# Inner product on C(R)

1. Dec 18, 2008

### Redukt

My apologies in advance for asking what (to me) looks like an extremely stupid question, but I just can't figure it out.

1. The problem statement, all variables and given/known data:
Where is this an inner product:
$$\int_{a}^{b}f(x)g(x) dx$$

a) on C[a,b]?
b) on C(R)?

The answer is that it is an inner product on a), but not on b) - apparently on b) the axiom of positivity fails. I do not understand how this is possible, since all functions that are C(R) are also C[a,b] - or have I just always misunderstood this notation? Does not C(R) mean "functions continuous on all of R"?

2. Relevant equations

This is what the answer key says: It fails on b) because $$\exists f \ne 0 : \Vert f \Vert^2 = 0$$

3. The attempt at a solution:

has mainly consisted of trying (unsuccessfully) to work backwards towards a counterexample. I don't know how to do this in any other, more general way, since the whole idea seems illogical to me. Please enlighten me, somebody?

2. Dec 18, 2008

### dirk_mec1

The inner product of the function $$f(x) = \sqrt{x}$$ is zero and this f is not the zero function.

3. Dec 18, 2008

### Vid

Any continuous function that is 0 on [a,b], but not anywhere else. sqrt(x) doesn't work for arbitrary a,b, but a simple piecewise function that looks similar to the absolute value function would.

4. Dec 18, 2008

### dirk_mec1

What are you talking about? The inproduct of sqrt(x) delivers x if you integrate this from -inf to +inf you'll get zero, right?

5. Dec 18, 2008

### Vid

It's not from -inf to inf. It's from a to b. The reason the integral from a to b isn't an inner product on C(R) is because there is an f in C(R) such that it's integral from a to b is zero, but f is not zero everywhere.

6. Dec 18, 2008

### dirk_mec1

Yes you're right I didn't read well enough. The integral has bounds form a to b.

7. Dec 19, 2008

### Redukt

I think this is occasion for a big *headdesk* (on my own behalf, obviously). How did I not see this? Thanks a lot. :)