Question

Would you be so kind help me please? It's about real analysis.

Answer 1

Let `f(x)` be defined and differentiable in the interval `I sub RR` and suppose `f'(x) >= 0` for every `x in I`.

Choose now two points `x_1,x_2 in I` with `x_1 < x_2`.

Based on Lagrange's theorem there must be a point `xi in (x_1,x_2)` such that:

`f'(xi )= (f(x_2)-f(x_1))/(x_2-x_1)`

But as `xi in I` then `f'(xi) >=0`, so:

`(f(x_2)-f(x_1))/(x_2-x_1) >=0`

and because `x_1 < x_2` so `x_2-x_1 >= 0`, then:

`f(x_2)-f(x_1) >= 0`

or:

`f(x_2) >= f(x_1)`

So we proved that if `f'(x) >=0` then:

`x_1 < x_2 => f(x_1) <= f(x_2)`

that is the function is increasing.

Now we prove the opposite implication: suppose the function is increasing so that:

`x_1 < x_2 => f(x_1) <= f(x_2)`

Taking any point `x in I`, as `f(x)` is differentiable, then the limit:

`lim_(h->0) (f(x+h)-f(x))/h = f'(x)`

exists.

Now for any `h >0` we have:

`x+ h > x`

which implies:

`f(x+h) >= f(x)`

and:

`(f(x+h)-f(x))/h >=0`

Then if we take the limit from the right, for the persistence of the sign we must have:

`lim_(h->0^+) (f(x+h)-f(x))/h >=0`

But when the limit exists, the limit from the right is equal to the limit, so:

`lim_(h->0) (f(x+h)-f(x))/h >=0`

which means:

`f'(x) >=0`