Convergence in norm operator If I have an operator valued functions $A(z):H_1\to H_2$ such that the following limit
$$\lim_{z\to z'}A(z)=A(z')$$
exists in the uniform topology of $B(H_1,H_2)$, that is
$$\Vert A(z)-A(z')\Vert_{H_1\to H_2}\to 0$$ when $z\to z'$ can I write
$$\Vert A(z)f-A(z')f\Vert_{H_2}\leq C\Vert f\Vert_{H_1}$$
with $C\to 0$ when $z\to z'$?
I remember you that
$$\Vert A\Vert_{H_1\to H_2}=\sup_{f\in H_1}\frac{\Vert Af\Vert_{H_2}}{\Vert f\Vert_{H_1}}$$

I assume your operators are linear since that operator norm makes sense only for that case, so we have by linearity and a classic bound
$$\Vert A(z)f-A(z')f\Vert_{H_2}\leq \Vert A(z)-A(z')\Vert_{H_1\to H_2} \Vert f\Vert_{H_1}$$
The classic bound can be deducted from definition of the operator norm:
$$\Vert A\Vert_{H_1\to H_2}=\sup_{f\in H_1 \, f \neq 0}\frac{\Vert Af\Vert_{H_2}}{\Vert f\Vert_{H_1}}$$
If you take any $f$ (the case $f:=0$ being trivial) you have
$$ \frac{\Vert Af \Vert_{H_2}}{\Vert f \Vert_{H_1}} \leq \sup_{f\in H_1 \, f \neq 0}\frac{\Vert Af\Vert_{H_2}}{\Vert f\Vert_{H_1}} = \Vert A\Vert_{H_1\to H_2}$$
and the inequality follows.
