Is the operator $T$ a bounded operator mapping $T: H^n([0,\pi]) \rightarrow H^{n-1}([0,\pi])$ ($H^n$ is the n-th Sobolev space with respect to $L^2$) or not? The operator itself is given by $T(f):=f'+\cos(2x)f.$ Somehow I feel that this is not true as these kind of differential operators often allow divergent sequences such as $f_n:=\sin(nx)$, but I could not find such a sequence in this case.
If anything is unclear, please let me know.
The differential operators like $T$ are not bounded from $H^n$ to $H^{n}$. But $T$ is bounded from $H^n$ to $H^{n-1}$, since the latter space requires one fewer derivative than the domain. Indeed, $$\|f'\|_{H^{n-1}}\le \|f\|_{H^n}$$ since the right hand side involves all of the terms that the left hand side involves (details vary depending on your precise definition of $H^n$).
The multiplication by $\cos 2x$ is bounded even from $H^n$ to $H^n$. Indeed, the $n$th derivative of the product $f(x)\cos 2x$ is of the form $$ \sum_{k=0}^n f^{(k)} \frac{d^{n-k}}{dx^{n-k}}(\cos 2x) $$ where the $L^2$ norm of every term is controlled by $\|f\|_{H^n}$.