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

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The discussion centers on the inner product space C[0,2] and the integral equality involving continuous functions. The inner product is defined as the integral of the product of functions and their conjugates over the interval [0,2]. The goal is to demonstrate that sqrt(2)||f|| is greater than or equal to the magnitude of the integral of f from 0 to 2. Participants consider the relevance of the Cauchy–Schwarz inequality in proving this relationship, with one user detailing their calculations to show the inequality holds. The conversation emphasizes the need for clarity in notation and understanding the implications of the integral in the context of the inner product space.
Poirot1
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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.
 
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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.
 
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