Here is an idea.
We have where , so on .
Hence f=0 on [a,b]
What do you think?
And I just want to make sure that proving there cannot be some point where . So suppose there is (at least) one point such that . Then since is continuous on , on some interval , so on the closed interval . Clearly since the function must achieve a minimum at a point on the interval by its continuity on a bounded interval, so for any partition , , so and since the function is zero everywhere else, then , which is a contradiction.
I think this seems a little overboard but I just want to make sure if this is correct or if there is an easier proof that shows cannot be positive at any point. Thanks.