Favard spaces are complete What seemed at first glance a straightforward verification had me completely lost, the definition I have of Favard space is:
Let $X$ be a Banach space and $(D(A),A)$ be the infinitesimal generator of a holomorphic semigroup $T(t)$ (note: we do not necessarily ask $D(A)$ to be dense). For $\alpha\in (0,1)$ we set the Favard space to be
$$\mathcal{F}_{\alpha}(A):=\left\{x\in X|\sup_{t\in (0,1]}\|t^{1-\alpha}AT(t)x\|<+\infty\right\}.$$
I managed (at least) to check that this is indeed a linear subspace of $X$ but when it comes to proving that it is closed I'm really unsure on why this should be the case. I've consulted basically every piece of literature on Interpolation Spaces and every single one of them does not prove this and says it is trivial.
The only helpful formula that comes to my mind is
$$\|AT(t)\|\leq \frac Ct,$$
for a certain constant $C$. But this doesn't really say a lot to me since it leads to
$$\|t^{1-\alpha}AT(t)\|\leq Ct^{-\alpha},$$
and the RHS becomes infinity when taking the $\sup$ on $(0,1]$.
 A: In general, the space $\mathcal F_\alpha(A)$ is not closed in $X$. I will explain why below, but let me first prove a true statement. The space $\mathcal F_\alpha(A)$ is complete when endowed with the norm $\lVert\cdot\rVert_\alpha=\lVert\cdot\rVert_X+\sup_{t\in (0,1]}\lVert t^{1-\alpha}A T(t)x\rVert$:
As mentioned in the question, that map
$$
p_t\colon X\to[0,\infty),\,x\mapsto\lVert t^{1-\alpha}AT(t)x\rVert
$$
is continuous for every $t\in (0,1]$. Thus $\sup_{t\in(0,1]}p_t\colon X\to [0,\infty]$ is lower semicontinuous as supremum of continuous maps.
Now if $(x_n)$ is Cauchy with respect to $\lVert\cdot\rVert_\alpha$, then it is in particular a Cauchy sequence in $X$. Hence there exists $x\in X$ such that $\lVert x-x_n\rVert_X\to 0$. Moreover, as $(\lVert x_n\rVert_\alpha)$ is bounded, we have
$$
\sup_{t\in (0,1]}p_t(x)\leq \liminf_{n\to\infty}\sup_{t\in (0,1]}p_t(x_n)\leq \liminf_{n\to\infty}\lVert x_n\rVert_\alpha<\infty.
$$
Thus $x\in \mathcal F_\alpha(A)$. Moreover,
$$
\lVert x-x_n\rVert_\alpha\leq\liminf_{m\to\infty}\lVert x_m-x_n\rVert_\alpha,
$$
which goes to zero as $n\to\infty$ since $(x_n)$ is a Cauchy sequence with respect to $\lVert\cdot\rVert_\alpha$.
Let us come back to the original claim If $\mathcal F_\alpha(A)$ were closed, this would imply that $\lVert\cdot\rVert_\alpha$ and $\lVert\cdot\rVert_X$ are equivalent by the open mapping theorem, or in other words, there exists a constant $C>0$ such that $\lVert\cdot\rVert_\alpha\leq C\lVert \cdot\rVert_X$.
But it is easy to find counterexamples to this inequality. For example, let $X=\ell^2(\mathbb N)$ and $(Ax)_n=-nx_n$ with maximal domain. For $t_n=(2n)^{-1}$ we have
$$
p_{t_n}(\delta_{2n})=\lVert (2n)^{-(1-\alpha)}e^{-}2n\cdot\delta_{2n}\rVert_2=\frac{(2n)^{\alpha}}e,
$$
which is clearly unbounded in $n$.
