2$ at $a=0$ is negative. Thus, there is a choice of $a$that makes $\|x-x(a)\|^2$
smaller than $\|x-\hat{x}\|^2$ which is a contradiction. On the otherhand, if $\langle x-\hat{x}$, $y-\hat{x}\rangle \leq 0$,
$$ \|x-y\|^2=\|x-\hat{x}\|^2+2\langle x-\hat{x},> \hat{x}-y\rangle+\|\hat{x}-y\|^2 \geq\|x-\hat{x}\|^2 $$