Can anyone see why it follows that $\Vert u_{j}-u_{l} \Vert^{2} = 2[\Vert u_{j} \Vert^{2}+ \Vert u_{l} \Vert^{2}] - \Vert u_{j} + u_{l} \Vert^{2} = 2[E(u_{j}) + E(u_{l})] - 4E(\frac{u_{j}+u_{l}}{2})$ in the proof above? Thanks
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The theorem is on page 40.
The first equality is just the parallelogram identity, valid for all norms arising from an inner product.
Next, for the purported equality
$$2[\Vert u_{j} \Vert^{2}+ \Vert u_{l} \Vert^{2}] - \Vert u_{j} + u_{l} \Vert^{2} = 2[E(u_{j}) + E(u_{l})] - 4E(\frac{u_{j}+u_{l}}{2}),$$
we observe that the linear part of $E$ cancels on the right hand side, so we are left with
$$\begin{align} 2[E(u_{j}) + E(u_{l})] - 4E(\frac{u_{j}+u_{l}}{2}) &= 2\left[\langle Au_j, u_j\rangle + \langle Au_l, u_l\rangle\right] - 4 \left\langle A\frac{u_j+u_l}{2}, \frac{u_j+u_l}{2}\right\rangle\\ &= 2\left[\langle Au_j, u_j\rangle + \langle Au_l, u_l\rangle\right] - \langle A(u_j+u_l), u_j+u_l\rangle\\ &= \left[\langle Au_j, u_j\rangle + \langle Au_l, u_l\rangle\right] - \langle Au_j,u_l\rangle - \langle Au_l,u_j\rangle\\ &= \langle A(u_j-u_l), u_j-u_l\rangle. \end{align}$$
However, that is generally not equal to $\lVert u_j - u_l\rVert_V^2$.
But, $A$ is assumed bounded, self-adjoint, and strictly positive (in the sense that there is a $c > 0$ with $\langle Au,u\rangle \geqslant c\lVert u\rVert^2$ for all $u$, presumably, since that is how "strictly positive is used in remark 2.11").
Therefore $(u \mspace{-3mu}\mid\mspace{-3mu} v) = \langle Au,v\rangle$ is an inner product on $V$ equivalent to the given inner product, and apparently $\lVert u\rVert$ denotes the norm arising from that inner product here (Note that elsewhere, the norm is indexed by the space, $\lVert\cdot\rVert_V$ resp. $\lVert\cdot\rVert_{V^\star}$).
So then the second equality is just the cancelling of the linear terms $-2\langle F,u\rangle$ on the right hand side.
Since, as noted above, the norms are equivalent, it follows that the minimising sequence $(u_j)_{j\in\mathbb{N}}$ is indeed a Cauchy sequence.