# Verifying an inner product

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 $$E$$

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

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,

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

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 $$|z|^{2} \geq 0)$$.