Let $1\leq X_n \leq N$ be a submartingale with respect to a filtration $(\mathcal{F}_n)_{n\geq 1}$, i.e., $\mathbb{E}[X_{n+1}\mid\mathcal{F}_{n}] \geq X_n$ and $N \geq 0$ is a scalar. Suppose $X_0 = 1$ and let $n_k$ denote the index at which the process doubles for the $k$-th time, i.e., $n_0 = 0$ and for $k \geq 1$, $$ n_k = \min\{n > n_{k-1}: X_n > 2X_{n-1}\}. $$ Let $\kappa_N$ be the set of all these indices until time $N$ and $K_N = |\kappa_N|$ be the total number of doublings until time $N$. Can we bound $\mathbb{E}[K_N]$?
My Approach: In the special case where $X_{n+1} \geq X_n$ monotonically, we claim that $K_N \leq 1 + \log N$. because otherwise, $$ X_{n_{K_N}} = \prod_{2 \leq n \leq n_{K_T}}\frac{X_{n}}{X_{n-1}} \cdot X_1 \geq \prod_{n \in \kappa_N, \\n \neq 1}\frac{X_{n}}{X_{n-1}} > 2^{K_N-1} > N, $$ which is a contradiction with $X_{n_{K_N}} \leq N$. Can we prove something similar in the more general case of submartingales?
Edit: Originally the question was with $0 \leq X_i \leq N$. But, now it is changed to $1 \leq X_i \leq N$ to avoid the counter example provided by @Peter Morfe.
Here is an explicit submartingale for which $E[K_N] \ge cN$ for a constant $c>0$. (We will not try to optimize this constant).
Suppose that $2^\ell-1 \le N <2^{\ell+1}-1$. Consider the Markov chain $\{X_t\}$ (a variant of the "greasy ladder") that from 1 jumps to 3, and for each $j \in [2,\ell-1]$ jumps from $2^j-1$ to either $1$ or $2^{j+1}-1$ with equal probability. The chain is started at $X_0=1$ and is absorbed at $2^\ell-1$. This chain is indeed a submartingale.
Let $\tau$ be the first time $t$ when $X_t=2^\ell-1$. Then $$P(\tau \ge N/8)\ge (1-2^{2-\ell})^{N/8-1} \ge 1/e$$ so $E[\tau \wedge N] \ge \frac{N}{8e}$. Let $I_t=1$ if $X_t \ge X_{t-1}$ and $I_t=0$ otherwise. Then $S_T=\sum_{t=1}^T (I_t-1/2)$ is a submartingale, so $$0 \le E[S_{\tau \wedge N}] \le E[K_N]- E[\tau \wedge N]/2 \le E[K_N]-\frac{N}{16e} \,.$$