In linear control theory, we consider a system of the form $$ \dot x = A\,x(t) + B\, u(t) \qquad \star $$ where $x(t)$ is the state function, $u(t)$ is the control function and $A,B$ are matrices.
We denote by $x(t,x_0,u)$ the solution at time $t$ starting from $x_0$ and associated to the control $u(t)$, and we have $$ x(t,x_0,u) = \exp(t\,A)\left(x_0+\int_0^T \exp(-s\,A)Bu(s) \; ds \right) $$ and the attainable set from $x_0$ at time $T$ is the set $$ A(x_0,T) = \cup_{u(\cdot)}\; x(t,x_0,u). $$
We can show that: $$ A(x_0,T)=\exp(At)x_0+A(0,T). $$
My question is the following.
Previously, I didn't specify the set where $u$ belongs. Typically we consider $u\in L^1([0,T],\mathbb{R}^m)$ for some $T>0$.
What if we take $u\in L^\infty([0,T],\mathbb{R}^m)$ ? $u\in L^2([0,T],\mathbb{R}^m)$ ?
What is the influence of the set where $u$ belongs ? Is that affect the attainable set ?
The standard assumption is that $u$ is piecewise continuous, which guarantees the existence and uniqueness of the solution to ($\star$) on any finite time interval. For a finite-horizon problem, $u$ doesn't need to be $L_p$, and in most cases it won't be. Consider for instance a step command, ramp command, or sinusoid. All of these are common inputs and none of them are $L_p.$ Most likely you are looking at infinite-horizon problems, in which you need integrals like
$$ J := \int_0^\infty (x^TQx + u^TRu)\ dt $$
to converge.
Concerning your question: if you restrict the input space, then in general you will change the reachable (attainable) set. This follows directly from the definition you gave plus the fact that if $A,B$ are sets, $x:A\rightarrow B$ is function, and $C\subset A$, then
$$ \bigcup_{c\in C}\{x(c)\} \subset\bigcup_{a\in A}\{x(a)\}. $$
Note: for $L_p-L_q$ inclusion theorems, see $L^p$ and $L^q$ space inclusion.