# 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?

Mar 20, 2018

$f \left(x\right) = \ln x + 1$

#### Explanation:

We split the inequality into 2 parts:
$f \left(x\right) - 1 \ge \ln x$ $\to$ (1)
$f \left(\frac{x}{e}\right) \le \ln x$$\to$ (2)

Let's look at (1):
We rearrange to get $f \left(x\right) \ge \ln x + 1$

Let's look at (2):
We assume $y = \frac{x}{e}$ and $x = y e$. We still satisfy the condition $y \in \left(0 , + \infty\right)$.$f \left(\frac{x}{e}\right) \le \ln x$
$f \left(y\right) \le \ln y e$
$f \left(y\right) \le \ln y + \ln e$
$f \left(y\right) \le \ln y + 1$
$y \in x$ so $f \left(y\right) = f \left(x\right)$.

From the 2 results, $f \left(x\right) = \ln x + 1$

Mar 20, 2018

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:

$f \left(x\right) = A \ln \left(x\right) + b$

Plugging in the conditions, this means
$A \ln \left(\frac{x}{e}\right) + b \le \ln x \le A \ln \left(x\right) + b - 1$
$A \ln \left(x\right) - A + b \le \ln x \le A \ln x + b - 1$

We can subtract $A \ln \left(x\right) + b$ from the entire equation to find
$- A \le \left(1 - A\right) \ln x - b \le - 1$

Flipping,
$1 \le \left(A - 1\right) \ln x + b \le A$

If we want this to be true for all x, we see that the upper bound is a constant and $\ln \left(x\right)$ 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 $A = b = 1$:

$f \left(x\right) = \ln \left(x\right) + 1$