1
$\begingroup$

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.

$\endgroup$
9
  • 1
    $\begingroup$ Where exactly are you stuck? Is it positive definiteness or linearity? $\endgroup$
    – Hans Engler
    Jun 18, 2016 at 15:43
  • $\begingroup$ 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. $\endgroup$
    – Rebellos
    Jun 18, 2016 at 15:45
  • 1
    $\begingroup$ 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. $\endgroup$
    – Hans Engler
    Jun 18, 2016 at 17:14
  • $\begingroup$ 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. $\endgroup$
    – Rebellos
    Jun 18, 2016 at 18:32
  • $\begingroup$ Can you show this in $C([0,1])$? If so, what is different here? $\endgroup$
    – Hans Engler
    Jun 18, 2016 at 18:42

1 Answer 1

Reset to default
2
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Sorry for not accepting at the first place, I just went through my non-reviewed questions and made sure I checked decent answers ! $\endgroup$
    – Rebellos
    Nov 20, 2018 at 19:04
  • 1
    $\begingroup$ @Rebellos : That's okay. I use this site to try to sharpen my mind, and to help others with questions at the same time. $\endgroup$
    – Disintegrating By Parts
    Nov 20, 2018 at 19:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .