Question

Please help solve this, I can't come up with a solution. The question is to find f? Given `f:(0,+oo)->RR` with `f(x/e)<=lnx<=f(x)-1 , x in (0,+oo)`

The answer is f(x) = lnx +1 but how do i prove it?

Answer 1

`f(x)=lnx+1`

Explanation:

We split the inequality into 2 parts:
`f(x)-1>=lnx` `->` (1)
`f(x/e)<=lnx` `->` (2)

Let's look at (1):
We rearrange to get `f(x)>=lnx+1`

Let's look at (2):
We assume `y=x/e` and `x=ye`. We still satisfy the condition `y in (0,+oo)`.`f(x/e)<=lnx`
`f(y)<=lnye`
`f(y)<=lny +lne`
`f(y)<=lny+1`
`y inx` so `f(y)=f(x)`.

From the 2 results, `f(x)=lnx+1`

Answer 2

Assume a form then use the bounds.

Explanation:

Based on the fact that we see that f(x) bounds ln(x), we might assume that the function is a form of ln(x). Let's assume a general form:

`f(x) = Aln(x) + b`

Plugging in the conditions, this means
`Aln(x/e) + b le lnx le Aln(x) + b - 1`
`Aln(x) - A + b le ln x le A ln x + b - 1`

We can subtract `Aln(x) + b` from the entire equation to find
`- A le (1-A)ln x - b le - 1`

Flipping,
`1 le (A-1)lnx + b le A`

If we want this to be true for all x, we see that the upper bound is a constant and `ln(x)` is unbounded, that term clearly must be 0. Therefore, A = 1, leaving us with

`1 le b le 1 implies b = 1`

So we have only the solution with `A = b = 1`:

`f(x) = ln(x) + 1`