Prove a Norm-Based Identity Involving Convex Combination

Question

Let $\textbf{x},\textbf{y} \in \mathbb{R}^n$ and $\theta \in [0,1]$. I would like to show that $$ \theta \|\textbf{x}\|^2 + (1-\theta) \|\textbf{y}\|^2 - \| \theta \textbf{x} + (1-\theta) \textbf{y} \|^2 = \theta (1-\theta) \|\textbf{x} - \textbf{y}\|^2 . $$

My try: Note that, by triangle inequality, we have $ \| \theta \textbf{x} + (1-\theta) \textbf{y} \|^2 \leq (\| \theta \textbf{x} \| + \| (1-\theta) \textbf{y} \|)^2 = \theta^2 \| \textbf{x} \|^2 + 2\theta (1-\theta) \| \textbf{x}\| \| \textbf{y}\| + (1-\theta)^2 \| \textbf{y} \|^2. $ This implies that \begin{align*} \theta \|\textbf{x}\|^2 + (1-\theta) \|\textbf{y}\|^2 - \| \theta \textbf{x} + (1-\theta) \textbf{y} \|^2 & \geq \theta \|\textbf{x}\|^2 + (1-\theta) \|\textbf{y}\|^2 - [\theta^2 \| \textbf{x} \|^2 + 2\theta (1-\theta) \| \textbf{x}\| \| \textbf{y}\| + (1-\theta)^2 \| \textbf{y} \|^2] \\ & = \theta \|\textbf{x}\|^2 (1-\theta) + (1-\theta) \|\textbf{y}\|^2 \theta - 2\theta (1-\theta) \| \textbf{x}\| \| \textbf{y}\| \\ & = \theta (1-\theta) [ \|\textbf{x}\|^2 + \|\textbf{y}\|^2 - 2 \| \textbf{x}\| \| \textbf{y}\| ] \\ & = \theta (1-\theta) [ \|\textbf{x}\|- \|\textbf{y}\|]^2 \\ \end{align*} Then I get stuck since the reversed triangle inequality works the other way around. Any comment/suggestion is appreciated.

Answer 1

There is no need to use triangle inequalities here. Just recall that $\lVert x\rVert^2 = \langle x,x\rangle$, and use the bilinearity of the scalar product: \begin{align*} \theta \lVert x\rVert^2+(1-\theta)\lVert y\rVert^2-\langle\theta x+(1-\theta)y,\theta x+(1-\theta)y\rangle & = \theta \langle x,x\rangle+(1-\theta)\langle y,y\rangle \\ & \qquad - \theta^2\langle x,x\rangle-(1-\theta)^2\langle y,y\rangle \\ & \qquad -2\theta(1-\theta)\langle x,y\rangle \\ & = \theta(1-\theta)(\lVert x\rVert^2+\lVert y\rVert^2-2\langle x,y\rangle)\\ & = \theta(1-\theta)\lVert x-y\rVert^2. \end{align*}

Answer 2

This is false for other $\ell_p$ norms. Try substituting in $x = e_1$ and $y = e_2$ and $\theta = \frac{1}{2}$, for example. You get a formula which you eventually simplifies to $p=2$.