Does one of these conditions for norms follow from the other?

Question

The two conditions are:

For all unit vectors $\mathbf{x}$ and $\mathbf{y}\hspace{-0.02 in}$, $\:$ if $\; \left|\left|\hspace{.03 in}\mathbf{x}\hspace{-0.05 in}+\hspace{-0.04 in}\mathbf{y}\right|\right| = 2 \;$ then $\: \mathbf{x} = \mathbf{y} \;$.
For all vectors $\mathbf{x}$ and $\mathbf{y}\hspace{-0.02 in}$, $\;$ if $\;\; \left|\left|\hspace{.03 in}\mathbf{x}\hspace{-0.05 in}+\hspace{-0.04 in}\mathbf{y}\right|\right| \: = \: \left|\left|\hspace{.02 in}\mathbf{x}\hspace{.02 in}\right|\right| + \left|\left|\hspace{.02 in}\mathbf{y}\right|\right| \;\;$ then $\mathbf{x}$ and $\mathbf{y}$ are colinear.

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Obviously, the lower of those conditions implies the upper of those conditions.
Does the lower of those conditions follow from the upper of those conditions?

Answer 1

Yes, the first condition implies the second.

Let's prove the contrapositive: If there are linearly independent $x,y$ with $\lVert x+y\rVert = \lVert x\rVert + \lVert y\rVert$, then there are two linearly independent unit vectors $u,v$ with $\lVert u+v\rVert = 2$.

Since the norm is homogeneous, we can assume that $\lVert x\rVert + \lVert y\rVert = 2$. If $\lVert x\rVert = \lVert y\rVert$, we are done. Otherwise, suppose without loss of generality $1-\eta = \lVert x\rVert < \lVert y\rVert = 1+\eta$. Then set $v = \frac{1}{1+\eta}y$ and $u = x + \frac{\eta}{1+\eta}y$. Then $u$ and $v$ are linearly independent (in particular different), $u+v = x+y$, hence $\lVert u+v\rVert = 2$, and $\lVert v\rVert = \frac{1}{1+\eta}\lVert y\rVert = 1$, as well as

$$\lVert u\rVert = \left\lVert x + \frac{\eta}{1+\eta}y\right\rVert \leqslant \lVert x\rVert + \frac{\eta}{1+\eta}\lVert y\rVert = (1-\eta) + \eta = 1,$$

while on the other hand $\lVert u\rVert \geqslant \lVert u+v\rVert - \lVert v\rVert = 2-1=1$, hence also $\lVert u\rVert = 1$.

Answer 2

If the norm arises from an inner product on some real Hilbert space, then yes.

To see this, note that $2 =||x+y|| = \langle x + y, x +y \rangle = ||x||^2 + 2Re\langle x, y \rangle + ||y||^2 = 2 + 2Re\langle x, y \rangle \implies Re\langle x,y \rangle = \langle x, y \rangle = 0$. So using, $\langle x, y \rangle = ||x|| ||y|| \cos \theta$ gives us that $\cos \theta = 0 \implies x,y $ are collinear.