inner product prove..

**transgalactic** · October 6th 2009, 02:15 PM

$<f,g>=3\int^{0}_{2}f(x)\overline{g(x)}dx +f(1)\overline{g(1)}\\$
$\overline{<f,g>}=\overline{3\int^{0}_{2}f(x)\overl ine{g(x)}dx +f(1)\overline{g(1)}}=3\int^{0}_{2}\overline{f(x)} g(x)dx +\overline{f(1)}g(1)=<g,f>\\$

i proved too that <f,f> =>0
how to prove <f,f>=0 if f=0
??

$<f,f>=3\int^{0}_{2}f(x)\overline{f(x)}dx +f(1)\overline{f(1)}=3\int^{0}_{2}{||f(x)||}^2dx +{||f(1)||}^2\\$
which is bigger that 0

so i disproved it
?

**tonio** · October 6th 2009, 03:36 PM

Originally Posted by transgalactic

$<f,g>=3\int^{0}_{2}f(x)\overline{g(x)}dx +f(1)\overline{g(1)}\\$
$\overline{<f,g>}=\overline{3\int^{0}_{2}f(x)\overl ine{g(x)}dx +f(1)\overline{g(1)}}=3\int^{0}_{2}\overline{f(x)} g(x)dx +\overline{f(1)}g(1)=<g,f>\\$

i proved too that <f,f> =>0
how to prove <f,f>=0 if f=0
??

$<f,f>=3\int^{0}_{2}f(x)\overline{f(x)}dx +f(1)\overline{f(1)}=3\int^{0}_{2}{||f(x)||}^2dx +{||f(1)||}^2\\$
which is bigger that 0

so i disproved it
?

Why the lower limit of the integral is 2 and upper one is zero? Anyway:

if <f,f> = 0 then both the integral and ||f(1)|| = 0 . Now, you didn't say but if we're working with continuous functions on [0,2] then a non-negative function's integral on [0,2] equals zero iff the function is zero, because otherwise, by continuity, there'd be an inteval (a,b) where f is not zero ==> IN{0,2}(||f(x)||^2)dx >= INT{a,b}(||f(x)||^2 dx > 0, by properties of Riemann's Integral.

If you're not working with continuous functions then the above indeed isn't an inner product

Tonio

**transgalactic** · October 6th 2009, 11:09 PM

it is a continues funtion
it is a complex function
i fixed the intervals its from 0 to 2
$<f,g>=3\int^{2}_{0}f(x)\overline{g(x)}dx +f(1)\overline{g(1)}\\$
$\overline{<f,g>}=\overline{3\int^{2}_{0}f(x)\overl ine{g(x)}dx +f(1)\overline{g(1)}}=3\int^{2}_{0}\overline{f(x)} g(x)dx +\overline{f(1)}g(1)=<g,f>\\$

i proved too that <f,f> =>0
how to prove <f,f>=0 if f=0
??

$<f,f>=3\int^{2}_{0}f(x)\overline{f(x)}dx +f(1)\overline{f(1)}=3\int^{2}_{0}{||f(x)||}^2dx +{||f(1)||}^2\\$
which is bigger that 0

and it equal 0 if f=0

so i proved it correctly

**tonio** · October 7th 2009, 04:10 AM

Originally Posted by transgalactic

it is a continues funtion
it is a complex function
i fixed the intervals its from 0 to 2
$<f,g>=3\int^{2}_{0}f(x)\overline{g(x)}dx +f(1)\overline{g(1)}\\$
$\overline{<f,g>}=\overline{3\int^{2}_{0}f(x)\overl ine{g(x)}dx +f(1)\overline{g(1)}}=3\int^{2}_{0}\overline{f(x)} g(x)dx +\overline{f(1)}g(1)=<g,f>\\$

i proved too that <f,f> =>0
how to prove <f,f>=0 if f=0
??

$<f,f>=3\int^{2}_{0}f(x)\overline{f(x)}dx +f(1)\overline{f(1)}=3\int^{2}_{0}{||f(x)||}^2dx +{||f(1)||}^2\\$
which is bigger that 0

and it equal 0 if f=0

so i proved it correctly

Well....yes and no (more "no" than "yes", imo). I think you still have to prove why if the integral of a non-negative real continuous function over certain interval I is zero then the function is zero on I.
Now f(x) may be complex but |f(x)| is real, and |f(x)| = 0 <==> f = 0.

Tonio

**transgalactic** · October 7th 2009, 06:24 AM

"why if the integral of a non-negative real continuous function over certain interval I is zero then the function is zero on I."
beause integral is a sum
and in here its a sum of positive values
so in order it to be 0
f has to be 0.

good enough ?

**tonio** · October 7th 2009, 10:22 AM

Originally Posted by transgalactic

"why if the integral of a non-negative real continuous function over certain interval I is zero then the function is zero on I."
beause integral is a sum
and in here its a sum of positive values
so in order it to be 0
f has to be 0.

good enough ?

Sorry but not even close: your argument didn't use continuity at all, but if the function is not continuous it still can be true that it is non-negative and its integral zero but the function is NOT zero.
Using upper and lower integrals (either by means of Darboux sums or of Riemann sums), and bounding the function properly, it can easily be shown that the integral of a continuous function positive over an interval has to be positive, and this is based in the following: if f is continuous in w and if f(w) > 0 then there exists a neighborhood I of w s.t. f(x) > 0 for all x in I.
This is not necessarily true at all if f is not cont. at w.

Tonio

**transgalactic** · October 7th 2009, 12:46 PM

but it is given that f is continues on [0,2]
you cant say "but if the function is not continuous"
so now my argument is correct?

**tonio** · October 7th 2009, 01:41 PM

Originally Posted by transgalactic

but it is given that f is continues on [0,2]
you cant say "but if the function is not continuous"
so now my argument is correct?

Still no because your argument doesn't use continuity at all it must, UNLESS it is already plain obvious to you guys that a continuous positive function is positivie in some interval of positive length...this is the theorem I mentioned in one of my messages.

Tonio

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