# Show that this inner product is positive definite

Tags:
1. Nov 5, 2015

### pondzo

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

I was able to do the second part of part a using integration by parts. But I am having no luck for the first part, proving that the inner product is positive definite. Pointers are appreciated!

2. Nov 5, 2015

### Orodruin

Staff Emeritus
Hint: what does it mean for an inner product to be positive definite?

3. Nov 5, 2015

### pondzo

Is this acceptable ?

$g(f,f)=\int_0^1 f(-p'f'-pf''+qf)dx\\~~~~~~~~~~~=\int_0^1-fp'f'-fpf''+qf^2dx\\~~~~~~~~~~~=\int_0^1-ff'p'dx-\int_0^1pff''dx+\int_0^1qf^2dx\\~~~~~~~~~~~=-[fpf']_0^1+\int_0^1p(f')^2dx+\int_0^1pff''dx-\int_0^1pff''dx+\int_0^1qf^2dx\\~~~~~~~~~~~=\int_0^1p(f')^2dx+\int_0^1pf^2dx$
Which is positive definite due to the square and p,q being positive definite and positive or zero.

4. Nov 5, 2015

### Orodruin

Staff Emeritus
Well, you do not have to do all of the partial integrations as this was the form the inner product was given on from the beginning, and the last p should be a q, but otherwise fine.

5. Nov 5, 2015

### pondzo

Oh of course, my mistake!

Do you think you would be able to give me some advice on part b ?

6. Nov 6, 2015

### geoffrey159

No, in the first part of the question, you are being asked to show that for all $f\in V_0$

(a) $g(f,f) \ge 0$
(b) $g(f,f) = 0 \Rightarrow f = 0$

Point (a) is easy, point (b) is more subtle and continuity plays a central role.

EDIT:
I realized I didn't help much in my post: there is a theorem that says that if $f: [a,b] \rightarrow \mathbb{R}$ is continuous, non-negative, and such that $\int_a^b f(x) \ dx = 0$, then $f = 0$

Last edited: Nov 6, 2015