Let $A$ be a bounded linear operator on a Banach space $X$. Can we show that for an arbitrary $n \in \mathbb{N}$ and $x \in X$ such that $\|x\|_X \geq 1$ we have that $$\|A^n x \| \leq \|Ax\|^n.$$
If $X$ is a Hilbert space and $A$ is a normal operator that I think that it is true.
By spectral theorem $A$ is unitarly equivalent to an operator of multiplication $M_{f}$ by a complex-valued essentially bounded measurable function $f$ on $L^2(\Omega, \mu)$, where $(\Omega, \mu )$ is a measure space. And for multiplication operator we have the following: \begin{align}\| M_f^n g\| =& \|f^n \cdot g\|= \mathrm{ess \ sup}_{z}|f(z)|^n |g(z)| \\ \leq& \mathrm{ess \ sup}_{z}|f(z)|^n|g(z)|^n = \|(f\cdot g)^n\|=\|f \cdot g\|^n=\|M_fg\|^n,\end{align} where $\|g\|=1$.
Do we have $$\|A^nx\| \geq \|Ax\|^n$$ for all $x$ such that $\|x\| < 1$.
This is also false for multiplication operators.
Consider
$$ \left (\begin{matrix} 10 & 0 \\ 0 & 0 \end{matrix}\right)$$
and $x = (\frac{1}{100}, \gamma)$, where $\gamma$ is chosen so that $\Vert x \Vert = 1$. Then $\Vert Ax \Vert^n= \frac{1}{10^n}$, but $\Vert A^n x\Vert = \frac{10^n}{100}$.