= Solution
Use the exact <line searches> from part (i). After minimizing along one coordinate, the corresponding <gradient> component is zero. Unless the optimum has been reached, the whole nonzero <gradient> therefore lies along the other coordinate, which is selected next. The chosen signed coordinate direction then has $\cos\theta_k=1$. A selected zero <gradient> component can be skipped; it already minimizes that coordinate.
The <Zoutendijk theorem> therefore gives $\nabla f(x_k)\to0$. The unique point with zero <gradient> is
$$
\boxed{x^*=\left(\frac{a+b}{2},\frac{a-b}{2}\right),\qquad f(x^*)=0.}
$$
Since $\nabla f(x)=H(x-x^*)$ and $H$ is invertible, <gradient> <convergence> implies $x_k\to x^*$. Positive definiteness makes this the unique <global minimum>.
<Convergence> also holds if “<line search>” is interpreted as any fixed-parameter <Wolfe conditions>[Wolfe search]. If coordinate $i$ is selected, its signed step $s$ satisfies $s=-\omega g_i/h$, where $h=c^{-2}+d^{-2}$ and $1-\eta\leq\omega\leq2(1-\sigma)$. Sufficient decrease gives $f(x_k)-f(x_{k+1})\geq\sigma(1-\eta)g_i(x_k)^2/h$. Summation shows that selected <gradient> components tend to zero and step lengths tend to zero. The other component was selected in the preceding step; <Lipschitz continuity> then makes it tend to zero as well. Thus the full <gradient> converges to zero. The simple one-step angle argument and sharp Q-linear factor below use exact coordinate minimization; arbitrary Wolfe choices need not have that same Euclidean one-step factor.
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