Please help with a general solution of a functional equation involving projections I saw the following claimed :
Let's say we have the functional equation
$f(R+S) = f(R) + f(S)$
where R and S are projections in a vector space, and f is a real valued function. Then its general solution is :
$f(R) = c\ Tr(R)$, where $c$ is some constant and Tr is the trace.
The reason given is that 

$Tr(R)$ is the only linear invariant that depends only on $R$

I'm not sure what that means, and how it shows that the general solution is what is claimed. Can anyone please clarify?
 A: In a Hilbert space $\mathcal{H}$, $R$ and $S$ are bounded operators
and hence are elements of $\mathcal{B(H)}$, the space of bounded operators
on $\mathcal{H}$. But then, assuming linearity, $f$ can be chosen to be a
linear functional on $\mathcal{B(H}$). Particular such functionals are the
states $\Phi (A)$, $A\in \mathcal{B(H)}$, positive functionals of norm 1
(https://en.wikipedia.org/wiki/State)(  https://en.wikipedia.org/wiki/Trace_class). These states
decompose into normal states, elements of the pre-dual of $\mathcal{B(H)}$
which is the trace class $\mathcal{B}_{1}(\mathcal{H})$ and more general
elements from the dual of $\mathcal{B(H)}$. A trace class state can be
written as
\begin{equation*}
\Phi (A)=\mathrm{Trace}\rho A,\;\rho \in \mathcal{B}_{1}(\mathcal{H}),\;\rho
>0,\;\mathrm{Trace}\rho =1.\;
\end{equation*}
which is the formula you mention.
There are some problems. In case the multiplicity of $R$ is infinite the
trace does not exist. Secondly, how do we exclude more general states?
