 Solution

Because $\hat{R}(\hat{h})\le \hat{R}(h^{\ast })$, adding and subtracting empirical risks and taking expectations leaves at most $E{\sup }_h(R(h)-\hat{R}(h))$. Symmetrization and the unit-Lipschitz contraction lemma for hinge loss bound this by $2{\mathcal{R}}_n(\mathcal{H})$.