Hi all..
i want to ask you about this question in real analysis..
"suppose that f continuous on [a,b] , that f(x) >= 0 (more or equal 0 ) for all x in [a,b] and that =0
prove that f(x)=0 for all x in [a,b] "
Suppose $\displaystyle \exists c \in [a,b]$ such that $\displaystyle f(c) > 0$. Then there exists a neighborhood $\displaystyle N$ of $\displaystyle c$ such that $\displaystyle f(x)>f(c)/2$ for all $\displaystyle x \in N \cap [a,b]$. (Prove this using the continuity of $\displaystyle f$ at $\displaystyle c$.) Then $\displaystyle 0 = \int_a^b f \geq \int_{N \cap [a,b]} f(c)/2 > 0 $ which is false.
I don't see how anything that you mention really has much to do with the problem. The solution I gave is quite standard : suppose the function is not identically zero, then it's bounded away from zero on a small subinterval, and hence its integral on this small subinterval is positive, and the integral on the whole interval is at least the integral on this small subinterval.