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Let $E$ be the linear space of all continuous complex-valued functions defined on the interval $(a,b)$ of the real line. Consider the inner product in $E$

$\langle f,g \rangle = \int_a^{b} f(x) \overline {g(x)} dx$

Answer

Linearity in the first argument is easy

Now, $\langle f,f \rangle = \int_a^b f(x) \overline{f(x)} dx = \int_a^b f(x)^2 dx \geq 0$ by Chebyshev inequality.

let $f(x)=0$ then $\int_a^b 0 dx =0= \langle f,f \rangle =0$

let $\langle f,f \rangle =0 \implies \int_a^b f(x)^2 dx =0 \implies f(x)=0$ a.e on $(a,b)$ thus

$\langle f(x),f(x) \rangle=0 \iff f(x)=0$.

For the conjugate symmetry,

$\langle f,g \rangle = \int_a^b f \overline g dx = \int_a^b \overline{\overline f} \overline g dx = \int_a^b \overline g \overline{\overline f} dx = \int_a^b \overline{g \overline{f}} dx$=$\overline{\langle g,f \rangle}$

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The only mistake is in writing $z\overline {z}=z^{2}$ and $z^{2}\geq 0$. For a complex number $z$ we have $z\overline {z}=|z|^{2} \geq0$. ($z^{2} \geq 0$ is not true but $|z|^{2} \geq 0)$.

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