MHB Is the Inner Product Space in C[0,2] Satisfied by the Integral Equality?

Poirot1
Messages
243
Reaction score
0
consider C[0,2], the set of continuous 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.
 
Physics news on Phys.org
What does g* mean?
 
conjugate
 
Poirot said:
consider C[0,2], the set of continuous 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)?
 
Opalg said:
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.
 
$$|\int_0^2 f(t)dt|=|2\bar{f}_1+2i\bar{f}_2|=\sqrt{4(\bar{f}_1)^2+4(\bar{f}_2)^2}=2\sqrt{(|\bar{f}|)^2}\leq 2\sqrt{\bar{|f|^2}}=2\sqrt{(\int_0^2 |f|^2 dt)/2}=\sqrt{2}||f||$$

where the bar is the average and $$f=f_1+if_2$$.

EDIT: I made a correction and some clarifications of notation.
 
Last edited by a moderator:

Similar threads

Replies
6
Views
2K
Replies
4
Views
117
Replies
1
Views
3K
Replies
8
Views
3K
Replies
4
Views
2K
Replies
8
Views
2K
Back
Top