i proved too that <f,f> =>0
how to prove <f,f>=0 if f=0
??
which is bigger that 0
so i disproved it
?
Why the lower limit of the integral is 2 and upper one is zero? Anyway:
if <f,f> = 0 then both the integral and ||f(1)|| = 0 . Now, you didn't say but if we're working with continuous functions on [0,2] then a non-negative function's integral on [0,2] equals zero iff the function is zero, because otherwise, by continuity, there'd be an inteval (a,b) where f is not zero ==> IN{0,2}(||f(x)||^2)dx >= INT{a,b}(||f(x)||^2 dx > 0, by properties of Riemann's Integral.
If you're not working with continuous functions then the above indeed isn't an inner product
Tonio
Well....yes and no (more "no" than "yes", imo). I think you still have to prove why if the integral of a non-negative real continuous function over certain interval I is zero then the function is zero on I.
Now f(x) may be complex but |f(x)| is real, and |f(x)| = 0 <==> f = 0.
Tonio
Sorry but not even close: your argument didn't use continuity at all, but if the function is not continuous it still can be true that it is non-negative and its integral zero but the function is NOT zero.
Using upper and lower integrals (either by means of Darboux sums or of Riemann sums), and bounding the function properly, it can easily be shown that the integral of a continuous function positive over an interval has to be positive, and this is based in the following: if f is continuous in w and if f(w) > 0 then there exists a neighborhood I of w s.t. f(x) > 0 for all x in I.
This is not necessarily true at all if f is not cont. at w.
Tonio