Show that this inner product is positive definite

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!

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

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.

Orodruin
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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.

Oh of course, my mistake!

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

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.
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 ##

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