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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 isWith $\delta =1-\mu$, a Taylor series of $(1-\delta {)}^{Y+1}$ givesConsequentlyThis 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

In a well-mixed closed-box model of galactic chemical evolution, the gas fraction of a galaxy also labels the permanently locked stellar mass: $d{M}_s=-M_{g,\mathrm{init}}d\mu$. 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 relationApplying the derivative to $f$ givesThus 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 $Y|\log \mu |\ll 1$; 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, ${\overline{Z}}_{\ast }({\mu }_f)=Z_{\odot }$ while the most recently formed stars have the maximum $z_f=1-{\mu }_f^Y$. At early consumption,so the corrected result isThe assumption here includes ${\delta }_f\ll 1$, equivalently $2Z_{\odot }/Y\ll 1$ 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 variablesThe 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 isIn the early-consumption regime $z_f\ll Y$, this becomes approximately uniform, $F\simeq 1/(Y{\delta }_f)$ 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 usingPositivity 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 $F_{\mathrm{literal}}\simeq 2/(Y{\delta }_f)$, maximum $Y{\delta }_f/2$ and mean $Y{\delta }_f/4$. Even this literal alternative gives maximum twice the population mean, never half. The distinction between birth abundance and cumulative mean resolves the calculation; the printed inverse maximum-to-mean ratio is false under either interpretation.