Question

Hello Friends,

This is a question concerning the proof that if domain of a function f is I in R1,
x0 is a local maximum and x0 is the only critical point then x0 is the global maximum of the function f.

It proves by the means of contradiction, that if x0 is not the global maximum then the function has another critical point
It assumes that there is a point y0 in I for which f(y0) > f(x0) and y0 > x0. But here I am quite lost;
How it is that f(y0) > f(x0), becuase the proof assumes that x0 is the local maximum (and tries to prove it's the global maximum as well)? If x0 is assumed to be the

Answer 1

When proving by contradiction you are assuming the opposite to be true, so in this case we assume that x0 is not the global max and that there is, in fact, a max greater than it at y0, therefore f(y0) > f(x0) by definition of it being a greater maximum. Thus f(y0) must have a critical point, by definition, but since there is only one critical point at x0 then y0 = x0, but then if f(y0) > f(x0) for y0=x0, it would fail the vertical line test which is a contradiction so our assumption that f(y0) > f(x0) is false, therefore f(x0) is the global max

Answer 2

Thanks JohnRocket for your answer. The way you've given the proof sounds clear. But the proof I have read assume that x0 and y0 are contained in the same interval (I) , x0 is the only critical point and the local maxima and yet f(y0) > f (x0). it then goes on that if x0 is not a global maxima then the function has another critical point: since x0 is the local maxima the function decreases to the right of x0 and since f (y0) > f(x0) the functoin should start increasing to the left of y0 so there exist another crtical point z0 between x0 and y0 but that contradicts the hypotheis that xo is the only critical point.
But my question is how can the assumption that x0 and y0 are contained in the same interval (I), x0 is the local maxima and still f(y0) > f(x0) be true. since if f(y0) > f(x0) then x0 does not remain the local maxima on which the proof is based. I hope i am able to explicate that my question concerns the x0 being a local maxima.
Could you please help.

Answer 3

It can't be true! And that's exactly what you're trying to show. It's the nature of proving by contradiction. You are proving that f(x0) is the global max. How do you do this? You assume that it's not the global max and attempt to prove that there is another max greater than it, ie f(y0) > f(x0). But as you go through proving that f(y0) > f(x0), you find something that doesn't make sense, a "contradiction". Therefore, you're original assumption, that there is a max that is greater can't be true and you can conclude that f(x0) is the max.

Answer 4

Thanks a lot JohnyRocket.