# does it defines inner product..

• Oct 1st 2009, 02:37 AM
transgalactic
does it defines inner product..
C[0,2]\\
f:[0,2]->c\\
$\displaystyle <f,g>=f(0)\overline{g(0)}+f(1)\overline{g(1)}+f(2) \overline{g(2)}$

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)+x f(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)

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
??
• Oct 5th 2009, 12:08 PM
tonio
Quote:

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

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)+x f(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)

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
??

If we denote the conjugate of a complex number z by z', then you have:

for a vector x (i.e., x is a function in C[0,2]):

<x,x> = x(0)x(0)' + x(1)x(1)' + x(2)x(2)' = |x(0)|^2 + |x(1)|^2 + |x(2)|^2

since zz' = |z|^2 for any complex number z, and thus you clearly have

<x,x> >= 0 .

Tonio
• Oct 6th 2009, 12:25 PM
transgalactic
i tried to prove step 2:

$\displaystyle <f,g>=f(0)\overline{g(0)}+f(1)\overline{g(1)}+f(2) \overline{g(2)}\\$
$\displaystyle k<f,g>=kf(0)\overline{g(0)}+kf(1)\overline{g(1)}+k f(2)\overline{g(2)}=<kf,g>\\$
$\displaystyle <f,kg>=kf(0)\overline{g(0)}+kf(1)\overline{g(1)}+k f(2)\overline{g(2)}?=\overline{k}<f,g>$

i dont know how in the end it proves it as
$\displaystyle \overline{k}<f,g>$
• Oct 6th 2009, 01:30 PM
tonio
Quote:

Originally Posted by transgalactic
i tried to prove step 2:

$\displaystyle <f,g>=f(0)\overline{g(0)}+f(1)\overline{g(1)}+f(2) \overline{g(2)}\\$
$\displaystyle k<f,g>=kf(0)\overline{g(0)}+kf(1)\overline{g(1)}+k f(2)\overline{g(2)}=<kf,g>\\$
$\displaystyle <f,kg>=kf(0)\overline{g(0)}+kf(1)\overline{g(1)}+k f(2)\overline{g(2)}?=\overline{k}<f,g>$

i dont know how in the end it proves it as
$\displaystyle \overline{k}<f,g>$

Remember that z'' = z and (Zw)' = z'w' (I use z' to denote complex conjugate), so:

<f, kg> = f(0)(kg(0))' + f(a) (kg(1))' + f(2)(hg(2))' = f(0)k'g(0)' + f(1)k'g(1)' + f(2)k'g(2)' = k'(f(0)g(0)' + f(1)g(1)' + f(2)g(2)' = k'<f,g>

Tonio
• Oct 6th 2009, 01:34 PM
transgalactic
thanks