$A$ is the infinitesimal generator of the $C_{0}$-semigroup $(T(t))_{t\ge 0}$ and $V$ is a one dimensional linear subspace of $X$.

I want to show that $V$ is $T(t)$-invariant $\iff$ $V=span{\{\phi\}}$ with $\phi$ an eigenvector of $A$.

By definition, $V$ is $T(t)$-invariant if $\forall t\ge 0$, $T(t)V\subset V$.

However, for finite dimensional systems, a (closed) subspace is $T(t)$-invariant $\iff$ it is $A$-invariant, for $T(t)=e^{At}$.

Since $V$ is one-dimensional, may I interpret this as a 'finite dimensional system' and use the above fact?


Because $V$ is a one-dimensional subspace, then $V=[\{ x\}]$ is the linear span of some non-zero vector $x$. Using $T(t)V \subset V$ gives a scalar function $\rho$ such that $$ T(t)x = \rho(t)x,\;\;\; t \ge 0. $$ Furthermore, $\rho(t+s)=\rho(t)\rho(s)$ follows from the semigroup property. Because $T(t)x$ is also $C_{0}$, then $\rho$ is continuous. So $\rho(t)=e^{a t}$ for some $a \ge 0$, which means that $x \in \mathcal{D}(A)$ and $$ Ax = \lim_{t\downarrow 0}\frac{\rho(t)-\rho(0)}{t}x=\rho'(0)x=ax. $$ The converse state is similarly proved. If $Ax=\lambda x$ for some $\lambda$, then $$ \frac{d}{dt}(e^{-\lambda t}T(t)x) = e^{-\lambda t}T(t)Ax-\lambda e^{-\lambda t}T(t)x = 0. $$ Thus $e^{-\lambda t}T(t)x = (e^{-\lambda t}T(t)x)|_{t=0}=x$, or $$ T(t)x = e^{\lambda t}x,\;\;\; t \ge 0. $$ So $T(t)V\subseteq V$, where $V=[\{x\}]$.


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