Is this statement about this function f true?

Question

Consider the following statement:

If f:[a,b] -->R (a and b in R) is continuous and f(a) < f(b), then [a, b] contains a subinterval where f is strictly increasing.

Is this true? So far, I could neither prove nor find a counter example.

If, instead of strictly increasing, we admit f only monotonically increasing, then does anything change?

White soax: The negation of the statement is not that f be strictly decreasing on all subintervals on [a,b], but just that there is no subinterval on [a, b] where f is strictly increasing. .

Answer 1

The statement is false.

Consider the Weierstrass nowhere-differentiable function (http://en.wikipedia.org/wiki/Weierstrass_function ).

The Weierstrass function is not differentiable anywhere, but it is continuous.

Thus, the restriction of the Weierstrass function to [a, b] is continuous but nowhere differentiable.

The Weierstrass function cannot be monotone on any subinterval of [a, b], because a monotone function is differentiable almost everywhere (as noted at http://en.wikipedia.org/wiki/Monotonic_function ) The Weierstrass function certainly is not differentiable almost everywhere on any subinterval of [a, b].

So, for a counterexample to your statement, we can select f to be the Weierstrass function, and select "a" and "b" to be any two points satisfying f(a) < f(b).

Answer 2

purposes are even whilst f(x) = f(-x), so as that regulations b out by using fact may no longer the definition of an excellent functionality. purposes are weird and wonderful whilst -f(x) = f(-x), so as that regulations c out by using fact may no longer the definition of a wierd functionality. so as meaning the main appropriate answer is a or d. only attempt them the two. BTW - on a i'm particular you advise -f(x) = f(-x), or you have 2 same products on the two area of an =, subsequently it would continuously be actual. observe no count what x is comparable to, f(x) = 7. yet as quickly as we throw a detrimental on one area, we ought to continuously throw a detrimental on the different to maintain them =, so -f(x) = -7 no count what x is. we could say x = 2. a. -f(x) = - f(2) = -7 and f(-x) = f(-2) = 7, so those are no longer equivalent to one yet another, a is fake d. f(x) = f(2) = 7 and f(-x) = f(-2) = 7, those are equivalent so the respond is d.

Answer 3

Write g(x,e)=f(x+e)-f(x). This is continuous under the product topology relative to the subset {(x,e) in R^2;a<=x<=b and a<=x+e<=b}, so the set A={(x,e);f(x+e)>f(x)} ={(x,e);0 =g^(-1)((0,infinity)) is open in this topology. Also the condition f(a) I'll add more later, time for work.

Unless I've made some mistake. Steiner, please check my work. :)

Answer 4

guess its pretty obvious rt..

if fa and a,b belongs to R.now it may not be monotonically increasing..since the increase in this interval may not be same as the increase in the other intervals.

Answer 5

Yes I think so, but as I am not sure, so I will change my mind and say no. On the other hand. Do you know, I used to be indecisive? But now I am not sure

Answer 6

Youve lost me. But well done u brainybox!! :)

Answer 7

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