# To Prove

1. Jun 26, 2005

### steven187

hello all

well iv been workin on this problem but no matter how many ideas cross my mind i cant prove it, iv tried playing around with the properties of continous functions on closed intervals but didnt get me anywhere

let f be a contiuous function on [a,b] which is non-negative, suppose that

$$\int_{a}^{b}f(x)dx=0$$
show that
$$f(x)=0 \ \forall x\epsilon [a,b]$$

this is probably the best of my proofs but i dont think its correct because i think i have to prove contradiction for all functions that are more than zero, please help any advice would be helpful

thank you

lets assume that f(x)>0 let f(x)=c>0 by FTC
$$\int_{a}^{b}f(x)dx=F(b)-F(a)=0$$
$$F(b)=F(a)$$
since f(x)>0 then F(x)=cx where c>=0
ca=cb
0=cb-ca
0=c(b-a) since x is an element of [a,b] for all a,b an element of R then c=0
therefore a contradiction and f(x)=0

2. Jun 26, 2005

### matt grime

if f is non-zero at some point in [a,b] it is non-zero in some interval around that point. and that's sufficient to show the answer.

if f(x)=c then it is a constant function. but you aren't told f(x) is constant, are you?

Last edited: Jun 26, 2005
3. Jun 26, 2005

### HallsofIvy

That certainly doesn't follow!

If, for any x, f(x)> 0 then there exist some small closed interval around that x on which f(x) is positive (since f is continuous). In fact, f must have a minimum on that interval which is positive. The integral of f on that small interval must be a positive number and, since f is non-negative the integral on the entire interval cannot be less than that positive number.

4. Jun 26, 2005

### saltydog

Why do I find this stuff so interesting?

I hear whispers of Cauchy in the hallways of Analysis.