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Inner product space

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Poirot

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Feb 15, 2012
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consider C[0,2], the set of continous functions from [0,2] to C.

The inner product is <f,g> = the integral of f(t)g(t)* from 0 to 2.


show that:

sqrt(2)||f|| is greater than or equal to the magnitude of the integral of f from 0 to 2, where ||.|| is the norm of f.
 

ModusPonens

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Jun 26, 2012
45
What does g* mean?
 
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Poirot

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Feb 15, 2012
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conjugate
 

Opalg

MHB Oldtimer
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Feb 7, 2012
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consider C[0,2], the set of continous functions from [0,2] to C.

The inner product is <f,g> = the integral of f(t)g(t)* from 0 to 2.


show that:

sqrt(2)||f|| is greater than or equal to the magnitude of the integral of f from 0 to 2, where ||.|| is the norm of f.
What have you tried so far? Can you think of results that might help here (Cauchy–Schwarz inequality perhaps, for a suitable choice of g)?
 
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Poirot

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Feb 15, 2012
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What have you tried so far? Can you think of results that might help here (Cauchy–Schwarz inequality perhaps, for a suitable choice of g)?
I've tried writing what each side is. I don't see how schwarz inequality is relevant. I'm interested in f, not g.
 

ModusPonens

Well-known member
Jun 26, 2012
45
\(\displaystyle |\int_0^2 f(t)dt|=|2\bar{f}_1+2i\bar{f}_2|=\sqrt{4(\bar{f}_1)^2+4(\bar{f}_2)^2}=2\sqrt{(|\bar{f}|)^2}\leq 2\sqrt{\bar{|f|^2}}=2\sqrt{(\int_0^2 |f|^2 dt)/2}=\sqrt{2}||f||\)

where the bar is the average and \(\displaystyle f=f_1+if_2\).

EDIT: I made a correction and some clarifications of notation.
 
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