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....

Hint: when does the triangle inequality hold as an equality? — dxiv, Feb 03 '18 at 00:34
when $\textbf{x} = 0$ or $\textbf{y} = 0$, but I don't see how that helps completely.... — D.C. the III, Feb 03 '18 at 00:43
@dc3rd Look at the one-dimensional case as a further hint. The triangle inequality implies the inequality in $3=|1+2|\leq |1|+|2|=3$, meanwhile $1=|1+(-2)|\leq |1|+|-2|=3$ — JMoravitz, Feb 03 '18 at 00:48
@dc3rd No. See Prove equality in triangle inequality for complex numbers for example. — dxiv, Feb 03 '18 at 00:49
@herbsteinberg incorrect, one would need the additional assumption that $|x|=|y|$ in order to conclude that $x=y$. Using my example above again $|1+2|=|1|+|2|$ despite $1\neq 2$. — JMoravitz, Feb 03 '18 at 01:04
It's called the triangle inequality for a reason. When does a triangle have its third leg exactly equal to the sum of its two other legs. Hint 2: The shortest distance between two points is a straight line. So if you visit a third point along the way that isn't colinear, can that be just as short? — fleablood, Feb 03 '18 at 01:25
I stand corrected. However, both vectors are of the same length, while my original statement should have said they were parallel. In this case they are the same, i.e. same length and parallel. — herb steinberg, Feb 04 '18 at 17:17

egreg · Answer 1 · 2018-02-03T10:10:07.660

3

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}

edited Feb 03 '18 at 10:10

answered Feb 03 '18 at 01:05

egreg

238,574

score 2 · Accepted Answer · answered Feb 03 '18 at 01:03

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 $x=y$

OnoL · Answer 3 · 2018-02-03T01:07:51.240

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 $x_i=y_i$.

score 0 · Answer 4 · answered Feb 03 '18 at 00:56

0

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

answered Feb 03 '18 at 00:56

bigfocalchord

5,597

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

4 Answers4