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Given Problem:

$f$ is a function that satisfies the 3 following properties:

  1. $f:\mathbb{N}\rightarrow\mathbb{N}$
  2. $\sqrt{f(x)}\ge\frac{f(x)+f(1)}{2}$ for some x within the given domain
  3. $\frac{f(n)}{f(1)} = 2n - (f(1))^2 , n\ge2$

It is required to find all such functions.

Clarification: Consider that $0\notin\Bbb{N}$

Given Solution:

The first condition states that the domain and codomain of $f$ is $\Bbb{N}$. So, AM-GM inequality can be applied here as all the numbers that we are working with are positive. Applying the Inequality, we can get: $$\sqrt{f(x)}\ge\frac{f(1)+f(x)}{2}\ge\sqrt{f(1)f(x)}$$

But since $f(1)\ge1$, we have $$\sqrt{f(x)}\le\sqrt{f(1)f(x)}$$ So, the inequalities given above are actually equalities. Hence, for the inequality to hold, we must have $f(1) = 1$

Substituting this in 3rd condition gives:$$f(n) = 2n-1, n\ge2$$

Note that $f(1) = 2\cdot1-1$

So, $$f(n)=2n-1, n\in\Bbb{N}$$

My Confusion:

The second property does not hold for $x=2$.

L.H.S. $=\sqrt{f(2)}$ $=\sqrt{2\cdot2-1}$ $=\sqrt{3}$ $=1.7320508...$

R.H.S. $=\frac{f(2)+f(1)}{2}$ $=\frac{3+1}{2}$ $=2$

So, we have L.H.S. $\lt$ R.H.S Which clearly violates the second property. Where is the mistake?

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  • $\begingroup$ Welcome to MSE. Please choose your tags with care. Your question is not about functional-analysis. $\endgroup$ – José Carlos Santos Apr 18 '18 at 10:03
  • $\begingroup$ It is said for some $x$ within the given domain, not for all. $\endgroup$ – Peter Melech Apr 18 '18 at 10:06
  • $\begingroup$ Note that since $ f ( 1 ) = 1 $, the second property holds for $ x = 1 $, and thus $ f ( x ) = 2 x - 1 $ is indeed a solution. $\endgroup$ – Mohsen Shahriari Apr 18 '18 at 19:10
  • $\begingroup$ Is the answer that no such function exists? Because that seems the case here $\endgroup$ – Naweed G. Seldon Apr 25 '18 at 19:23
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I suppose the answer is that no such function exists, if you notice your inequality with $f(1) \geq 1$ gives us

$$\sqrt{f(x)}\ge\frac{f(1)+f(x)}{2}\ge\sqrt{f(1)f(x)} \geq \sqrt{f(x)}$$ So, not only do we get $$f(1)f(x) = f(x)\implies f(1) = 1$$ We also get $$\sqrt{f(x)}=\frac{f(1)+f(x)}{2}=\frac{1+f(x)}{2} \tag{1}$$

Now, we have arrived at $$f(n) = 2n-1$$ Now, substituting that in $(1)$ we see $$\sqrt{2n-1}=\frac{2n-1 + 1}{2}=n \implies n^2 - 2n +1=0 \implies n=1$$

So, no such function exists

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