= Solution
Put $d_i=\dim K_i$, including $d_0=0$. The <restriction> of $T^i$ to $K_{i+1}$ has <kernel> $K_i$, so the <rank-nullity theorem> gives
$$
\dim T^i(K_{i+1})=d_{i+1}-d_i.
$$
Likewise $T^{i-1}|_{K_i}$ has <kernel> $K_{i-1}$ and hence image <dimension> $d_i-d_{i-1}$. But $T(K_{i+1})\subseteq K_i$, which implies
$$
T^i(K_{i+1})=T^{i-1}(T(K_{i+1}))\subseteq T^{i-1}(K_i).
$$
Comparing <dimensions> yields the required concavity of the <kernel dimensions of powers of a nilpotent map>:
$$
\boxed{d_{i+1}-d_i\leq d_i-d_{i-1},\qquad 2d_i\geq d_{i-1}+d_{i+1}\quad(1\leq i<m)}.
$$
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