For two polynomials, f,g
If I defined inner product of these two polys to be $$(f\cdot g) = \left|\int_0^1 f(x)g(x)dx\right|$$ (does this satisfy inner product?
I think it is not.
$x\cdot y=y\cdot x$ ok
$x\cdot(y+z)=x\cdot y+x\cdot z$ it fails to satisfy
$c(x\cdot y)=(cx)\cdot y$ it fails to satisfy
$x\cdot x>0$ if $x\neq0$
This is what I concluded and I have a proof for each case that does not satisfy it.
I just want to make sure that I did check it right, or Anybody tell me the answer?
Actually, $(u|v)=\int_0^1 u(x)v(x)dx$ is an inner product on polynomials (and also on continuous functions, for example), and Gram-Schmidt orthogonalization with it yields Legendre polynomials if you integrate on $[-1,1]$, and shifted Legendre polynomials if you integrate on $[0,1]$.
However, with the absolute value, it's not an inner product, and the reason is that the absolute value is not linear, as you point out.
It's enough to show counterexamples with constant polynomials, and with your inner product $(u|v)=\left|\int_0^1 u(x)v(x)dx\right|$:
$$(1|-1)=(1|1)=1\neq-(1|1)=-1$$ $$(1|2-1)=(1|1)=1\neq (1|2)+(1|-1)=3$$
The other two axioms or however easily proved true.