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Prove that $C[a,b]$ is a inner product space..

Please help me to prove the following axioms or can anybody send me a link which includes these proofs.

1.conjugate symmetry property

2.inner product of $f$,$f=0$ iff $f=0$

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    Although this is a very basic and standard inner product space, it'd be a good idea you *provide* the function you want to prove is an inner product (integral and stuff...?). All the necessary stuff usually follows pretty smoothly from the basic properties of Riemann integrals.2012-10-18
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    yeah.proving these properties are very basic.but the second property needs some analysis part.so i need to write them correctly.that's why i want anyone to help me.the function is inner product of f,f=integral of f(t)g(t)(bar) form a to b.2012-10-18
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    From the "bar" thing I gather you're talking about (analytic?) *complex* functions of a real variable, for which the symbol $\,C[a,b]\,$ is not, I think, usually used: this symbol is used more widely for continuous real functions defined on the closed interval $\,[a,b]\,$ . Is this correct?2012-10-18
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    yes,i'm talking about the complex inner product on C[a,b] with the integral defined above.bar is the conjugate of the function.2012-10-18
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    If proving the properties is "very basic" then why are you having to ask somebody else to do it?2012-10-18

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$$\langle\,f\,,\,g\,\rangle:=\int_a^b f(t)\overline{g(t)}dt=\int_a^b \overline{\overline{f(t)}g(t)}\,dt=\overline{\int_a^b g(t)\overline{f(t)}}\,dt=\overline{\langle\,g\,,\,f\,\rangle}$$

$$\langle\,f\,,\,f\,\rangle=0\Longleftrightarrow \int_a^b |f(t)|^2\,dt=0\stackrel{\text{by continuity}}\Longleftrightarrow |f(t)|^2=0\,\,\text{on}\,\,[a,b]\Longleftrightarrow f(t)=0$$

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    thank you very much DonAntonio..I really appreciate your feed back!!!2012-10-18