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?

2 Answers
Mar 20, 2018

#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#

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(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 #