[proofplan] We prove the duality by identifying a single formal Cauchy kernel in two bases. First we expand the kernel $\prod_{i,j}(1-x_i y_j)^{-1}$ as $\sum_\lambda h_\lambda(x)m_\lambda(y)$ by grouping monomials in the $y$-variables by their exponent partitions. Then we expand the same kernel in the power-sum basis as $\sum_\lambda z_\lambda^{-1}p_\lambda(x)p_\lambda(y)$. The second expansion is the reproducing kernel for the Hall [inner product](/page/Inner%20Product), so comparing the two expressions forces $\langle h_\lambda,m_\mu\rangle=\delta_{\lambda,\mu}$. [/proofplan] [step:Define the Cauchy kernel in two independent alphabets] Let $x = (x_1,x_2,\dots)$ and $y = (y_1,y_2,\dots)$ be two independent countable alphabets. We work in the completed graded [tensor product](/page/Tensor%20Product) \begin{align*} \widehat{\operatorname{Sym}_{\mathbb{Q}}(x) \otimes_{\mathbb{Q}} \operatorname{Sym}_{\mathbb{Q}}(y)}, \end{align*} where completion is by total homogeneous degree. Define the Cauchy kernel $K(x,y)$ by \begin{align*} K(x,y) := \prod_{i \geq 1}\prod_{j \geq 1}(1-x_i y_j)^{-1}. \end{align*} This product is well-defined as a formal [power series](/page/Power%20Series) because, in every fixed total degree, only finitely many factors contribute. [/step] [step:Expand the Cauchy kernel in the complete and monomial bases] For each fixed index $j \geq 1$, the generating function for complete homogeneous symmetric functions gives \begin{align*} \prod_{i \geq 1}(1-x_i y_j)^{-1} = \sum_{r \geq 0} h_r(x)y_j^r. \end{align*} Multiplying over all $j \geq 1$, we obtain \begin{align*} K(x,y) = \prod_{j \geq 1}\left(\sum_{r \geq 0} h_r(x)y_j^r\right). \end{align*} For a finitely supported sequence of nonnegative integers $a=(a_1,a_2,\dots)$, write $y^a := \prod_{j \geq 1} y_j^{a_j}$ and define $h_a(x) := \prod_{j \geq 1} h_{a_j}(x)$, where factors with $a_j=0$ equal $h_0(x)=1$. Then the coefficient of $y^a$ in the preceding product is $h_a(x)$. Let $\lambda$ be the partition obtained by rearranging the positive entries of $a$ in weakly decreasing order. Since multiplication in $\operatorname{Sym}_{\mathbb{Q}}$ is commutative, $h_a(x)=h_\lambda(x)$. Grouping all monomials $y^a$ whose positive exponent multiset is $\lambda$ gives the monomial symmetric function $m_\lambda(y)$. Therefore \begin{align*} K(x,y)=\sum_{\lambda} h_\lambda(x)m_\lambda(y), \end{align*} where the sum ranges over all partitions $\lambda$. [guided] The point of this step is to see why the complete homogeneous functions appear on the $x$-side and the monomial functions appear on the $y$-side. Fix one $y$-variable, say $y_j$. By definition, $h_r(x)$ is the sum of all monomials of total degree $r$ in the $x$-alphabet, so its ordinary generating function is \begin{align*} \sum_{r \geq 0}h_r(x)y_j^r = \prod_{i \geq 1}(1-x_i y_j)^{-1}. \end{align*} Multiplying this identity over all $j \geq 1$ gives \begin{align*} K(x,y) = \prod_{j \geq 1}\left(\sum_{r \geq 0} h_r(x)y_j^r\right). \end{align*} Now choose a finitely supported sequence $a=(a_1,a_2,\dots)$ of nonnegative integers. The monomial $y^a=\prod_{j\geq 1}y_j^{a_j}$ is obtained by choosing the term $h_{a_j}(x)y_j^{a_j}$ from the $j$th factor for each $j$. Hence its coefficient is \begin{align*} \prod_{j \geq 1}h_{a_j}(x). \end{align*} If $\lambda$ is the partition formed by rearranging the nonzero entries of $a$, then this coefficient is exactly $h_\lambda(x)$, because $h_\lambda$ means the product of the $h_{\lambda_i}$ and multiplication is commutative. Finally, the monomial symmetric function $m_\lambda(y)$ is the sum of all distinct monomials $y^a$ whose nonzero exponent multiset is $\lambda$. Thus grouping the expansion according to the partition shape of the exponent sequence gives \begin{align*} K(x,y)=\sum_{\lambda} h_\lambda(x)m_\lambda(y). \end{align*} [/guided] [/step] [step:Expand the same kernel in the power-sum basis] Using the formal identity $\log(1-t)^{-1}=\sum_{r\geq 1}t^r/r$, applied degree by degree, we compute \begin{align*} \log K(x,y) &= \sum_{i,j \geq 1}\log(1-x_i y_j)^{-1} \\ &= \sum_{i,j \geq 1}\sum_{r\geq 1}\frac{x_i^r y_j^r}{r} \\ &= \sum_{r\geq 1}\frac{1}{r}\left(\sum_{i\geq 1}x_i^r\right)\left(\sum_{j\geq 1}y_j^r\right) \\ &= \sum_{r\geq 1}\frac{p_r(x)p_r(y)}{r}. \end{align*} Exponentiating gives \begin{align*} K(x,y) = \prod_{r\geq 1}\exp\left(\frac{p_r(x)p_r(y)}{r}\right). \end{align*} Expanding each exponential factor yields \begin{align*} K(x,y) &= \prod_{r\geq 1}\sum_{m\geq 0}\frac{1}{m!}\left(\frac{p_r(x)p_r(y)}{r}\right)^m \\ &= \sum_{\lambda} z_\lambda^{-1}p_\lambda(x)p_\lambda(y), \end{align*} because choosing $m_r(\lambda)$ copies of the $r$th factor contributes \begin{align*} \prod_{r\geq 1}\frac{1}{m_r(\lambda)!r^{m_r(\lambda)}}=z_\lambda^{-1}. \end{align*} [/step] [step:Use the power-sum expansion as the reproducing kernel for the Hall inner product] Let $f\in \operatorname{Sym}_{\mathbb{Q}}$ be arbitrary, and write its power-sum expansion as \begin{align*} f(x)=\sum_{\lambda} a_\lambda p_\lambda(x), \end{align*} where $a_\lambda\in\mathbb{Q}$ and all but finitely many $a_\lambda$ are zero. Define \begin{align*} R(x,y):=\sum_{\lambda} z_\lambda^{-1}p_\lambda(x)p_\lambda(y). \end{align*} Using bilinearity of the Hall inner product in the $x$-variables and the defining relation $\langle p_\nu,p_\lambda\rangle=z_\nu\delta_{\nu,\lambda}$, we obtain \begin{align*} \langle f(x), R(x,y)\rangle_x &= \left\langle \sum_{\nu}a_\nu p_\nu(x), \sum_{\lambda}z_\lambda^{-1}p_\lambda(x)p_\lambda(y)\right\rangle_x \\ &= \sum_{\nu,\lambda}a_\nu z_\lambda^{-1}\langle p_\nu(x),p_\lambda(x)\rangle p_\lambda(y) \\ &= \sum_{\nu,\lambda}a_\nu z_\lambda^{-1}z_\nu\delta_{\nu,\lambda}p_\lambda(y) \\ &= \sum_{\nu}a_\nu p_\nu(y) \\ &= f(y). \end{align*} Thus $R(x,y)$ is the reproducing kernel for the Hall inner product. Since the previous step showed $R(x,y)=K(x,y)$, the Cauchy kernel $K(x,y)$ has the same reproducing property. [/step] [step:Compare coefficients in the monomial basis to obtain the duality] Using the complete-monomial expansion of $K(x,y)$ and applying the reproducing property to $f(x)=m_\mu(x)$, where $\mu$ is a fixed partition, gives \begin{align*} m_\mu(y) &= \langle m_\mu(x),K(x,y)\rangle_x \\ &= \left\langle m_\mu(x),\sum_{\lambda}h_\lambda(x)m_\lambda(y)\right\rangle_x \\ &= \sum_{\lambda}\langle m_\mu,h_\lambda\rangle m_\lambda(y). \end{align*} The monomial symmetric functions $(m_\lambda)_\lambda$ form a basis of $\operatorname{Sym}_{\mathbb{Q}}$, so equality of coefficients in this basis gives \begin{align*} \langle m_\mu,h_\lambda\rangle=\delta_{\lambda,\mu}. \end{align*} The Hall inner product is symmetric because its defining Gram matrix in the power-sum basis satisfies \begin{align*} \langle p_\alpha,p_\beta\rangle=z_\alpha\delta_{\alpha,\beta} =z_\beta\delta_{\beta,\alpha} =\langle p_\beta,p_\alpha\rangle. \end{align*} Therefore \begin{align*} \langle h_\lambda,m_\mu\rangle=\langle m_\mu,h_\lambda\rangle=\delta_{\lambda,\mu}. \end{align*} Since $(h_\lambda)_\lambda$ and $(m_\lambda)_\lambda$ are bases indexed by the same set of partitions, this proves that they are dual bases with respect to the Hall inner product. [/step]