[proofplan]
We prove the identity degree by degree. In each homogeneous degree $n$, the partitions of $n$ form a finite index set, so the change-of-basis coefficients from $\{h_\lambda\}$ to $\{u_\lambda\}$ and from $\{m_\lambda\}$ to $\{v_\lambda\}$ are finite square matrices. The duality condition says that these matrices are inverse transposes, and therefore the degree-$n$ part of $\sum_\lambda u_\lambda(x)v_\lambda(y)$ collapses to $\sum_{\lambda\vdash n}h_\lambda(x)m_\lambda(y)$. Summing the homogeneous identities over all $n\geq 0$ gives the claimed equality in the completed [tensor product](/page/Tensor%20Product).
[/proofplan]
[step:Reduce the completed identity to homogeneous degree pieces]
For each integer $n\geq 0$, let $\mathcal{P}_n$ denote the finite set of partitions of $n$. Since $\operatorname{Sym}$ is graded and $u_\lambda,v_\lambda\in \operatorname{Sym}^{|\lambda|}$, the degree-$n$ component of
\begin{align*}
\sum_{\lambda} u_\lambda(x)v_\lambda(y)
\end{align*}
is
\begin{align*}
\sum_{\lambda\in \mathcal{P}_n} u_\lambda(x)v_\lambda(y).
\end{align*}
Likewise, the degree-$n$ component of the Cauchy kernel expansion is
\begin{align*}
\sum_{\lambda\in \mathcal{P}_n} h_\lambda(x)m_\lambda(y).
\end{align*}
Thus it is enough to prove, for every $n\geq 0$,
\begin{align*}
\sum_{\lambda\in \mathcal{P}_n} u_\lambda(x)v_\lambda(y)
=
\sum_{\lambda\in \mathcal{P}_n} h_\lambda(x)m_\lambda(y).
\end{align*}
[/step]
[step:Write both homogeneous bases in the standard dual bases]
Fix $n\geq 0$. Since $\{h_\alpha\}_{\alpha\in \mathcal{P}_n}$ is a $K$-basis of $\operatorname{Sym}^n$, for each $\lambda\in\mathcal{P}_n$ there are unique coefficients $A_{\lambda\alpha}\in K$, indexed by $\alpha\in\mathcal{P}_n$, such that
\begin{align*}
u_\lambda=\sum_{\alpha\in\mathcal{P}_n} A_{\lambda\alpha}h_\alpha.
\end{align*}
Since $\{m_\beta\}_{\beta\in \mathcal{P}_n}$ is also a $K$-basis of $\operatorname{Sym}^n$, for each $\mu\in\mathcal{P}_n$ there are unique coefficients $B_{\mu\beta}\in K$, indexed by $\beta\in\mathcal{P}_n$, such that
\begin{align*}
v_\mu=\sum_{\beta\in\mathcal{P}_n} B_{\mu\beta}m_\beta.
\end{align*}
Let $A=(A_{\lambda\alpha})_{\lambda,\alpha\in\mathcal{P}_n}$ and $B=(B_{\mu\beta})_{\mu,\beta\in\mathcal{P}_n}$ denote the corresponding finite square matrices over $K$.
[/step]
[step:Translate duality into an inverse transpose relation]
For $\lambda,\mu\in\mathcal{P}_n$, bilinearity of the Hall [inner product](/page/Inner%20Product) and the defining duality $(h_\alpha,m_\beta)=\delta_{\alpha\beta}$ give
\begin{align*}
(u_\lambda,v_\mu)
&=
\left(\sum_{\alpha\in\mathcal{P}_n} A_{\lambda\alpha}h_\alpha,\sum_{\beta\in\mathcal{P}_n} B_{\mu\beta}m_\beta\right)\\
&=
\sum_{\alpha,\beta\in\mathcal{P}_n} A_{\lambda\alpha}B_{\mu\beta}(h_\alpha,m_\beta)\\
&=
\sum_{\alpha\in\mathcal{P}_n} A_{\lambda\alpha}B_{\mu\alpha}.
\end{align*}
The assumed duality $(u_\lambda,v_\mu)=\delta_{\lambda\mu}$ therefore says
\begin{align*}
\sum_{\alpha\in\mathcal{P}_n} A_{\lambda\alpha}B_{\mu\alpha}=\delta_{\lambda\mu}.
\end{align*}
Equivalently,
\begin{align*}
AB^\top=I.
\end{align*}
Since $A$ and $B^\top$ are finite square matrices, this implies also
\begin{align*}
B^\top A=I.
\end{align*}
In entries, for all $\alpha,\beta\in\mathcal{P}_n$,
\begin{align*}
\sum_{\lambda\in\mathcal{P}_n} B_{\lambda\beta}A_{\lambda\alpha}=\delta_{\beta\alpha}.
\end{align*}
[guided]
The purpose of this step is to convert the abstract duality condition into a concrete matrix identity. Fix $\lambda,\mu\in\mathcal{P}_n$. We have already written $u_\lambda$ in the $h$-basis and $v_\mu$ in the $m$-basis:
\begin{align*}
u_\lambda=\sum_{\alpha\in\mathcal{P}_n} A_{\lambda\alpha}h_\alpha,
\qquad
v_\mu=\sum_{\beta\in\mathcal{P}_n} B_{\mu\beta}m_\beta.
\end{align*}
Using bilinearity of the Hall inner product, we compute
\begin{align*}
(u_\lambda,v_\mu)
&=
\left(\sum_{\alpha\in\mathcal{P}_n} A_{\lambda\alpha}h_\alpha,\sum_{\beta\in\mathcal{P}_n} B_{\mu\beta}m_\beta\right)\\
&=
\sum_{\alpha,\beta\in\mathcal{P}_n} A_{\lambda\alpha}B_{\mu\beta}(h_\alpha,m_\beta).
\end{align*}
Now the defining property of the Hall inner product in these bases is exactly
\begin{align*}
(h_\alpha,m_\beta)=\delta_{\alpha\beta}.
\end{align*}
Therefore every term with $\alpha\neq\beta$ vanishes, and the double sum reduces to
\begin{align*}
(u_\lambda,v_\mu)=\sum_{\alpha\in\mathcal{P}_n} A_{\lambda\alpha}B_{\mu\alpha}.
\end{align*}
The hypothesis says that $(u_\lambda,v_\mu)=\delta_{\lambda\mu}$, so
\begin{align*}
\sum_{\alpha\in\mathcal{P}_n} A_{\lambda\alpha}B_{\mu\alpha}=\delta_{\lambda\mu}.
\end{align*}
This is precisely the matrix equation
\begin{align*}
AB^\top=I.
\end{align*}
Why does this give the identity in the opposite order? Because $\mathcal{P}_n$ is finite, so $A$ and $B^\top$ are finite square matrices. A left inverse of a finite square matrix is also a right inverse. Hence
\begin{align*}
B^\top A=I.
\end{align*}
Writing this entrywise gives, for every $\alpha,\beta\in\mathcal{P}_n$,
\begin{align*}
\sum_{\lambda\in\mathcal{P}_n} B_{\lambda\beta}A_{\lambda\alpha}=\delta_{\beta\alpha}.
\end{align*}
This is the exact cancellation needed when we expand $\sum_\lambda u_\lambda(x)v_\lambda(y)$.
[/guided]
[/step]
[step:Expand the kernel in the dual homogeneous bases]
Using the two change-of-basis expansions and rearranging only finite sums, we obtain
\begin{align*}
\sum_{\lambda\in\mathcal{P}_n} u_\lambda(x)v_\lambda(y)
&=
\sum_{\lambda\in\mathcal{P}_n}
\left(\sum_{\alpha\in\mathcal{P}_n} A_{\lambda\alpha}h_\alpha(x)\right)
\left(\sum_{\beta\in\mathcal{P}_n} B_{\lambda\beta}m_\beta(y)\right)\\
&=
\sum_{\alpha,\beta\in\mathcal{P}_n}
\left(\sum_{\lambda\in\mathcal{P}_n} A_{\lambda\alpha}B_{\lambda\beta}\right)
h_\alpha(x)m_\beta(y)\\
&=
\sum_{\alpha,\beta\in\mathcal{P}_n}
\delta_{\alpha\beta}h_\alpha(x)m_\beta(y)\\
&=
\sum_{\alpha\in\mathcal{P}_n} h_\alpha(x)m_\alpha(y).
\end{align*}
This proves the desired identity in homogeneous degree $n$.
[/step]
[step:Sum the homogeneous identities in the completed tensor product]
The preceding argument holds for every integer $n\geq 0$. Therefore, in each homogeneous degree $n$, the degree-$n$ component of $\sum_\lambda u_\lambda(x)v_\lambda(y)$ equals the degree-$n$ component of $\Omega(x,y)$. Since $\widehat{\operatorname{Sym}\otimes\operatorname{Sym}}$ is the completion in which equality is determined degree by degree, we conclude
\begin{align*}
\sum_{\lambda} u_\lambda(x)v_\lambda(y)
=
\sum_{\lambda} h_\lambda(x)m_\lambda(y)
=
\Omega(x,y).
\end{align*}
This is the claimed kernel expansion in the dual homogeneous bases $\{u_\lambda\}$ and $\{v_\lambda\}$.
[/step]
