Bound the norm of the partial trace of an operator on a Hilbert space Let $H=H_1 \otimes H_2$ a composite Hilbert space and let $A, B$ bounded linear operators on $H$, and we can assume they are trace class. Let $A_2$ we denote the operator on $H_2$ obtained by taking the partial trace on $H_1$, i.e. $A_2 = \mathrm{tr}_1 (A) \in B(H_2)$. Assume that we know there is some $\varepsilon > 0$ so that $$ ||A-B||_{B(H)} < \varepsilon. $$ Can we deduce a similar bound for $$ ||A_2 - B_2 ||_{B(H_2)} ? $$

Notes:
1) The finite dimensional case $\dim (H_1) = n_1$, $\dim (H_2) = n_2$, $\dim (H) = n$ all finite would be fine (but if you have a hint or source for an answer in the general case I'd appreciate!).
2) (My approach) Choose the Frobenius (or Hilbert-Schmidt) norm, $$ ||A||_{B(H)} = \sqrt{\mathrm{tr}(A^\dagger A)}$$ and let $C=A-B$. We know that $ ||I_1 \otimes C_2 ||_{B(H)} = ||I_1||_{B(H_1)} ||C_2||_{B(H_2)}$ (cf. Reed Simon Vol. 1 page 299), then 
$$
\begin{array}{rcl}
||C_2||_{B(H_2)} &=& \frac{1}{\sqrt{n_1}} ||I_1 \otimes C_2||_{B(H)} \\
&=& \frac{1}{\sqrt{n_1}} ||C - (I_1 \otimes C_2) - C ||_{B(H)} \\
&\leq& \frac{\varepsilon}{\sqrt{n_1}} + \frac{1}{\sqrt{n_1}} ||C - (I_1 \otimes C_2)||_{B(H)} 
\end{array}
$$
and expect to bound the term on the right by $\alpha ||C||_{B(H)}$. To see what is going on I put
$$
C = C^{(1)}_1 \otimes C^{(2)}_1 + C^{(1)}_2 \otimes C^{(2)}_2 
$$
(taking enough terms of these form should be fine to represent an arbitrary $C$). A calculation gives
$$
||C - (I_1 \otimes C_2)||_{B(H)} = ||(C^{(1)}_1 - (\mathrm{tr} C^{(1)}_1) I_1)\otimes C_1^{(2)} + (C^{(1)}_2 - (\mathrm{tr} C^{(1)}_2) I_1)\otimes C_2^{(2)}||_{B(H)}
$$
how close this is to $||C||_{B(H)}$?
 A: There is a bound on the norm of partial trace of operator that depends on the type of norm and the size of the Hilbert space you traced out. For the Schatten $p$-norm, which is defined for $p\ge1$ as
$$\|A\|_p=\left[\text{Tr}(|A|^p)\right]^{1/p}$$
one can show a bound, due to Rastegin, where
$$\|\text{Tr}_1(A)\|_p \le [\dim(H_1)]^{(p-1)/p} \|A\|_p$$
This bound can be made dimension-independent by choosing $p=1$, in which case the norm is the trace norm. But if you choose $p=2$, which gives the Frobenius norm, you'll have a dimension-dependent bound.
A: (a colleague just helped me to arrive at the answer. Moreover, it seems to me that it carries over to the infinite dimensional space so long as the partial trace is well defined). 
Let $ C = \sum_{ij} \alpha_{ij} e_i \otimes e_j,$ where $e_i$ is a matrix with only one non-zero entry equal to one, so we have this basis for the complete space. We calculate 
$$
\begin{array}{rcl}
||C_2||_{(B(H_2),||\cdot||_F)}^2 &=& ||\sum_{ij} \alpha_{ij} \mathrm{Tr}(e_i) e_j ||_{(B(H_2),||\cdot||_F)}^2 \\
&=& \sum_{ij} |\alpha_{ij}|^2  |\mathrm{Tr}(e_i)|^2 \\
&\leq& \sum_{ij} |\alpha_{ij}|^2 = ||C||_{(B(H),F)}^2
\end{array}
$$
where we used that $\mathrm{Tr}(e_i)$ is either one or zero. We conclude that $||C_2|| \leq \varepsilon$. 
