# Thread: how to prove this inner product

1. ## how to prove this inner product

f:[0,2]-> C
C[0,2]
$=\int_{1}^{2}f(x)\overline{g(x)}dx$

i was told that its not an inner product
because the integral interval is from 1 to 2
so we dont know what on 0 to 1 interval
why its interfering?
what could happen that could make it a non inner product
?

2. Originally Posted by transgalactic
f:[0,2]-> C
C[0,2]
$=\int_{1}^{2}f(x)\overline{g(x)}dx$

i was told that its not an inner product
because the integral interval is from 1 to 2
so we dont know what on 0 to 1 interval
why its interfering?
what could happen that could make it a non inner product
?

It may be that <f,f> = INT{1,2} ( |f(x)|^2) dx = 0 without f(x) = 0 on all [0,1]

Tonio

3. so from 0 to 1 it could be a graph which integral is not 0
but from 1 to 2 its integral is 0
correct?

4. Originally Posted by transgalactic
so from 0 to 1 it could be a graph which integral is not 0
but from 1 to 2 its integral is 0
correct?

Yup. For example, let f(x) = -x + 1 in [0,1] and f(x) = 0 in [1,2]. This function is cont. on [0,2] and <f,f> = 0, but f =/= 0 .

Tonio