Solution

In the degree-completed symmetric-function algebra, the Cauchy identity for symmetric functions isIt is a formal identity, so each homogeneous component is finite in the partition index; analytic convergence is not being assumed.

Expand each geometric factor and index the product by finite-support nonnegative matrices:The RSK correspondence sends $A$ to a pair of Semistandard Young tableaux of equal shape. Its content identities turn the matrix weight into the product of the recording tableau's $x$ weight and the insertion tableau's $y$ weight. Summing independently over both tableaux of each shape proves the identity.

Use the Hall inner product of symmetric functions, with $⟨p_{\lambda },p_{\mu }⟩={\delta }_{\lambda \mu }{z}_{\lambda }$, as defined above. The left side is the reproducing kernel ${\sum }_{\lambda }{p}_{\lambda }(x)p_{\lambda }(y)/z_{\lambda }$. Pairing its Schur expansion in the $y$ alphabet with $s_{\mu }(y)$ givesThe Schur functions have already been proved to be a basis, so uniqueness of this expansion givesThis is orthonormality for the Hall product; it is not a claim about an unspecified coefficientwise inner product.

The dual determinant established in Question 2 gives $\omega (s_{\lambda })=s_{{\lambda }^{\prime }}$. Apply the symmetric-function involution in the $y$ alphabet. Since $\omega (H_y(t))=E_y(t)$, it transforms each factor $H_y(x_i)$ into $E_y(x_i)$. The dual Cauchy identity is

For arbitrary coefficients $f_r$ in a commutative ring, use the integral symmetric-function algebra and its free complete generators, specializing $h_r\mapsto f_r$. This works over any such coefficient ring; a literal homomorphism from the rational algebra $\Lambda$ also presupposes a $\mathbb{Q}$-algebra structure on the target. Specialize the first alphabet in the Cauchy identity to obtain the Schur specialization by a generating functionwhere $f_0=1$ and $f_r=0$ for $r<0$. Indeed, the left side is a product of the specialized $H(t_i)$, and the Jacobi–Trudi identity specializes to the displayed determinant. For an arbitrary infinite $F$ the result lies in the degree completion; its coefficients in each degree are well defined.