Triangle Equality in a normed linear space

Question

The following statement is true or false:

If $x, y$ are elements of a normed linear space, then $$\|x + y\| = \|x\| + \|y\| \iff x = 0\ \text{or}\ y = tx$$ for some $t ≥ 0$.

What I have tried is as follows:

It is clear that if $x = 0\ \text{or}\ y = tx $ then the equality will hold. But for the converse part, let \begin{align*} \|x+y\|& =\|x\|+\|y\|\\ \implies \|x+y\|^2 & =(\|x\|+\|y\|)^2\\ \end{align*} After that I stuck. Also in the following article Characterization of the norm triangle equality, I have read that in the case of a strictly convex normed linear space $V$, the equality $\|x + y\| = \|x\| + \|y\|$ holds for nonzero vectors $x,y ∈ V$ if and only if $\frac{x}{\|x\|} = \frac{y}{\|y\|} $.

Thank you for the help.

Answer 1

Take $\|(x,y)\| = |x|+|y|$. Then $\|(1,0)\| + \|(0,1)\| = \|(1,1)\|$.

Answer 2

The example of the $\ell^1$-norm on $\mathbb R^2$ illustrating the failure of the proposition in general by user @copper.hat is very nice. In this answer I'll give a substitute that is true in inner product spaces.

Let $(X,\langle\cdot,\cdot\rangle)$ be a real or complex inner product space. (We use $\Re z$ to denote the real part of $z\in\mathbb C$.)

Proposition. Suppose $e_1,\dots,e_n$ in $X$ are unit vectors and $|e_1 + \dots + e_n| = n$. Then $e_1=e_2=\dots=e_n$.

Proof. We use induction on $n$. The case $n = 2$ is the most important, the case $n = 1$ being trivial. In case $n = 2$, consider that $$ 4 = \langle e_1+e_2,e_1+e_2\rangle = 2 + 2\Re\langle e_1,e_2\rangle. $$ Therefore, $\langle e_1-e_2,e_1-e_2\rangle$ is equal to $$ 2 - 2\Re\langle e_1,e_2\rangle = 0. $$ Suppose the claim is true for some $n\ge 2$, and consider the case $n+1$. By assumption and the triangle inequality, $$ n+1=|e_1+\dots+e_{n+1}|\le |e_1+\dots+e_n|+1\le n+1, $$ so all inequalities are equalities, and in particular, $$ |e_1+\dots+e_n| = n. $$ By induction, all $e_1=e_2=\dots=e_n$. Similarly, $e_2=\dots=e_{n+1}$, hence all $e_1=e_2=\dots=e_{n+1}$. $\square$

Note this is not a case of the "all horses are the same color" fallacy of induction because we verified the case $n=2$ as the base of our induction.