[proofplan] We expand $f$ in the Schur basis and use the Schur expansion of the Cauchy kernel. Since the Hall [inner product](/page/Inner%20Product) is taken only in the $X$ variables, each coefficient is computed by pairing $f[X]$ against $s_\lambda[X]$. The orthonormality of the Schur functions collapses the resulting sum to the original Schur expansion of $f[Y]$. [/proofplan] [step:Expand the symmetric function in the Schur basis] Let $\mathcal{P}$ denote the set of integer partitions. Since the Schur functions form a $\mathbb{Q}$-basis of $\operatorname{Sym}_{\mathbb{Q}}$, there exist a finite subset $A \subset \mathcal{P}$ and coefficients $a_\mu \in \mathbb{Q}$, indexed by $\mu \in A$, such that \begin{align*} f[X]=\sum_{\mu \in A} a_\mu s_\mu[X]. \end{align*} The same symmetric function evaluated in the alphabet $Y$ is therefore \begin{align*} f[Y]=\sum_{\mu \in A} a_\mu s_\mu[Y]. \end{align*} [/step] [step:Pair the Schur expansion of $f$ with the Cauchy kernel] Using the Cauchy expansion \begin{align*} \Omega[X,Y]=\sum_{\lambda \in \mathcal{P}}s_\lambda[X]s_\lambda[Y], \end{align*} and linearity of $\langle\cdot,\cdot\rangle_X$ in the $X$ variables, we obtain \begin{align*} \langle f[X],\Omega[X,Y]\rangle_X &= \left\langle \sum_{\mu \in A} a_\mu s_\mu[X], \sum_{\lambda \in \mathcal{P}}s_\lambda[X]s_\lambda[Y] \right\rangle_X \\ &= \sum_{\lambda \in \mathcal{P}} \sum_{\mu \in A} a_\mu \langle s_\mu[X],s_\lambda[X]\rangle_X s_\lambda[Y]. \end{align*} This manipulation is well-defined because the expansion of $f[X]$ has finite support in the Schur basis, so for each $\lambda$ only finitely many coefficients from $f[X]$ are involved. [guided] The inner product is taken only in the alphabet $X$, so every factor $s_\lambda[Y]$ is a scalar coefficient with respect to that pairing. We first write both inputs in the Schur basis: \begin{align*} f[X]=\sum_{\mu \in A}a_\mu s_\mu[X], \qquad \Omega[X,Y]=\sum_{\lambda \in \mathcal{P}}s_\lambda[X]s_\lambda[Y]. \end{align*} Here $A$ is finite because $f$ is an ordinary symmetric function, not an infinite formal series. Therefore bilinearity of the Hall inner product in the $X$ variables gives \begin{align*} \langle f[X],\Omega[X,Y]\rangle_X &= \sum_{\lambda \in \mathcal{P}} \left\langle \sum_{\mu \in A}a_\mu s_\mu[X], s_\lambda[X] \right\rangle_X s_\lambda[Y] \\ &= \sum_{\lambda \in \mathcal{P}} \sum_{\mu \in A} a_\mu \langle s_\mu[X],s_\lambda[X]\rangle_X s_\lambda[Y]. \end{align*} The point of this expansion is that the Hall inner product was chosen precisely so that the Schur basis is orthonormal. Thus the kernel $\Omega[X,Y]$ is set up to extract the Schur coefficients of $f[X]$ and reproduce the same linear combination in the alphabet $Y$. [/guided] [/step] [step:Use Schur orthonormality to collapse the coefficient sum] By the defining orthonormality of the Schur basis for the Hall inner product, \begin{align*} \langle s_\mu[X],s_\lambda[X]\rangle_X=\delta_{\mu\lambda}. \end{align*} Substituting this into the previous expression gives \begin{align*} \langle f[X],\Omega[X,Y]\rangle_X &= \sum_{\lambda \in \mathcal{P}} \sum_{\mu \in A} a_\mu \delta_{\mu\lambda} s_\lambda[Y] \\ &= \sum_{\mu \in A} a_\mu s_\mu[Y] \\ &= f[Y]. \end{align*} This proves the reproducing identity for every $f \in \operatorname{Sym}_{\mathbb{Q}}$. [/step]