[proofplan]
We expand $f$ and $g$ in the Schur basis with nonnegative coefficients. Multiplying the two expansions reduces the problem to understanding products $s_\lambda s_\mu$. The [Littlewood-Richardson rule](/theorems/5183) expresses each such product as a nonnegative integer combination of Schur functions, so each coefficient in the Schur expansion of $fg$ is a finite sum of nonnegative [real numbers](/page/Real%20Numbers).
[/proofplan]
[step:Expand the two Schur-positive functions in the Schur basis]
Let $d,e \in \mathbb{N} \cup \{0\}$ be the homogeneous degrees of $f$ and $g$, respectively. Let $\mathcal{P}_d$ denote the finite set of partitions of $d$, and let $\mathcal{P}_e$ denote the finite set of partitions of $e$.
Since $f$ and $g$ are Schur-positive, there exist coefficients $a_\lambda \in [0,\infty)$ for $\lambda \in \mathcal{P}_d$ and coefficients $b_\mu \in [0,\infty)$ for $\mu \in \mathcal{P}_e$ such that
\begin{align*}
f &= \sum_{\lambda \in \mathcal{P}_d} a_\lambda s_\lambda, \\
g &= \sum_{\mu \in \mathcal{P}_e} b_\mu s_\mu.
\end{align*}
The sums are finite because there are only finitely many partitions of a fixed nonnegative integer.
[/step]
[step:Multiply the Schur expansions and apply the Littlewood-Richardson rule]
Multiplying the two finite sums in the ring $\Lambda_{\mathbb{R}}$ gives
\begin{align*}
fg
&= \left(\sum_{\lambda \in \mathcal{P}_d} a_\lambda s_\lambda\right)
\left(\sum_{\mu \in \mathcal{P}_e} b_\mu s_\mu\right) \\
&= \sum_{\lambda \in \mathcal{P}_d} \sum_{\mu \in \mathcal{P}_e}
a_\lambda b_\mu\, s_\lambda s_\mu.
\end{align*}
For partitions $\lambda \in \mathcal{P}_d$ and $\mu \in \mathcal{P}_e$, the Littlewood-Richardson rule for Schur functions states
(citing a result not yet in the wiki: Littlewood-Richardson rule for Schur functions) that
\begin{align*}
s_\lambda s_\mu
=
\sum_{\nu \in \mathcal{P}_{d+e}} c_{\lambda\mu}^{\nu} s_\nu,
\end{align*}
where $c_{\lambda\mu}^{\nu}$ is the number of Littlewood-Richardson tableaux of shape $\nu/\lambda$ and content $\mu$. Hence
\begin{align*}
c_{\lambda\mu}^{\nu} \in \mathbb{N} \cup \{0\}
\end{align*}
for every $\lambda \in \mathcal{P}_d$, $\mu \in \mathcal{P}_e$, and $\nu \in \mathcal{P}_{d+e}$.
Substituting this expansion into the product gives
\begin{align*}
fg
&= \sum_{\lambda \in \mathcal{P}_d} \sum_{\mu \in \mathcal{P}_e}
a_\lambda b_\mu
\left(\sum_{\nu \in \mathcal{P}_{d+e}} c_{\lambda\mu}^{\nu} s_\nu\right) \\
&= \sum_{\nu \in \mathcal{P}_{d+e}}
\left(
\sum_{\lambda \in \mathcal{P}_d}
\sum_{\mu \in \mathcal{P}_e}
a_\lambda b_\mu c_{\lambda\mu}^{\nu}
\right)s_\nu.
\end{align*}
[guided]
We want to prove that $fg$ has a Schur expansion with nonnegative coefficients. Since $f$ and $g$ are already Schur-positive, the only possible obstruction is that multiplying Schur functions might introduce negative coefficients. The Littlewood-Richardson rule is precisely the theorem that controls this multiplication.
Starting from the Schur expansions,
\begin{align*}
f &= \sum_{\lambda \in \mathcal{P}_d} a_\lambda s_\lambda, \\
g &= \sum_{\mu \in \mathcal{P}_e} b_\mu s_\mu,
\end{align*}
where $a_\lambda \geq 0$ and $b_\mu \geq 0$, distributivity in $\Lambda_{\mathbb{R}}$ gives
\begin{align*}
fg
&= \sum_{\lambda \in \mathcal{P}_d}
\sum_{\mu \in \mathcal{P}_e}
a_\lambda b_\mu\, s_\lambda s_\mu.
\end{align*}
Now fix partitions $\lambda \in \mathcal{P}_d$ and $\mu \in \mathcal{P}_e$. The product $s_\lambda s_\mu$ is homogeneous of degree $d+e$, so its Schur expansion involves only Schur functions $s_\nu$ with $\nu \in \mathcal{P}_{d+e}$. The Littlewood-Richardson rule for Schur functions
(citing a result not yet in the wiki: Littlewood-Richardson rule for Schur functions) gives
\begin{align*}
s_\lambda s_\mu
=
\sum_{\nu \in \mathcal{P}_{d+e}} c_{\lambda\mu}^{\nu} s_\nu,
\end{align*}
where $c_{\lambda\mu}^{\nu}$ counts Littlewood-Richardson tableaux of shape $\nu/\lambda$ and content $\mu$. Because it is a cardinality, each $c_{\lambda\mu}^{\nu}$ is a nonnegative integer:
\begin{align*}
c_{\lambda\mu}^{\nu} \in \mathbb{N} \cup \{0\}.
\end{align*}
Substituting this formula into the product expansion yields
\begin{align*}
fg
&= \sum_{\lambda \in \mathcal{P}_d}
\sum_{\mu \in \mathcal{P}_e}
a_\lambda b_\mu
\left(\sum_{\nu \in \mathcal{P}_{d+e}} c_{\lambda\mu}^{\nu} s_\nu\right) \\
&= \sum_{\nu \in \mathcal{P}_{d+e}}
\left(
\sum_{\lambda \in \mathcal{P}_d}
\sum_{\mu \in \mathcal{P}_e}
a_\lambda b_\mu c_{\lambda\mu}^{\nu}
\right)s_\nu.
\end{align*}
The interchange of sums is valid because all indexing sets $\mathcal{P}_d$, $\mathcal{P}_e$, and $\mathcal{P}_{d+e}$ are finite.
[/guided]
[/step]
[step:Check that every resulting Schur coefficient is nonnegative]
For each $\nu \in \mathcal{P}_{d+e}$, define the real number
\begin{align*}
A_\nu
:=
\sum_{\lambda \in \mathcal{P}_d}
\sum_{\mu \in \mathcal{P}_e}
a_\lambda b_\mu c_{\lambda\mu}^{\nu}.
\end{align*}
Each factor in every summand is nonnegative:
\begin{align*}
a_\lambda \geq 0,\qquad b_\mu \geq 0,\qquad c_{\lambda\mu}^{\nu} \geq 0.
\end{align*}
Therefore $A_\nu \geq 0$ for every $\nu \in \mathcal{P}_{d+e}$.
The product has the Schur expansion
\begin{align*}
fg = \sum_{\nu \in \mathcal{P}_{d+e}} A_\nu s_\nu
\end{align*}
with all coefficients $A_\nu$ nonnegative. Hence $fg$ is Schur-positive.
[/step]
