Are $f$ satisfying $|f(y)| = |f(x+y) - f(x)|$ additive? (This question is inspired by this question, and in particular the comment by Charlie Cunningham.)

Let $(V, \|\cdot\|)$ be a normed real vector space. Let $f: V \to V$ be a function satisfying, for all $x, y \in V$,
  $$
\|f(y)\| = \|f(x+y) - f(x)\|.
$$
  Does it follow that $f$ is additive, i.e., that $f(x + y) = f(x) + f(y)$ for all $x, y \in V$?

This seems like it should be a fairly easy question, but I haven't quite figured it out yet. Some fairly trivial observations:


*

*Taking $y=0$ gives us $\|f(0)\| = \|f(x) - f(x)\| = 0$, so $f(0) = 0$.

*Taking $y=-x$ gives us $\|f(-x)\| = \|-f(x)\|$.

*If we could prove that $f(-x) = -f(x)$, we could take $y=-2x$ to prove that $\|f(2y)\| = 2\|f(y)\|$. 

*Using the equation for both $y$ and $-y$ gives us
$$
\|f(x+y) - f(x)\| = \|f(x-y) - f(x)\|.
$$

*It does not matter that $f$ is an endofunction (it could go $f: V \to W$ too) for the answer of the question, but this was a simpler formulation.

 A: No.  
In case of maps $V \to W$ a counterexample for $\mathbb R \to \mathbb R^2$ with the $l^\infty$ norm is found in
Väisälä, Jussi, A proof of the Mazur-Ulam theorem, Am. Math. Mon. 110, No. 7, 633-635 (2003). ZBL1046.46017.  
(and I didn't check, but maybe that example is already in the 1932 paper of Mazur & Ulam)
Here, I will adapt that example to get one mapping a space to itself.

Example
$(V , \|\cdot\|)$ is the sequence space $c_0$.  That is, elements of $V$ are infinite sequences $\mathbf x = (x_1,x_2,x_3,\cdots)$ of real numbers such that $\lim_{j\to\infty} x_j = 0$.  The norm is
$$
\|\mathbf x\| = \max\big\{|x_1|,|x_2|,|x_3|,\cdots\big\}.
$$
It is well-known that this $V$ is a Banach space.  
The map $f : V \to V$ is defined as follows.  If $\mathbf x = (x_1,x_2,x_3,\cdots)$, then $f(\mathbf x) = (|x_1|, x_1, x_2, x_3,\cdots)$.   
We claim that $f$ satisfies $\|f(\mathbf x+\mathbf y) - f(\mathbf y)\| =
\|f(\mathbf x)\|$ for all $\mathbf x, \mathbf y \in V$.
First, $\|f(\mathbf x)\| = \|\mathbf x\|$.  Indeed,
$$
\|\mathbf x\| = \max\big\{|x_1|,|x_2|,|x_3|,|x_4|,\cdots\big\}
=\max\big\{|x_1|,|x_1|,|x_2|,|x_3|,|x_4|,\cdots\big\} = \|f(\mathbf x)\| .
$$
Next, to compute $\|f(\mathbf x+\mathbf y) - f(\mathbf y)\|$, note that by the triangle inequality $\big||x_1+y_1|-|y_1|\big| \le |x_1|$.  So
\begin{align}
f(\mathbf x+\mathbf y) - f(\mathbf y) &=
\big(|x_1+y_1|, x_1+y_1, x_2+y_2, x_3+y_3,\cdots\big) -
\big(|y_1|,y_1,y_2,y_3,\cdots \big)
\\ &=
\big(|x_1+y_1|-|y_1|, x_1, x_2, x_3,\cdots\big)
\\
\|f(\mathbf x+\mathbf y) - f(\mathbf y)\| &= 
\max\big\{\big||x_1+y_1|-|y_1|\big|, |x_1|, |x_2|, |x_3|,\cdots\big\}
\\&=\max\big\{|x_1|, |x_2|, |x_3|,\cdots\big\} = \|\mathbf x\| = \|f(\mathbf x)\|
\end{align}
This completes the proof that $\|f(\mathbf x+\mathbf y) - f(\mathbf y)\| =
\|f(\mathbf x)\|$.  

Now a counterexample shows that $f$ is not additive:
\begin{align}
\mathbf x &= (1,0,0,0,\cdots)\\
\mathbf y &= (-1,0,0,0,\cdots)\\
\mathbf x + \mathbf y &= (0,0,0,0,\cdots)\\
f(\mathbf x) &= (1,1,0,0,0,\cdots)\\
f(\mathbf y) &= (1,-1,0,0,0,\cdots)\\
f(\mathbf x) + f(\mathbf y) &= (2,0,0,0\cdots)\\
f(\mathbf x + \mathbf y) &= (0,0,0,0,0,\cdots)\\
f(\mathbf x + \mathbf y) &\ne f(\mathbf x) + f(\mathbf y)
\end{align}
