Show that $f$ is a Cauchy function if $|f(x + y)| = |f(x)| + |f(y)|$.

Question

Let $f \colon \mathbb{R} \to \mathbb{R}$ be a solution of the functional equation $$|f(x + y)| = |f(x)| + |f(y)| \quad \forall x,y \in\mathbb{R}\text.$$ Show that $f$ is an additive function.

Answer 1

You have $|f(0+0)| = |f(0)|+|f(0)|$ this imply $f(0) = 0$

But you also have $\forall x \in \mathbb{R}$

$$ |f(x)| + |f(-x)| = |f(0)| = 0$$

Hence $\forall x \in \mathbb{R} f(x) = 0 $ and $f$ is the null function

Unless you made a copy mistake...

Answer 2

Take $y = -x$: we have $|f(0)| = |f(x)| + |f(-x)|$. In particular $|f(x)| \le |f(0)|$, so $f$ is bounded. And since $|f(nx)| = n |f(x)|$ for all positive integers $n$, this implies $f = 0$.