Trouble understanding proof using inner product

Question

I am having trouble understanding why a proof holds.

Specifically why can't the statement:

$$ \langle x_0 - x_1 , x_0 \rangle =0 $$

be reversed to say

$$ \langle x_1 - x_0 , x_0 \rangle =0 $$

Thereby helping us to complete the square for: $\left\langle {{x}_{0}},{{x}_{0}} \right\rangle -\left\langle {{x}_{1}},{{x}_{1}} \right\rangle $

and show:

$$ \langle x_1 , x_1 \rangle \le \langle x_0, x_0 \rangle $$

The author defines his inner product using

• $\left\langle {{x}_{1}},{{x}_{2}} \right\rangle =\left\langle {{x}_{2}},{{x}_{1}} \right\rangle $
• $ \left\langle {{x}_{1}},a{{x}_{2}}+b{{x}_{3}} \right\rangle =a\left\langle {{x}_{1}},{{x}_{2}} \right\rangle +b\left\langle {{x}_{1}},{{x}_{3}} \right\rangle $
• $\left\langle {{x}_{1}},{{x}_{2}} \right\rangle \ge 0;\left\langle {{x}_{1}},{{x}_{1}} \right\rangle =0$ iff $x_1=0$.

Finally a linear transformation $T$ is the adjoint transformation associated with linear transformation $L$ if for all $x$, $y$

$${{\left\langle y,Lx \right\rangle }_{y}}={{\left\langle Ty,x \right\rangle }_{x}}$$

and is denoted by $L^*$

Here is the theorem and proof:

Answer 1

We have \begin{align}\langle x_1-x_0,x_1-x_0\rangle&=\langle x_1,x_1\rangle+\langle x_0,x_0\rangle -\langle x_0,x_1\rangle -\langle x_1,x_0\rangle\\ &=\langle x_1,x_1\rangle+\langle x_0,x_0-x_1\rangle+\langle x_0-x_1,x_0\rangle-\langle x_0,x_0\rangle\\&=\langle x_1,x_1\rangle+0+0-\langle x_0,x_0\rangle\ge 0,\end{align} whence the required result follows: $\langle x_1,x_1\rangle \ge\langle x_0,x_0\rangle$. This means that the two-point open-loop control obtained from the controllability Grammian (which is your $LL^*$) is energy optimal, i.e. it has the minimum possible norm among all controls solving the came two-point problem.

I do not see though, how can you arrive at the opposite by reversing the expression in the scalar product...