I have seen expressions involving the inner product where the adjoint operator shows up, and I cannot figure out why/how the adjoint operator shows up. For example, say I have a function $f$ like this:
$f(x)=\frac{1}{2}||Ax-b||^2$
When $x=x+\Delta x$ I have seen this rewritten as the following:
$f(x+\Delta x)=\frac{1}{2}||Ax-b||^2+<A^*(Ax-b),\Delta x>+\frac{1}{2}||A\Delta x||^2$
So, I tried to work this out myself to figure out how the self adjoint operator $A^*$ shows up in there, but I the closest I can get is the following:
1) I re-express it as $f(x)=\frac{1}{2}||Ax-b+A\Delta x||^2$
2) Then take the square expansion (FOIL):
$f(x+\Delta x)=\frac{1}{2}||Ax-b||^2+<Ax-b,A\Delta x>+\frac{1}{2}<A\Delta x, A\Delta x>$
But now from here I do not see how we can get a self-adjoint operator to show up. The closest that I can get, is if I use the property of conjugate symmetry (see https://en.wikipedia.org/wiki/Inner_product_space#Elementary_properties)
which gets me $<Ax-b, A\Delta x>=A^*<Ax-b, \Delta x>$, which is close, but doesn't quite achieve this : $f(x+\Delta x)=\frac{1}{2}||Ax-b||^2+<A^*(Ax-b),\Delta x>+\frac{1}{2}||A\Delta x||^2$.
How do I achieve that expression?
I'll assume we are working with a symmetric inner product over a real vector space.
Consider $$\|Au+v\|^2.$$ This equals $$\left<Au+v,Au+v\right> =\left<Au,Au\right>+\left<Au,v\right>+\left<v,Au\right> +\left<v,Au\right>$$ since the inner product is additive in both arguments. By symmetry, this is $$\|A u\|^2+2\left<v,Au\right> +\|v\|^2$$ and by adjointness this is $$\|A u\|^2+2\left<A^*v,u\right>+\|v\|^2.$$
In your application, take $v=Ax-b$ and $u=\Delta x$.
For complex inner product spaces and Hermitian inner products one gets a somewhat different outcome.