# Inner products

1. Nov 25, 2006

### daniel_i_l

1) C[-1,2] is a space of all continues functions f: [-1,2] -> C (complex)
Is:
$$<f,g> = \int_{-1}^{2}|f(t) + g(t)|dt$$
an inner product of C[-1,2]?
I think that the answer is no because:
$$<f+g, h> \neq <f,h> + <g,h>$$
for some f and g. this can happen when all the functions are positive and so:
|f(t) + h(t) + g(t)| doesn't equal |f(t) + h(t)| + |g(t) + h(t)|

2)
V is a space of all real functions with defined double derivatives in the interval
$$[-\pi, \pi]$$
we define:
$$<f,g> = f(-\pi)g(-\pi) + \int_{-\pi}^{\pi}f''(x)g''(x)dx$$
is <f,g> an inner product of V?
also here i think that the answer is no because <f,f> can equal zero even if all of f isn't 0, this can happen if f(-pi) is zero and the double derivative is 0 everywere (contiues slope).
am i correct?
Thanks.

Last edited: Nov 25, 2006
2. Nov 25, 2006

### matt grime

2) there is nothing in the definition of inner product that means <f,f>=0 if and only if f=0.

Wht is the definition of an inner product? do these satisfy, or not satisfy the definitions?

3. Nov 25, 2006

### daniel_i_l

Isn't that the 3rd property of the inner product? look here for example:
http://planetmath.org/encyclopedia/InnerProduct.html [Broken]

did i misunderstand something?

and is (1) correct?
Thanks.

Last edited by a moderator: May 2, 2017
4. Nov 25, 2006

### matt grime

Sorry, my mistake - I was thinking of a bilinear pairing, not an inner product.

So, you have the definitions, and the counter examples: what was the question?

Last edited: Nov 25, 2006
5. Nov 25, 2006

### daniel_i_l

I just wanted to make sure that my counter examples were correct.
Thanks