Show that if $\left\|\frac{\textbf{x} + \textbf{y}}{2}\right\| = 1$ then $\textbf{x} = \textbf{y}$

Question

Suppose that $\textbf{x}$ and $\textbf{y}$ are unit vectors in $\mathbb{R}$. Show that if $\|\frac{\textbf{x} + \textbf{y}}{2}\| = 1$ Then $\textbf{x} = \textbf{y}$

Attempt:

the only manipulation I could think of to arrive near a solution is:

$$\|\textbf{x + y}\| \leq \|\textbf{x}\| + \|\textbf{y}\| = 2$$

From here I got a partial explanation from my prof, but I can't seem to put the rest together....

Answer 1

Organizing thoughts and correcting statements:

$\|\frac{x+y}{2}\}=1\implies \|x+y\|=2$

Then, by the triangle inequality and the assumption that $x$ and $y$ are unit vectors we have:

$2=\|x+y\|\leq \|x\|+\|y\|=1+1=2$

and so the triangle inequality happened to be strict equality in this case.

In $\Bbb R^n$ equipped with the usual euclidean norm, the triangle inequality is an equality if and only if the triangle formed by $x,y,x+y$ is degenerate, i.e. of area zero, which further implies that $x$ and $y$ lie along the same ray. Phrased differently yet again and looking at the case that $x\neq 0$ and $y\neq 0$, this implies that $x=cy$ for some real scalar $c>0$.

The only way for two unit vectors to both be in the same direction is if they are in fact the same vectors in the first place, hence $x=y$

Answer 2

$$ ||\frac{x+y}{2}||=1 \leftrightarrow ||x+y|| = 2 \leftrightarrow (x+y)^2 = 4 \leftrightarrow (x^2-2xy+y^2) = 0 \leftrightarrow (x-y)^2=0 \leftrightarrow x=y$$

Answer 3

Let $\mathbf x=(x_1,\dots, x_n), \mathbf y=(y_1,\dots, y_n)\in \Re^n$. Then $$1=\left\|\frac{\mathbf x+\mathbf y}{2}\right\|=\frac 12\sqrt{\sum_{i=1}^n (x_i+y_i)^2},$$ which implies that $$4=\sum_{i=1}^n x_i^2+\sum_{i=1}^ny_i^2+2\sum_{i=1}^nx_iy_i=\|\mathbf x\|^2+\|\mathbf y\|^2+2\sum_{i=1}^nx_iy_i=2+2\sum_{i=1}^n x_iy_i.$$ So $$\sum_{i=1}^n x_iy_i=1=\sum_{i=1}^n\frac{x_i^2+y_i^2}{2}.$$ But by the Arithmetic-Geometric inequality, this holds if and only if $x_i=y_i$.

A simpler proof uses the polarization formula: $$\frac 14 (\|\mathbf x+\mathbf y\|^2-\|\mathbf x-\mathbf y\|^2)=\mathbf x\cdot \mathbf y=\frac 12(\|\mathbf x+\mathbf y\|^2-\|\mathbf x\|^2-\|\mathbf y\|^2),$$ from which we can solve that $\|\mathbf x-\mathbf y\|=0$.

Answer 4

Parallelogram identity: $$ \|\mathbf{x}+\mathbf{y}\|^2+\|\mathbf{x}-\mathbf{y}\|^2= 2\|\mathbf{x}\|^2+2\|\mathbf{y}\|^2 $$ that, in your case, becomes $$ 4+\|\mathbf{x}-\mathbf{y}\|^2=2+2 $$

Note. The statement is only true if the Euclidean norm is assumed. Indeed, with the $1$-norm, we have \begin{gather} \|(1,0)\|_1=1\\ \|(0,1)\|_1=1\\ \|(1,0)+(0,1)\|_1=2 \end{gather}