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\begin{enumerate}
\item[\bf 2.14 b)]
Let $y$ be given as in the hint.
From part (a), we have (wlog) $\forall k$ 
\begin{eqnarray*}
||x^{k+1} -y||^2 - ||x^k - y||^2+2\alpha^kf(x^k) - (\alpha^k||\nabla
f(x^k)||)^2 <&2\alpha^kf(x^k) - 2\alpha^k\delta\\
\mbox{or equivalently}\\
||x^{k+1} -y||^2 - ||x^k - y||^2 + 2\alpha^k\delta < (\alpha^k||\nabla
f(x^k)||)^2
\end{eqnarray*}
Suppose that the sequence $\{x^k\}$ is unbounded, i.e. wlog $||x^k||
\rightarrow \infty.$ Wlog we can assume $\alpha^k >0.$
Then wlog 
\[  ||x^{k+1} -y||^2 - ||x^k - y||^2 > 0. \]
(By taking a subsequence if needed.) Now, dividing by $\alpha^k$ we get
\[
\frac{||x^{k+1} -y||^2 - ||x^k - y||^2}{\alpha^k}
 + 2\delta < \alpha^k(||\nabla
f(x^k)||)^2, \,\, \forall k \geq \bar k.\\
\]
The right-hand side converges to 0, while the left-hand side is strictly
positive, a contradiction.

Therefore, we can assume that $x^k \rightarrow x^*.$ We need only show
that $x^*$ is a stationary point, by convexity. Suppose not, i.e.
suppose that $||\nabla f(x^k)|| \geq \epsilon > 0, \forall k.$ Then,
\[
\alpha^k \epsilon^2 \leq \alpha^k ||\nabla f(x^k)||^2 \leq
f(x)^{k+1}-f(x^k).
\]
Summing yields
\[
\epsilon^2 \sum_1^{\infty}\alpha^k \leq f(x^*) - f(x^1)
\]
a contradiction. Therefore, $x^*$ is a stationary point and so a global
minimum of the convex function.

\item[\bf 2.14 c)]
As in  part b, we get
\[
||x^{k+1} -y||^2 - ||x^k - y||^2 + 2\alpha^k\delta < (\alpha^k||\nabla
f(x^k)||)^2
\]
or equivalently
\[
||x^{k+1} -y||^2 - ||x^k - y||^2  < (\alpha^k||\nabla
f(x^k)||)^2 - 2\alpha^k\delta \leq (s^k)^2.
\]
This implies, as above in part b, that the sequence $\{x^k\}$ is
bounded. As above, if $||\nabla f(x^k) || \geq \epsilon > 0,$
\[
s^k \epsilon \leq \alpha^k ||\nabla f(x^k)||^2 \leq
f(x)^{k+1}-f(x^k).
\]
Summing yields
\[
\epsilon \sum_1^{\infty}s^k \leq f(x^*) - f(x^1),
\]
a contradiction again.
\end{enumerate}
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