# how to prove this inner product

• Oct 6th 2009, 03:57 PM
transgalactic
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
?
• Oct 6th 2009, 04:30 PM
tonio
Quote:

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
• Oct 7th 2009, 12:04 AM
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?
• Oct 7th 2009, 05:12 AM
tonio
Quote:

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