= Solution
{c}
Fix $x$ and put $g=\nabla_{\mathcal B}f(x)$, the <block coordinate gradient>. For $t\geq0$, define the one-variable <function> $q(t)=f(x-tg)$. By the <chain rule>,
$$
q'(t)=-\nabla f(x-tg)^Tg=-\nabla_{\mathcal B}f(x-tg)^Tg.
$$
The last equality holds because the coordinates of $g$ outside $\mathcal B$ are zero. The <fundamental theorem of calculus> and the <Cauchy-Schwarz inequality> therefore give
$$
\begin{aligned}
f(x-\alpha g)-f(x)
&=-\alpha\|g\|^2-\int_0^\alpha\big(\nabla_{\mathcal B}f(x-tg)-g\big)^Tg\,dt\\
&\leq-\alpha\|g\|^2+\int_0^\alpha\|\nabla_{\mathcal B}f(x-tg)-g\|\,\|g\|\,dt.
\end{aligned}
$$
For $\alpha\geq0$, the supplied condition (6), applied at the fixed starting point $x$ with parameter $t$, bounds the <integrand> by $L_{\mathcal B}t\|g\|^2$. Integration yields
$$
\boxed{f(x-\alpha\nabla_{\mathcal B}f(x))\leq f(x)-\alpha\left(1-\frac{L_{\mathcal B}\alpha}{2}\right)\|\nabla_{\mathcal B}f(x)\|^2.}
$$
This argument needs only the given directional <Lipschitz continuity> and continuous <differentiability>; <convexity> is not needed for this part. If the <block coordinate gradient> is zero, the inequality is equality.
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