= Solution
On the physical positive-energy branch, put $\omega=q^0+\xi E_P$. The massless outgoing <parton> obeys
$$
E_{k'}^2-\omega^2=-(\xi P+q)^2=-q^2-2\xi\nu=2\nu(x-\xi).
$$
Since $E_{k'}=\omega>0$ on the support and $\nu>0$, the <positive-energy on-shell delta function identity> gives
$$
\frac{\delta(\omega-E_{k'})}{E_{k'}}=2\delta(E_{k'}^2-\omega^2)=\frac{\delta(x-\xi)}{\nu}.
$$
Insert this into the <parton model> sum, and write
$$
S(x)=\sum_f Q_f^2\left[q_f(x)+\bar q_f(x)\right].
$$
The <parton distribution functions> $q_f,\bar q_f$ are number densities in momentum fraction. The contributing tensor becomes
$$
W_H^{\mu\nu}\doteq S(x)\left[\frac{x}{\nu}P^\mu P^\nu-\frac12g^{\mu\nu}\right].
$$
Comparing with part (b), under the same <leptonic tensor> contraction, gives
$$
F_1(x)=\frac12 S(x),\qquad F_2(x)=xS(x),\qquad
\boxed{F_2(x,q^2)=2xF_1(x,q^2).}
$$
This is the <Callan-Gross relation> for massless <partons> of <spin angular momentum> $1/2$ at leading order. The <longitudinal deep-inelastic structure function> is $F_L=F_2-2xF_1=0$ in this approximation. Target-mass effects and radiative <Quantum chromodynamics> corrections can change the relation. The leading <parton model> also gives <Bjorken scaling>: at this level the <deep-inelastic structure functions> depend on $x$ alone.
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