Let $X$ be a Banach space and $\{A(t)\}_{t\in [0,T]}$ a family of closed and densely defined operators with constant domain $D(A(t))=Y$. As the operators are closed, $\|v\|_Y:=\|v\|+\|A(0)v\|$ is a Banach norm for $Y$.
Suppose that $t\mapsto A(t)v\in C^1([0,T],X)$ for each $v\in Y$, then prove that $A(t)\in B(Y,X)$ (is a bounded operator) and $t\mapsto A(t)$ is continuos in the operator $B(Y,X)$ norm.
What I've tried:
$A(t):(Y\|\cdot\|_Y)\longrightarrow (X,\|\cdot\|) $ are still closed so bounded by the closed graph theorem. We also have the bound:
$$\|A(t+h)v-A(t)v\|\leq h\sup_{t\in [0,T] }\|A'(t)v\| $$
Are $A'(t):Y\longrightarrow X$ also bounded operators? If so we can apply the uniform boundedness principle to deduce that $\sup_t \|A'(t)\|_{Y\to X}<+\infty$ and: $$\|A(t+h)-A(t)\|_{Y\to X}\leq h\sup_{t\in [0,T] }\|A'(t)\|_{Y\to X} \overset{h\to 0}{\longrightarrow} 0$$
PS: I don't know neither if less hypothesis such as $A(t)v$ just continuous or only differentiable are enough to prove it.