Inner product on Complex Space - Proof

Studying an introduction to Hermitan inner products and complex spaces, I've found my self stuck to deal with a rather than classic example of an inner product.

The complete exercise goes as follows :

Let $V =C([0,1])$, the complex space of the continuous functions : $[0,1] \to \mathbb C$.

Show that the following is an inner product :

$\langle x, y\rangle = \int_0^1f(x)\overline{g(x)}dx$

Now if V becomes the vector space of the bounded and partly continuous functions $\big($the bounded functions $f:[0,1] \to \mathbb C$ for which exists a finite sequence $0 = a_0 < a_1 < \dots < a_n = 0$ with $f$ continuous in every $(a_i,a_{i+1})$ $\big)$ does the upper form continue to be an inner product ?

Do I need to show the properties of the Hermitan inner product ? If so, how do I proceed ? I don't even know where to start with this and it seems an elementary example since everywhere it's just used as an inner product and the proof is never asked for. I'd appreciate any help.

• Where exactly are you stuck? Is it positive definiteness or linearity? – Hans Engler Jun 18 '16 at 15:43
• I know there are 3 things. The interchangable factor, the linearity and the positive definiteness. I don't even know how to start any of them, probably the positive definiteness is the easiest since the integral of a perfect square will always be bigger or equal to zero. – Rebellos Jun 18 '16 at 15:45
• Go over the corresponding proofs for $C[(0,1])$ and see where exactly there might be a difficulty. when you try to extend these arguments to $V$. Narrow down the difficulty as much as you can. – Hans Engler Jun 18 '16 at 17:14
• I've shown that $<f,f>=0 \Leftrightarrow f = 0$ and that $<\overline{g,f}>=<f,g>$ and $<f,f> > 0$ and $<af,g>=a<f,g>$ but i cannot show the linearity with the addition. – Rebellos Jun 18 '16 at 18:32
• Can you show this in $C([0,1])$? If so, what is different here? – Hans Engler Jun 18 '16 at 18:42

If you allow the functions in your space to be discontinuous at even one point, then you can consider the function $f$ that is $1$ at such a point, and $0$ everywhere else on the interval. Then $\int_0^1 |f|^2dt=0$ even though $f$ is not identically $0$. So that's the basic problem: you don't get positive definiteness.