TITLE: Describe the set of all elements $x,y \in H$, such that $\|x+y\|=\|x\|+\|y\|$.
QUESTION [6 upvotes]: The question is as follows:
Let $H$ be a Hilbert space. Describe the set of all elements $x,y \in H$, such that $\|x+y\|=\|x\|+\|y\|$.
$\textbf{An idea:}$
The set of all such elements will be in a unit sphere that contains a line segment $[x,y]$ where $x,y \in H$ and $x \neq y.$
Such elements are linearly independent, because suppose they are dependent and say $y = \beta x$ for some $\beta \in \mathbb{C}$. Then we have $1 = \|ax + (1-a)\beta x \| = \|a + (1-a)\beta\|$. Then for $a = 0$ we get $|\beta|=1$ and for $a = \frac{1}{2}$ we get $|1+\beta|=2$ which implies that $\beta = 1$ and so $x=y$, which is contradiction.
Can you please let me know if I am wrong?
And if I am wrong? Can you please let me know what is the correct answer?
Thanks!

REPLY [3 votes]: It holds for the set
$$\{(x,y) \in H^2 : x = 0 \text{ or } y = \lambda x \text{ for some } \lambda \ge 0\} = \{(x,y) \in H^2 : x = \lambda y \text{ or } y = \lambda x \text{ for some } \lambda \ge 0\}$$
Indeed, squaring both sides gives:
$$\|x+y\|^2 = (\|x\| + \|y\|)^2 = \|x\|^2 + \|y\|^2 + 2\|x\|\|y\|$$
We have:
$$\|x+y\|^2 = \langle x + y, x + y \rangle = \|x\|^2 + \langle x, y \rangle + \langle y, x \rangle + \|y\|^2 = \|x\|^2 + \|y\|^2 + 2\operatorname{Re}\langle x, y\rangle$$
So we conclude $\operatorname{Re}\langle x, y\rangle = \|x\|\|y\|$.
In general this holds:
$$\left|\operatorname{Re}\langle x, y\rangle\right| \le \left|\langle x, y\rangle\right| \stackrel{CSB}{\le} \|x\|\|y\| $$
so in our case we can deduce $\left|\langle x, y\rangle\right| = \|x\|\|y\|$. Equality in CSB holds if and only if $x, y$ are colinear, meaning there exists $\lambda \in \mathbb{C}$ such that $y = \lambda x$, or $x = 0$.
However, that is not sufficient. Assume $x \ne 0$.
Using the original formula for $-y$ we have:
$$\|x - y\| = \|x\| + \|-y\| = \|x\| + \|y\| = \|x + y\|$$
Squaring gives:
$$\|x\|^2 + \|y\|^2 - \langle x, y\rangle - \langle y, x \rangle = \|x - y\|^2 = \|x + y\|^2 = \|x\|^2 + \|y\|^2 + \langle x, y\rangle + \langle y, x \rangle$$
so $\langle x, y\rangle = - \langle y, x \rangle$
We have
$$\overline{\lambda}\|x\|^2 = \langle x, \lambda x\rangle = \langle x, y\rangle = - \langle y, x \rangle = \langle \lambda x, y\rangle = \lambda\|x\|^2$$
which means $\lambda \in \mathbb{R}$.
Plugging in back to the original formula yields:
$$ |1+\lambda|\|x\| = \|x + \lambda x\|= \|x + y\| = \|x\| + \|y\| = \|x\| + \|\lambda x\| = (1 + |\lambda| )\|x\|$$
So $|1+\lambda| = 1 + |\lambda|$.
Squaring both sides gives:
$$1 + 2\lambda + \lambda^2 = |1+\lambda|^2 = (1 + |\lambda|)^2 = 1 + 2|\lambda| + |\lambda|^2$$
so $\lambda = |\lambda|$ which implies $\lambda \ge 0$.
This is certainly necessary, and you can check that this is also the sufficient condition.