Continue to Site

# does it defines inner product..

Status
Not open for further replies.

#### transgalactic

##### Banned
does it defines inner product..(analisys)

C[0,2]\\
f:[0,2]->c\\
$<f,g>=f(0)\overline{g(0)}+f(1)\overline{g(1)}+f(2)\overline{g(2)}$

i dont know how to solve the complement.
i tried to prove there easier version without the complement first.

<f,g>=f(0)g(0)+f(1)g(1)+f(2)g(2)\\
A.<g,f>=g(0)f(0)+g(1)f(1)+g(2)f(2)=f(0)g(0)+f(1)g(1)+f(2)g(2)=<f,g>
B.x<f,g>=x(f(0)g(0)+f(1)g(1)+f(2)g(2))=xf(0)g(0)+xf(1)g(1)+xf(2)g(2)=<xf,g>=<f,xg>

C.<x+y,g>=(x(0)+y(0))g(0)+(x(1)+y(1))g(1)+(x(2)+y(2))g(2)=<x,g>+<y,g>

step d:
<x,x>=x(0)x(0)+x(1)x(1)+x(2)x(2)

i dont know how to prove that
<x,x> greater or equal 0
i dont know the values of x
??

Last edited:

#### skyhawk

##### New Member
If x(0) is real, then x(0)x(0) > 0 because the square of any real number is positive, unless the number is zero.

Likewise x(1)y(1) > 0 and x(2)x(2) > 0 so <x,x> > 0 unless
x(0) = x(1) = x(2) = 0 in which case <x,x> = 0.

Last edited:

#### transgalactic

##### Banned
how do you know that x(0) is real
i only know that x is continues [0,2]
?

Last edited:

#### skyhawk

##### New Member
If x isn't real then you need the form of inner product that uses the complement (complex conjugate). Any number times its complex conjugate is greater than or equal to zero.

#### transgalactic

##### Banned
ok i tried to make the original question into easier one but i made it harder

actually the original question is:
$<f,g>=f(0)\overline{g(0)}+f(1)\overline{g(1)}+f(2)\overline{g(2)}$

how to tackle the original one
?

Last edited:
Status
Not open for further replies.

Replies
4
Views
7K
Replies
4
Views
3K
Replies
3
Views
2K
Replies
9
Views
2K
Replies
5
Views
4K