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Is it true that every algorithm with runtime complexity of $T(n)=\Omega(n)$ satisfies that $T(n)=\Theta(f(n))$ for some convex function $f$?

All the examples that I could think of satisfy the above mentioned requirement. For example, all polynomials are convex on $\mathbb N_+$. Additionally, $n\log(n)$ is convex from $n=2$ and on, so it's asymptotically equivalent to convex function.

Dudi Frid
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    The polynomial $x^2-4x+4$ is not convex on $\Bbb N_+$. You meant to say all polynomials with positive leading coefficient are eventually convex on $\Bbb N_+$. Or all polynomials with nonnegative coefficient are convex on $\Bbb N_+$. – John L. May 04 '19 at 16:27

2 Answers2

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Is it true that every algorithm with runtime complexity of $T(n)=\Omega(n)$ satisfies that $T(n)=\Theta(f(n))$ for some convex function $f$?

No. A simple example is $$T(n)=\begin{cases}n & \text{ when } n \text{ is odd,}\\ n^2 & \text{ when } n \text{ is even.} \end{cases}$$

What if we also require $T(n)$ be strictly increasing?

The answer is still no. A simple example is

$$T(n)=\begin{cases}2^{(n+1)^2} & \text{ when } n \text{ is odd,}\\ 2^{n^2}+1 & \text{ when } n\text{ is even.} \end{cases}$$

Since $\displaystyle\lim_{k\to\infty}\frac{T(2k)}{T(2k-1)}=1$ and $\displaystyle\lim_{k\to\infty}\frac{T(2k+1)}{T(2k)}=\infty,$ there is no convex function $f$ such that $T(n)=\Theta(f(n))$.

Exercise 1. What if we also require both $T(n)$ be strictly increasing and $T(n+1)/T(n)$ be bounded?

Exercise 2. Let $T_1$ and $T_2$ be two increasing functions such that $T_2(n)\not=O(T_1(n))$. Show that there is a function $T(n)$ that is bounded between $f_1$ and $f_2$ such that there is no convex function $f(n)$ such that $T(n)=\Theta(f(n))$.

John L.
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No. Consider $T(n) = n$ if $n$ is even, $T(n) = 2^n$ if $n$ is odd.

D.W.
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