$L^p$ boundedness of a function In a Banach space $X$, let $T:X\rightarrow X$ be a bounded operator and $f:[0,a]\rightarrow X$ a measurable function. Assume that $Tof\in L^p([0,a],X)$, is $f\in L^p([0,a],X)$?
Thank you.
 A: Not necessarily:


*

*If $Tg=0$ for all $g$, $f$ can be anything measurable.

*If $Tg=g(a)$ is evaluation to a constant, $f$ can again be anything measurable.

*For a more specific example, $T: f(x) \mapsto xf(x) $ is bounded (by $a$), and if 
$$f(x)=\begin{cases} x^{-1/p} & x>0 \\ 0&  x=0\end{cases},$$ 
then 
$$Tf = \begin{cases} 1 & x>0 \\ 0&  x=0\end{cases},$$
so $Tf \in L^p$ but $f \notin L^p$.

A: Following gives you a category of examples where $T \neq 0$
Let $T : X \to X$ be a bounded linear map which map in which $Ker (T) \neq \{0\}$
Now consider any function $g:[0,\frac{1}{2}a]\rightarrow Ker(T)$ such that  $g \notin L^p ([0,\frac{1}{2}a], Ker(T))$ (take $g$ so that $\int \|g\|_{X} = + \infty$.) 
Now define $f:[0,a]\rightarrow X$ with $f=g$ on  $[0,\frac{1}{2}a]$,   and $f=0$ on  $(\frac{1}{2}a, a].$ Observe that $f \notin L^p([0,a], X)$
Therefore you can check that $Tof (t) = 0 \quad  \forall t \in [0,a]$ which shows $Tf\in L^p([0,a],X)$.
This gives you a hint about a necessary condition such that $f\in L^p([0,a],X).$
