Are NP problems lower bounded by exponential order of growth?

My understanding of P. vs NP is quite limited. I can understand P refers to an algorithm with an upper bound (big O) with order of growth $n^c$ for some constant c and variable n. My question is, do NP problems have a hypothesized lower bound order of growth (big Omega) of $c^n$ for deterministic machines? I can't find this stated anywhere and I'm trying to understand if this is something that is assumed or not.

Thanks.

First of all, every problem in P is in NP, and so we need to change your question to involve NP-complete problems rather than just NP problems.

The Exponential Time Hypothesis states that 3SAT requires exponential time $c^m$ for some constant $c > 1$, where $m$ is the number of variables. The Strong Exponential Time Hypothesis states that $k$SAT requires exponential time $c_k^m$, where $c_k \to 2$.

However, your conjecture is unconditionally false. Consider the language $L$ consisting of all pairs $\langle x,y \rangle$ such that $x$ is a 3SAT instance on $m$ variables (in particular, $m \leq |x|$), and $y = 0^{|x|^{100}}$. The language $L$ is clearly NP-complete, but it can be solved in time $2^m \mathit{poly}(n) = 2^{n^{1/100}} \mathit{poly}(n)$, where $n$ is the input size.

The Exponential Time Hypothesis does imply that every NP-complete problem requires time $c^{n^\alpha}$, for some $c>1$ and $\alpha>0$. This kind of running time is sometimes called subexponential time, but unfortunately this expression has several different interpretations, depending on context.

StackExchange Computer Science Q#129897, answer score: 16