Describe the set of all elements $x,y \in H$, such that $\|x+y\|=\|x\|+\|y\|$. 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!
 A: (Consider first a real Hilbert space $H$. The complex case follows below.)
Squaring both sides of the equality we get
$$
(1) \qquad \|x\| \, \|y\| = \langle x , y \rangle .
$$
On the other hand, the Cauchy-Schwarz inequality gives
$$
\langle x , y \rangle \leq \|x\| \, \|y\|
$$
with equality if and only if $x=\lambda y$ or $y = \lambda x$ for some $\lambda \geq 0$.
Hence equality in (1) holds only for this kind of vectors.
If $H$ is a complex Hilbert space, the above proof needs a small modification.
Namely, reasoning as above we get that the required equality holds if and only if
$$
(2) \qquad \text{Re}\langle x , y \rangle = \|x\| \, \|y\|.
$$
Clearly, (2) holds if $x=0$ or $y=0$.
Assume now that $x, y\neq 0$, and consider the normalized vectors
$\xi := x/\|x\|$, $\eta := y / \|y\|$.
Since $\|\xi\| = \|\eta\| = 1$, equality (2) (for nonzero vectors) is
equivalent to
$$
(3) \qquad \text{Re}\langle \xi , \eta \rangle = 1.
$$
we have that
$$
\|\xi - \eta\|^2 = 2 ( 1 - \text{Re}\langle \xi , \eta \rangle)
$$
hence (3) holds if and only if $\xi = \eta$, i.e., if and only if $x$ is a positive multiple of $y$. 
A: 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.
