= Solution
Let $h$ be the standard Cartan element of the <sl2 Lie algebra>, with $[h,e]=2e$ and $[h,f]=-2f$. For a finite-dimensional representation, its <formal character> records its <weight multiplicities>:
$$
\boxed{\chi_V(q)=\sum_{j\in\mathbb Z}(\dim V_j)q^j,\qquad
V_j=\{v:hv=jv\}.}
$$
The <classification of finite-dimensional sl2 representations> and the <Weyl complete reducibility theorem> give $V=\bigoplus_{m\ge0}a_mL_m$, with finitely many nonzero $a_m$. Each $L_m$ has <weights> $m,m-2,\ldots,-m$, each of multiplicity one. Therefore
$$
\chi_V(q)=\sum_{m\ge0}a_m(q^m+q^{m-2}+\cdots+q^{-m}),
\qquad c_j=\sum_{\substack{m\ge |j|\\m\equiv j\ (2)}}a_m.
$$
In particular $c_j=c_{-j}$, and for $j\ge0$ one has $c_j-c_{j+2}=a_j\ge0$. This proves <parity unimodality of an sl2 character>: \b[within each parity, the coefficients increase towards zero and decrease away from zero]. For a representation whose <weights> have a single parity, deleting the intervening zero coefficients gives a symmetric unimodal sequence.
The parity convention is necessary for the printed assertion. If unimodality means the full coefficient sequence at every consecutive integer exponent, it is false: the standard two-dimensional representation has character $q+q^{-1}$, whose coefficients at exponents $-1,0,1$ are $1,0,1$. The statement established above is unimodality on cosets of the <root lattice> for $\mathfrak{sl}_2$, and is the version needed in part ii. Infinite-dimensional representations need not even have Laurent-polynomial characters.
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