Increasing function $f: \mathbb{N} \to \mathbb{R}^{\ge 0}$ such that $f(3^n) = 2^n$ and $\forall n \exists c : f(n+c)+f(n-c) \ge 2f(n).$

Question

Can we build $f: \mathbb{N} \to \mathbb{R}^{\ge 0}$ which is increasing and satisfies the following criterions?

1. $f(3^n) = 2^n$ for all $n.$
2. There exists $N$ such that for all $n \ge N,$ there exists some $c$ (which may depend on $n$) such that $f(n+c)+f(n-c) \ge 2f(n)$

I introduced the first criterion to make it easier to come up with the function on paper. I made a table using graph paper listing $n$ and $f(n)$ for $n = 1, \dots, 33$ and then put $1,2,4,8,\dots$ under the powers of $3.$ Now I'm having difficulties filling out this table so that the second criterion is satisfied

Remark: Even if this task is possible, $N$ might be insanely large. However, trying to beg for a larger $N$ is just delaying the inevitable. In any case, I can just let the first $N$ terms be something like $\epsilon, 2\epsilon \dots, \epsilon N,$ but the real challenge begins after I've put down those terms because from then on, there is no violation of the second criterion. With that being said, you might as well try to see if a low value of $N$ works.

1|2|3|4|5|6|7|8|9|10|11|12|13|14|15|16|17|18|19|20|21|22|23|24|25|26|27|28|29|30|31|32|33

1|?|2|?|?|?|?|?|4|?|?|?|?|?|?|?|?|?|?|?|?|?|?|?|?|?|8|?|?|?|?|?|?|

Update: An arithmetic progression between the powers of two almost works. It only fails right on the powers of $2.$ So maybe there's some way to modify this idea that works.

Answer 1

No, such $f$ can't exist. Consider an increasing function on $\mathbb N$ satisfying (1), and take any $N \in \mathbb N$.

If $3^m \le x < 3^{m+1}$ we have $2^m \le f(x) < 2^{m+1}$, so for any constant $\alpha > 0$, $f(x) - \alpha x \to -\infty$ as $x \to \infty$. Take $\alpha$ so $f(N) - N \alpha > f(x) - x\alpha$ for all $x < N$, i.e. $$\alpha < \min_{x < N} \frac{f(N) - f(x)}{N-x}.$$ Then the maximum of $f(x) - \alpha x$ for all $x \in \mathbb N$ is attained at some $x_0$ with $x_0 \ge N$. Moreover by choosing $\alpha$ appropriately we can ensure this maximum is unique (because there are only countably many $\alpha$ for which there is a tie). Now for any $c > 0$, adding $f(x_0) - \alpha x_0 > f(x_0 - c) - \alpha (x_0 - c)$ and $f(x_0) - \alpha x_0 > f(x_0 + c) - \alpha (x_0 + c)$ we get $2 f(x_0) - 2 \alpha x_0 > f(x_0 - c) + f(x_0 + c) - 2 \alpha x_0$, i.e. $2 f(x_0) > f(x_0 - c) + f(x_0 + c)$, contradicting (2).