CG Optimisation Function

Question

I need to prove the following statement:

Given the unique minimiser $\overline{x}$ of the function $f(x) := \frac{1}{2}<Ax,x> + <b,x> + c$ for some $c \in \mathbb{R}, b \in \mathbb{R}^n$ and a symmetric positive definite matrix $A$. Prove that $f(x) = f(\overline{x}) + \frac{1}{2}\vert\vert x-\overline{x} \vert\vert_A^2$ holds for all $x \in \mathbb{R}^n$

So far I got to this point: $\begin{align*} f(\overline{x}) + \frac{1}{2}\vert\vert{x - \overline{x}}\vert\vert_A^2 &= \frac{1}{2}\overline{x}^TA\overline{x} + \overline{x}^Tb + c + \frac{1}{2}((x-\overline{x})^TA(x-\overline{x})) \\ &= \frac{1}{2}\overline{x}^TA\overline{x} + \overline{x}^Tb + c + \frac{1}{2}(x^TAx - x^TA\overline{x} - \overline{x}^TAx + \overline{x}^TA\overline{x}) \\ &= \frac{1}{2}\overline{x}^TA\overline{x} + \overline{x}^Tb + c + \frac{1}{2}x^TAx - x^Tb + \frac{1}{2}\overline{x}^TA\overline{x} \\ &= \overline{x}^TA\overline{x} + \overline{x}^Tb + c + \frac{1}{2}x^TAx - x^Tb \\ &= \frac{1}{2}x^TAx - x^Tb + c + 2\overline{x}^Tb \end{align*}$

How can I get $<b,x>$ from $2\overline{x}^Tb - x^Tb$?

Answer 1

For the function $f(x) = \frac{1}{2}< Ax,x > + < b,x > + c$ the equality $f(x) = f(\overline{x}) + \frac{1}{2}\vert\vert x - \overline{x}\vert\vert_A^2$ can not hold. It holds only for the function $f(x) = \frac{1}{2} < Ax,x > - < b,x > + c$.

Answer 2

There is probably an error in your third equality, because you cannot get from $2\overline{x}^Tb - x^Tb$ to $<b,x>$ (since $\overline{x}$ depends on $A$). It goes well until the expression right before that:

$$\overline{x}^TA\overline{x} + \overline{x}^Tb + c + \frac{1}{2}x^TAx - x^TA\overline{x},$$

but I do not see what the next step was that you took. If you plug in the expression for $\overline{x}$ (you can explicitly write the minimizer in terms of $A$ and $b$), you should arrive at the solution.