[proofplan]
We first prove the Chebyshev product identity $T_m(x)T_n(x) = \frac{1}{2}[T_{|m-n|}(x) + T_{m+n}(x)]$ by using the substitution $x = \cos\theta$ to reduce it to the trigonometric product-to-sum formula $\cos(m\theta)\cos(n\theta) = \frac{1}{2}[\cos((m-n)\theta) + \cos((m+n)\theta)]$. We then derive the general product formula for Chebyshev expansions by multiplying the two series term by term, applying the pairwise identity, and collecting terms by output index $k$.
[/proofplan]
[step:Prove the product identity $T_m T_n = \frac{1}{2}[T_{|m-n|} + T_{m+n}]$ via the cosine product-to-sum formula]
Let $x \in [-1, 1]$ and write $x = \cos\theta$ for $\theta \in [0, \pi]$. By definition, $T_m(\cos\theta) = \cos(m\theta)$ for all integers $m \geq 0$. Therefore
\begin{align*}
T_m(x) T_n(x) = \cos(m\theta)\cos(n\theta).
\end{align*}
Applying the product-to-sum identity for cosines:
\begin{align*}
\cos(m\theta)\cos(n\theta) = \frac{1}{2}\bigl[\cos((m-n)\theta) + \cos((m+n)\theta)\bigr].
\end{align*}
Since $\cos$ is an even function, $\cos((m-n)\theta) = \cos(|m-n|\theta) = T_{|m-n|}(\cos\theta) = T_{|m-n|}(x)$, and $\cos((m+n)\theta) = T_{m+n}(x)$. Therefore
\begin{align*}
T_m(x) T_n(x) = \frac{1}{2}\bigl[T_{|m-n|}(x) + T_{m+n}(x)\bigr]
\end{align*}
for all $x \in [-1, 1]$ and all $m, n \geq 0$.
[guided]
The proof reduces the Chebyshev product identity to the elementary trigonometric identity $\cos\alpha\cos\beta = \frac{1}{2}[\cos(\alpha - \beta) + \cos(\alpha + \beta)]$, which is itself a consequence of the addition formulas $\cos(\alpha \pm \beta) = \cos\alpha\cos\beta \mp \sin\alpha\sin\beta$ (adding the two gives $\cos(\alpha-\beta)+\cos(\alpha+\beta) = 2\cos\alpha\cos\beta$).
The step where $\cos((m-n)\theta)$ becomes $T_{|m-n|}(x)$ uses the even parity of $\cos$: if $m \geq n$, then $\cos((m-n)\theta) = T_{m-n}(x)$ directly. If $m < n$, then $\cos((m-n)\theta) = \cos((n-m)\theta) = T_{n-m}(x) = T_{|m-n|}(x)$. In either case, the result is $T_{|m-n|}(x)$.
[/guided]
[/step]
[step:Derive the general product formula by expanding $f(x)g(x)$ and collecting terms]
Given Chebyshev expansions $f(x) = \sum_{m=0}^{\infty} \breve{f}_m T_m(x)$ and $g(x) = \sum_{n=0}^{\infty} \breve{g}_n T_n(x)$, multiply term by term:
\begin{align*}
f(x)g(x) = \sum_{m=0}^{\infty}\sum_{n=0}^{\infty} \breve{f}_m \breve{g}_n \, T_m(x) T_n(x).
\end{align*}
Applying the product identity from the first step:
\begin{align*}
f(x)g(x) = \frac{1}{2}\sum_{m=0}^{\infty}\sum_{n=0}^{\infty} \breve{f}_m \breve{g}_n \bigl[T_{|m-n|}(x) + T_{m+n}(x)\bigr].
\end{align*}
We collect the coefficient of $T_k(x)$ for each $k \geq 0$. The term $T_k(x)$ arises in two ways:
1. From $T_{m+n}(x)$ when $m + n = k$, contributing $\frac{1}{2}\sum_{\substack{m, n \geq 0 \\ m+n = k}} \breve{f}_m \breve{g}_n$.
2. From $T_{|m-n|}(x)$ when $|m - n| = k$, contributing $\frac{1}{2}\sum_{\substack{m, n \geq 0 \\ |m-n| = k}} \breve{f}_m \breve{g}_n$.
Therefore
\begin{align*}
f(x)g(x) = \frac{1}{2}\sum_{k=0}^{\infty} T_k(x)\left(\sum_{\substack{m,n \geq 0 \\ m+n=k}} \breve{f}_m \breve{g}_n + \sum_{\substack{m,n \geq 0 \\ |m-n|=k}} \breve{f}_m \breve{g}_n\right).
\end{align*}
[guided]
To see why these are the only contributions to $T_k(x)$, note that each pair $(m, n)$ in the double sum produces two Chebyshev terms: $T_{m+n}(x)$ and $T_{|m-n|}(x)$. The term $T_k(x)$ appears from $T_{m+n}(x)$ when $m + n = k$ (a finite number of pairs: $m = 0, n = k$ through $m = k, n = 0$), and from $T_{|m-n|}(x)$ when $|m - n| = k$ (an infinite set of pairs: $(m, n) = (k, 0), (k+1, 1), (k+2, 2), \ldots$ and similarly with $m$ and $n$ swapped).
The rearrangement of the double series is justified because the Chebyshev coefficients of $f$ and $g$ are assumed to give convergent expansions, so $\sum_m |\breve{f}_m| < \infty$ and $\sum_n |\breve{g}_n| < \infty$ (or the product series converges in the appropriate sense).
[/guided]
[/step]
