#Calculus Proof
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Well if h is a positive continuous function on [a,b] such that it’s integral is 0 on [a,b] what can you say of h ?
It's a weighted function?
It’s 0 on [a,b], any positive continuous function on [a,b] which has 0 integral is 0
Oh right yea
But f(x) need not be a positive continuous function alone right?
It could be intersecting the x-axis at some point in the interval and the integral could be 0 too
If it’s positive continuous and has 0 integral then it is 0
That is true. But it hasn't been mentioned anywhere that f(x) is positive continuous.
Not f
you know that the integral of f(x)g(x) is 0 for all g
Yes
what about when ||g=f||
Then integral of f(x)^2 would be 0
Yes
.
If there exists a point in [a, b] where f(x) is not 0, then f(x)^2 would be greater than 0 at that point.
Which means f(x) has to be 0 at any point in the interval?
well yeah f(x)^2=0 => f(x)=0
But what if g is not equal to f?
We're taking a specific case to prove it and generalizing it to the entire interval.
Is there a generic solution to it?
Well there is nothing else to prove, you know that the integral of f(x)g(x) is 0 for any continuous function g, if it’s true for any function g then it’s true for g=f so you have that the integral of f(x)^2 is 0, with what I said before this implies that f(x)^2=0 for all x ie f(x)=0 for all x
Got it then
Thanks @wide tendon
Np
If it is alright with you, could I contact you in private for a couple more problems of a similar structure?
I'm having a bit of a trouble wrapping my head around those
I’m sorry but I’m kinda busy because of uni, but you can make help posts, other people may help
And I’ll comment if I have time
Alright then. Thanks again for the help.
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