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Hi guys, there's a problem to which me and my pals just can't seem to get an answer that is congruent the answer on the back of the book. The homwork is due in two days and I just want to make sure the book is truly wrong. The question's simple enough: given f,g continuous on the circle ([-pi,pi)), is
[tex](f,g)=\int_0^{\pi} fg^*dt[/tex]
an inner product on the space of continuous functions defined on the circle?
We're saying yes because the 3 properties are verified:
i) (af+bg,h) = a(f,h)+b(g,h), as if evident by the properties of the integral.
ii) [tex](f,g)^* = \left( \int_0^{\pi} gf^*dt \right)^* = \int_0^{\pi} (gf^*)^*dt = \int_0^{\pi} fg^*dt = (g,f)[/tex]
iii) [tex](f,f) = \int_0^{\pi} ff^*dt = \int_0^{\pi} |f|^2 dt \geq 0[/tex]
(since |f| >= 0 and = 0 <==> f=0)
Hence all 3 properties are verified. Any objection?
[tex](f,g)=\int_0^{\pi} fg^*dt[/tex]
an inner product on the space of continuous functions defined on the circle?
We're saying yes because the 3 properties are verified:
i) (af+bg,h) = a(f,h)+b(g,h), as if evident by the properties of the integral.
ii) [tex](f,g)^* = \left( \int_0^{\pi} gf^*dt \right)^* = \int_0^{\pi} (gf^*)^*dt = \int_0^{\pi} fg^*dt = (g,f)[/tex]
iii) [tex](f,f) = \int_0^{\pi} ff^*dt = \int_0^{\pi} |f|^2 dt \geq 0[/tex]
(since |f| >= 0 and = 0 <==> f=0)
Hence all 3 properties are verified. Any objection?