How to prove that frame functions on Hilbert spaces are additive?

Question

Let $\newcommand{\calH}{\mathcal{H}}\newcommand{\eff}{\operatorname{Eff}(\calH)}\calH$ be some separable Hilbert space, and denote with $\eff$ the set of effects on $\calH$, that is, the set of positive operators $E$ such that $0\le E\le I$. Define on this space the frame functions as the maps $f:\eff\to[0,1]$ such that $$\sum_kf\left( E_k\right) = 1$$ for every set $\{E_k\}_k\subset\eff$ such that $\sum_k E_k=I$ (I'm taking this definition from Caves et al. (2005)).

It turns out that such frame functions are linear. In particular, they are additive on the set of effects. I'm trying to get a better understanding of why this is the case.

Given any pair of effects $E_1,E_2\in\eff$ such that $E_1+E_2\le I$, I can easily see the additivity: defining $E_3=I-E_1 - E_2$ we see that $E_3\in\eff$ and $E_1+E_2+E_3=I$, and thus $$1 = f(E_1 + E_2 + E_3) = f(E_1) + f(E_2) + f(E_3) = f(E_1 + E_2) + f(E_3),$$ implying $f(E_1 + E_2) = f(E_1) + f(E_2)$. This also seems the argument made in the paper above (see Eq. (4)).

However, the claim seems to be that frame functions are more generally additive on any pair of effects, and thus in particular also when $E_1+E_2\le I$ does not hold. How is the statement proven in this more general case?

Answer 1

Let $\newcommand{\calH}{\mathcal{H}}\newcommand{\eff}{\operatorname{Eff}(\calH)} f$ be a frame function.

Here's a breakdown of the necessary steps to prove linearity of any frame function $f$ on its domain $\eff$:

1. For any $E_1,E_2\in\eff$ such that $E_1+E_2\in\eff$, $f(E_1+E_2)=f(E_1)+f(E_2)$.

   Proof: Define $E_3\equiv I-E_1-E_2$. Then $E_3\in\eff$. Moreover $f(E_1+E_2)+f(E_3)=f(E_1)+f(E_2)+f(E_3)=1$, thus $f(E_1+E_2)=f(E_1)+f(E_2)$.

2. For any $E\in\eff$ and positive integer $k\ge0$, $f(E/k)=f(E)/k$.

   Proof: $E/k\in\eff$ and $k(E/k)\in \eff$, thus $$f(E) = f(\underbrace{E/k+\cdots +E/k}_k) = k f(E/k).$$

3. For any $E_1,E_2\in\eff$, $f(E_1+E_2)=f(E_1)+f(E_2)$.

   Proof: $E_1+E_2\le 2I$, thus $(E_1+E_2)/2\in\eff$, and the conclusion follows from the two previous points: $$f(E_1+E_2) = f(2(E_1+E_2)/2)=2(f(E_1)/2+f(E_2)/2)= f(E_1)+f(E_2).$$

4. For any $q\in\mathbb{Q}\cap[0,\infty]$ and $E\in\eff$, $f(qE)=qf(E)$.

   Proof: Point (3) shows that $f(kE)=k f(E)$ for any positive integer $k$, and point (2) that $f(E/k)=f(E)/k$. Together, these amount to $f(qE)=q f(E)$ for any positive rational $q$.

5. For any $E_1,E_2\in\eff$ such that $E_1\le E_2$, $f(E_1)\le f(E_2)$.

   Proof: From the assumptions follows that $E_2-E_1\in\eff$. Thus $$f(E_2) = f((E_2-E_1)+E_1) = f(E_2-E_1)+f(E_1) \ge f(E_1).$$

6. For any $\alpha\ge0$ and $E\in\eff$, $f(\alpha E)=\alpha f(E)$.

   Proof: Let $(p_k)$ be an increasing sequence converging to $\alpha$ from below, and $(q_k)$ a decreasing sequence converging to $\alpha$ from above. Then $q_k E\le \alpha E\le p_k E$, and from our previous points we have $$p_k f(E) \le f(\alpha E) \le q_k f(E).$$ Going to the limit we conclude that $\alpha f(E)=f(\alpha E)$.

One can then go a step further to extend $f$ as a linear function defined over the full set of Hermitian operators $\mathrm{Herm}(\calH)$. For the purpose, we observe that any $H\in\mathrm{Herm}(\calH)$ can be written as $H=\alpha_+ H_+-\alpha_- H_-$ for some $\alpha_\pm\ge0$ and $H_\pm\in\eff$. We then define $\tilde f:\mathrm{Herm}(\calH)\to\mathbb R$ as $$\tilde f(H) = \alpha_+ f(H_+) - \alpha_- f(H_-).$$ Such $\tilde f$ is thus a linear functional on $\mathrm{Herm}(\calH)$. As such, it can be represented as an inner product, which is essentially the point of (Busch's version of) Gleason's theorem.

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I then also found that essentially the same argument is presented in (Busch 2003).