= Solution
There is a genuine inconsistency in this part. For any nonnegative <metallicity distribution function>, its maximum is at least its mean. A population of positive mean solar <metallicity> cannot have a maximum equal to half that mean. For example, a uniform abundance distribution on $[0,2Z_\odot]$ has mean $Z_\odot$ and maximum $2Z_\odot$. Thus the printed maximum-to-mean ratio cannot be proved. The following calculation explains the supplied function, gives the consistent distribution, and also treats the literal birth-metallicity interpretation separately.
Let $f(\mu)$ denote the supplied abundance function. Algebraically it is
$$
f(\mu)=\frac{Y-(Y+1)\mu+\mu^{Y+1}}{(Y+1)(1-\mu)}.
$$
With $\delta=1-\mu$, a <Taylor series> of $(1-\delta)^{Y+1}$ gives
$$
f(1-\delta)=\frac{Y\delta}{2}+\frac{Y(1-Y)\delta^2}{6}+O(\delta^3).
$$
Consequently
$$
\boxed{f(1-\delta)\simeq\frac{Y\delta}{2}.}
$$
This is an early-consumption expansion: its fractional correction is $(1-Y)\delta/3+O(\delta^2)$. Small yield alone does not make the expression linear throughout most of the gas-consumption history. At fixed $0<\mu<1$, the small-$Y$ expansion instead is
$$
f(\mu)=Y\left[1+\frac{\mu\log\mu}{1-\mu}\right]+O(Y^2).
$$
In a well-mixed <closed-box model of galactic chemical evolution>, the <gas fraction of a galaxy> also labels the permanently locked stellar mass: $dM_s=-M_{g,\rm init}\,d\mu$. Newly formed stars inherit the current <gas-phase metallicity> $Z_g(\mu)$. Therefore the mass-weighted mean <stellar metallicity> satisfies the <cumulative and birth stellar metallicities> relation
$$
\overline Z_*(\mu)=\frac1{1-\mu}\int_\mu^1 Z_g(u)\,du,\qquad
Z_g(\mu)=-\frac{d}{d\mu}[(1-\mu)\overline Z_*(\mu)].
$$
Applying the derivative to $f$ gives
$$
\boxed{\overline Z_*(\mu)=f(\mu)\quad\Longleftrightarrow\quad Z_{\rm birth}(\mu)=Z_g(\mu)=1-\mu^Y.}
$$
Thus the supplied function has the natural interpretation of a cumulative mean, not an individual star's birth abundance. Direct integration of $1-u^Y$ verifies that it gives exactly $f$. This is the <saturating closed-box enrichment law>: it corresponds to effective <stellar yield> $Y(1-Z_g)$, with $Z_g$ measured as a fraction. The usual constant-yield trace-metal model instead gives $Z_g=-Y\log\mu$. Both agree to first order when $Y|\log\mu|\ll1$; their exact formulae should not be identified outside that regime.
Let the final gas fraction be $\mu_f>0$ and $\delta_f=1-\mu_f$. In the consistent mean-metallicity interpretation, $\overline Z_*(\mu_f)=Z_\odot$ while the most recently formed stars have the maximum $z_f=1-\mu_f^Y$. At early consumption,
$$
\overline Z_*\simeq\frac{Y\delta_f}{2},\qquad z_f\simeq Y\delta_f,
$$
so the corrected result is
$$
\boxed{Z_{*,\max}\simeq2Z_\odot.}
$$
The assumption here includes $\delta_f\ll1$, equivalently $2Z_\odot/Y\ll1$ in this approximation. Exactly, $z_f=1-\mu_f^Y$ and $Z_\odot=f(\mu_f)$ determine the maximum implicitly; it remains larger than the mean, but its ratio to the mean need not equal two after substantial consumption.
For the mass-weighted <metallicity distribution function>, stars formed while the gas fraction falls by $-d\mu$ have probability $dP=-d\mu/\delta_f$. Writing their individual abundance as $z=1-\mu^Y$ gives $\mu=(1-z)^{1/Y}$ and the <change of variables>
$$
\boxed{dP=F(z)\,dz,\qquad F(z)=\frac{(1-z)^{1/Y-1}}{Y(1-\mu_f)},\quad 0<z<1-\mu_f^Y.}
$$
The density is zero outside this interval. Its cumulative probability is $[1-(1-z)^{1/Y}]/(1-\mu_f)$, so it integrates to one; integrating $zF(z)$ reproduces $f(\mu_f)$. A star-count distribution has the same form if the <initial mass function> and survival selection give a fixed number of observed stars per unit locked mass. Luminosity-weighted samples require different weights.
For trace abundances, $(1-z)^{1/Y-1}\simeq e^{-z/Y}$ to leading order at fixed $z/Y$. The corresponding ordinary <closed-box metallicity distribution> is
$$
F_{\rm trace}(z)=\frac{e^{-z/Y}}{Y(1-\mu_f)},\qquad 0<z<-Y\log\mu_f.
$$
In the early-consumption regime $z_f\ll Y$, this becomes approximately uniform, $F\simeq1/(Y\delta_f)$ on $[0,Y\delta_f]$. Its mean is half its upper endpoint, again producing the corrected maximum $2Z_\odot$.
If the supplied $f(\mu)$ is instead retained literally as the individual birth abundance, set $q=f(\mu)$ rather than $z=1-\mu^Y$. The observation-time population mean is then $\delta_f^{-1}\int_{\mu_f}^1 f(u)\,du$, not $f(\mu_f)$. Its exact distribution can be specified parametrically using
$$
\frac{df}{d\delta}=\frac{1-\mu^Y-f(\mu)}{1-\mu}>0,\qquad
\boxed{F_{\rm literal}(f(\mu))=\frac{1-\mu}{(1-\mu_f)[1-\mu^Y-f(\mu)]},\quad \mu_f<\mu<1.}
$$
Positivity follows because $f$ is the average of the increasing function $1-(1-\delta)^Y$. Its support is $0<q<f(\mu_f)$, and normalization follows directly from $dP=d\delta/\delta_f$. At early consumption it is uniform with $F_{\rm literal}\simeq2/(Y\delta_f)$, maximum $Y\delta_f/2$ and mean $Y\delta_f/4$. Even this literal alternative gives maximum twice the population mean, never half. \b[The distinction between birth abundance and cumulative mean resolves the calculation; the printed inverse maximum-to-mean ratio is false under either interpretation.]
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