$\newcommand\rank[1]{\lvert#1\rvert}$Let $\Bbb{P}$ be a 1-differential poset with a unique bottom element $\emptyset \in \Bbb{P}$. ~~With some minor abuse in terminology,~~ The
Plancherel ~~ measure~~ state $\varphi_\mathrm{P}$ is the real-valued function defined for each $x \in \Bbb{P}$ by
\begin{equation}
\varphi_\mathrm{P}(x) \mathrel{:=} \, {1 \over {\, \rank x!} }
\dim \bigl(\emptyset, x \bigr),
\end{equation}
where $\rank x= n$ is the rank of the element, $\dim\bigl(\emptyset, x \bigr)$
is the number of saturated chains $x_0 \lhd \, \dotsb \lhd \, x_n$
in $\Bbb{P}$ starting at $x_0 = \emptyset$ and ending at $x_n = x$,
and $x \lhd \, y$ signifies the covering relation in the poset.
The Plancherel measure is related to $\varphi_\mathrm{P}$ as the former can be expressed as the mapping $\Bbb{P} \ni x \mapsto \varphi_\mathrm{P}(x) \cdot \dim \bigl(\emptyset, x \bigr)$.
Clearly
$\varphi_\mathrm{P}$ is *non-negative* (indeed, it's strictly positive),
it is *normalised* (i.e. $\varphi_\mathrm{P}(\emptyset) = 1$), and it
is *harmonic* in the sense that
\begin{equation}
\varphi_\mathrm{P}(x) = \, \sum_{x \lhd \, y} \, \varphi_\mathrm{P}(y)
\end{equation}

**Question:** Is the Plancherel state $\varphi_\mathrm{P}$ always *minimal* in the sense that
whenever $\varphi_\mathrm{P} = s\phi_1 + (1-s)\phi_2$ for some parameter $0 < s < 1$ and for some pair
of non-negative, normalised, harmonic functions $\phi_1, \phi_2:
\Bbb{P} \longrightarrow \Bbb{R}$ then $\varphi_\mathrm{P} = \phi_1 = \phi_2$? If the assertion is not true in general, why is it true, for example, in the case of the Young–Fibonacci Lattice $\Bbb{P} = \Bbb{YF}$ ?

regards, ines.

P.S. This question is a follow-up to Minimal harmonic functions on a poset, which was recently posted on Math StackExchange.

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