On a function from a group with values lying in an inner product space Let $(G,+)$ be a group and $V$ be an inner product space (over $\mathbb R$ , or $ \mathbb C $ ) ; let $f:G \to V$ be a function such that $||f(x+y)||\ge ||f(x)+f(y)|| , \forall x,y\in G$ , then how to prove that $f(x+y)=f(x)+f(y),\forall x,y \in G$ ? 
If $(S,+)$ is a semigroup and $V$ be an inner product space (over $\mathbb R$ , or $ \mathbb C $ ) ; let $f:S \to V$ be a function such that $||f(x+y)||= ||f(x)+f(y)|| , \forall x,y\in S$ , then I can prove that  $f(x+y)=f(x)+f(y),\forall x,y \in S$ . But I don't know how to do the the aforementioned problem.
Please help . Thanks in advance 
 A: First we'll need a general fact about inner product spaces. Suppose $u$, $v$, $w$ elements of an inner product space satisfying the three inequalities
$$
\tag{$1$}
\|u\|\geq \|v+w\|,\,\,\,\, \|v\|\geq \|u+w\|,\,\,\, \|w\|\geq \|u+v\|.
$$
Then it follows that $u+v+w=0$. To see this, write $s=u+w+v$. In terms of $s$, the inequalities can be written (resp.)
$$
\mathrm{Re}\, \langle s,2u-s\rangle\geq 0,\,\,\,\, \mathrm{Re}\, \langle s,2v-s\rangle\geq 0,\,\,\,\,  \mathrm{Re}\, \langle s,2w-s\rangle\geq 0.
$$
Sum these inequalities to get $\mathrm{Re}\, \langle s,-s\rangle\geq 0$, which implies $s=0$.
We'll also need a fact about $f$. In your inequality, take $x=y=0$ to find $f(0)=0$, and take $y=-x$ to find $f(-x)=-f(x)$, so that also $\|f(-x)\|=\|f(x)\|$.
Now, for arbitrary $x$, $y\in G$, set $u=f(x)$, $v=f(y)$, and $w=-f(x+y)=f(-(x+y))$. For this choice of $u$, $v$, $w$, the three inequalities $(1)$ comes from the functional inequality for $f$ via the substitutions (resp.)
$$
\begin{eqnarray*}
&x&\mapsto y,\,\,\,&y&\mapsto -(x+y),\\
&x&\mapsto-(x+y),\,\,\,&y&\mapsto x,\\
&x&\mapsto x\,\,\,&y&\mapsto y.
\end{eqnarray*}
$$
We conclude that $u+v+w=f(x)+f(y)-f(x+y)=0$.
