differentiability/holomorphicity of family of bounded operators Edit: It seems I made a mistake in the statements on differentiability.  I will replace weak differentiable implies strong differentiable with weak continuously differentiable implies strongly continuous.
Suppose we have a family $\{B(t)\}$ of bounded operators from a Banach space $X$ to a Banach space $Y$, with $t$ in the interval $(a,b)$.  If $t\mapsto B(t)x$ is continuously differentiable for all $x$ in $X$, is $t\mapsto B(t)$ continuous in the norm topology on $B(X,Y)$?  (*)
A similar question about holomorphicity can be asked for a family of bounded operators with complex parameter.
I easily found proofs that for functions of the reals (resp. complexes) taking values in a Banach space, weakly continuously differentiable (resp. holomorphic) implies strongly continuous (resp. holomorphic) (**).  I tried to apply this to prove * but don't know enough about the dual of B(X,Y).  I'm not sure if * is true by a general principle using ** or if the differentiable and holomorphic versions must be proven from scratch in different ways.
I want to know the differentiable case because it is used without proof (and is supposedly obvious) in Pazy's semigroups book in the study of hyperbolic evolution equations.  It seems like he's hinting at an incorrect proof by using the uniform boundedness principle on {dB(t)/dt}, which would assume that B(t) is differentiable in B(X,Y) in the first place.  In another part of the book * is claimed to be obvious.
 A: 
If $t\mapsto B(t)x$ is (Frechet) differentiable for all $x$ in $X$, is $t\mapsto B(t)$ (Frechet) differentiable in the norm topology on $B(X,Y)$?

I think you don't have to specify "Fréchet" when dealing with functions of one real variable $t$. Anyway, the answer seems to be negative. Let $X=Y=L^1[0,1]$. Given $x\in X$ and $t\in\mathbb R$, define $B(t)x$ as a function on $[0,1]$, namely
$$B(t)x(s) = \begin{cases} sx(s),\quad &s>t \\ 0, & s<t\end{cases} \tag1$$
(If $t\le 0$, the definition reads as $B(t)x(s)=sx(s)$ for all $s\in [0,1]$). 
For every $x$, 
$$\frac{B(t)x-B(0)x}{t}(s) = \begin{cases} 0,\quad &s>t \\   -(s/t)\, x(s), \quad & s<t\end{cases} \tag2$$
Bounding the $L^1$ norm of (2) by $\int_0^t |x(s)|\,d(s)$, we conclude that 
$$\lim_{t\to 0}\frac{B(t)x-B(0)x}{t} =0 \tag3$$
On the other hand, the limit $t^{-1}(B(t)-B(0))$ does not converge to $0$ in the operator norm. Indeed, no matter how small (positive) $t$ we take, plugging the unit norm vector
$$x(s)=\begin{cases} 0,\quad &s>t \\ t^{-1}, & s<t \end{cases} \tag4$$
into (2) yields
$$\|t^{-1}(B(t)x-B(0)x)\|_{L^1} = \int_0^t \frac{s}{t^2}\,ds = \frac12 \tag5$$
Thus, $\|t^{-1}(B(t)-B(0))\|\ge 1/2$.
