= Solution
Let $|\mathcal O\rangle$ be the state associated with a scalar <conformal primary operator> of <scaling dimension> $\Delta$. In radial quantization,
$$
K_a|\mathcal O\rangle=0,\qquad
M_{ab}|\mathcal O\rangle=0,\qquad
D|\mathcal O\rangle=i\Delta|\mathcal O\rangle,
$$
and $P_a^\dagger=-K_a$. The norm of a level-one <conformal descendant> is
$$
\langle\mathcal O|(-K_a)P_b|\mathcal O\rangle
=-\langle\mathcal O|[K_a,P_b]|\mathcal O\rangle
=2\Delta\,\delta_{ab}.
$$
Unitarity first gives $\Delta\geq0$. If $\Delta=0$, every $P_a|\mathcal O\rangle$ is null, so the local operator is translation invariant and belongs to the identity conformal family. Excluding the identity therefore gives $\Delta>0$.
Now consider the scalar level-two descendant $P^2|\mathcal O\rangle$. The <conformal algebra> and the scalar-primary conditions give
$$
[K_a,P^2]|\mathcal O\rangle
=-2(2\Delta-d+2)P_a|\mathcal O\rangle.
$$
Applying the second $K_a$ and summing over $a$ yields
$$
\|P^2|\mathcal O\rangle\|^2
=\langle\mathcal O|K^2P^2|\mathcal O\rangle
=8d\Delta\left(\Delta-\frac{d-2}{2}\right)
\langle\mathcal O|\mathcal O\rangle.
$$
Positivity of this norm, together with $\Delta>0$, proves the scalar <conformal unitarity bound>
$$
\boxed{\Delta\geq\frac{d-2}{2}}.
$$
At equality the level-two descendant is null; in position space this is the free scalar equation of motion.
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