Skew Schur functions by Hall adjointness

Suppose $d\ge 0$. The index set $\{\nu :\nu ⊢d\}$ is finite by [F2]. For each such $\nu$, [F1] gives $s_{\mu }{s}_{\nu }\in {\mathrm{\Lambda }}^{\mathrm{\mid }\mu \mathrm{\mid }+d}={\mathrm{\Lambda }}^{\mathrm{\mid }\lambda \mathrm{\mid }}$, so its Hall pairing with $s_{\lambda }$ is defined and integral by [F3]. The displayed finite sum is therefore a well-defined element of ${\mathrm{\Lambda }}^d$ by [F1] and [F4]. If $d<0$, the separately specified zero is well-defined in $\mathrm{\Lambda }$.

Let $\rho$ be any partition. If $d\ge 0$ and $\mathrm{\mid }\rho \mathrm{\mid }=d$, orthonormality in [F4] gives $⟨s_{\lambda \mathrm{/}\mu },s_{\rho }{⟩}_H=⟨s_{\lambda },s_{\mu }{s}_{\rho }{⟩}_H$ by extracting the $s_{\rho }$ coefficient in the defining sum. If $\mathrm{\mid }\rho \mathrm{\mid }\mathrm{\ne }d$, the left side is zero by [F1] and [F3], while the right side is zero because $s_{\mu }{s}_{\rho }$ has degree $\mathrm{\mid }\mu \mathrm{\mid }+\mathrm{\mid }\rho \mathrm{\mid }\mathrm{\ne }\mathrm{\mid }\lambda \mathrm{\mid }$. If $d<0$, then $\mathrm{\mid }\mu \mathrm{\mid }+\mathrm{\mid }\rho \mathrm{\mid }>\mathrm{\mid }\lambda \mathrm{\mid }$, so both sides again vanish. Thus the adjointness identity holds for every partition $\rho$; it determines the element uniquely because the Schur functions are an orthonormal basis in each degree. For $\mu =\mathrm{\varnothing }$, [F5] gives $s_{\mathrm{\varnothing }}=1$, and the definition yields $s_{\lambda \mathrm{/}\mathrm{\varnothing }}=s_{\lambda }$; in particular $s_{(1)\mathrm{/}\mathrm{\varnothing }}=s_{(1)}$. When $\mathrm{\mid }\lambda \mathrm{\mid }=\mathrm{\mid }\mu \mathrm{\mid }$, it gives $s_{\lambda \mathrm{/}\mu }=1$ if $\lambda =\mu$ and zero otherwise.