# Integrals and proof by contradiction

Assume f is continuous on [a,b]. Assume also that the integral from a to b of f(x)g(x)dx=0 for every function g that is continuous on [a,b]. Prove that f(x)=0 for all x in [a,b].

I try a proof by contradiction, but I am a little unsure of the logical negation. I came up with assume the first two statements are true but f(x)=/0 for any x in [a,b].

### Re: Integrals and proof by contradiction

So, then would I have to prove that the orthogonal functions g(x) is a proper subset of the continuous functions g(x)?

### Re: Integrals and proof by contradiction

Well my friend, if is not so, how can do there exists the integral?

### Re: Integrals and proof by contradiction

I'm sorry, I must not be following you right. Could you perhaps show an example of a function g(x) that \int_a^b f(x)g(x)dx=0 if f(x)\neq 0 for all x in [a,b].

### Re: Integrals and proof by contradiction

Here is a nice concrete example of the scenario you requested.

Let a=-1, b=1, f(x)=1, and g(x)=x. Then \int_a^b f(x)g(x)=\int_{-1}^1 x=0.

### Re: Integrals and proof by contradiction

that f(x) must = 0?

### Re: Integrals and proof by contradiction

I really dont get this but can you at least say if my logical negation for the contradiction was correct if i were to solve it by contradiction?

### Re: Integrals and proof by contradiction

I want to find a g(x) such that \int_a^b f(x)g(x)=0 implies that f(x)=0.

I wanted to answer all of nheit's questions regarding my most recent post all in one place.

I am puzzled that this question was even asked:

"I want to find a g(x) such that \int_a^b f(x)g(x)=0 implies that f(x)=0."

I specifically told you to let g(x)=f(x). In other words, there is not necessarily a function g that works for *every single f*, but, given a function f, you are allowed to choose g so that it depends on f (in this case, identical with f).

"that f(x) must = 0?"

Yes, but why?

"I really dont get this but can you at least say if my logical negation for the contradiction was correct if i were to solve it by contradiction?"

Yes, your logical negation is correct. On the other hand, to be honest, I am baffled that someone who understands the concept of proof by contradiction is unable to read a sketch of a straightforward proof and fill in the gaps.

I said, "Note that the graph of y=(f(x))^2 does not go below the x axis. (Why?)" This is because, no matter what x is, (f(x))^2 must be nonnegative.

Next, I said, "On the other hand, by the given property of f, we have \int_a^b (f(x))^2=0. What does this tell you about the graph of y=(f(x))^2?" Remember, the integral represents area. Also note that (f(x))^2 is continuous on [a,b] since f is. If a continuous function does not go below the x axis and the area under the curve is 0, what must the function be? How does this help you conclude that f(x)=0 (rather than just guessing that it helps)?

Ok, the reason why I started with contradiction is my ta gave us the hint that we should. However, the proof he used was very similar to yours. His contradiction came when f(x) \neq 0 and \int_a^b (f(x))^2dx>0. Thanks for all your help.

### Re: Integrals and proof by contradiction

Hi,
well, I suggest this idea: Suppose f(x)\neq 0 (use \neq = not equal, for =/). Then g(x) must be orthogonal to f(x) in [a,b]. But
\int_a^b f(x)g(x)dx=0 for every continuous g(x) and not just only for those orthogonal functions g(x).

### Re: Integrals and proof by contradiction

I would not resort to proof by contradiction for this problem. This is one of those "so simple it's hard" problems.

The "big idea" is to let g(x)=f(x). Note that the graph of y=(f(x))^2 does not go below the x axis. (Why?) On the other hand, by the given property of f, we have \int_a^b (f(x))^2=0. What does this tell you about the graph of y=(f(x))^2?

### Re: Integrals and proof by contradiction

since we make g(x)=f(x),then
Integral from a to b f(x)g(x) dx
=Integral from a to b f(x)^2 dx
>=min{f(x)^2}*(b-a)>=0
by the topic,Integral from a to b f(x)^2 dx=0,
The equal sign is satisfied only in the condition of f(x)=0.