Why is it important to know if a frame is a Parseval frame? I understand that a Parseval frame is one in which both upper and lower frame bounds equal 1. What's the main advantage to having this be the case? Or, more specifically, if I'm constructing a frame for use in signal or image analysis, let's say, then why would I care if it was Parseval or not?
The best answer I can think of is that it's useful to have the canonical dual frame, so you can more easily compute the (least-squares) coordinate representation of vectors wrt the frame. Finding the canonical dual frame might be computationally difficult, but if your frame is Parseval then the frame is self dual, so you automatically have the dual frame.
Related to this is that the reconstruction operator is a coisometry when the frame is Parseval.
Is that it, or is there something I'm overlooking?
Thanks!
 A: You're right that a Parseval frame is its own canonical dual, and this reduces the computational requirements for reconstructing a signal represented against the frame.
Additionally, if $A \le B$ are your optimal frame bounds, the condition number for the frame is $\sqrt{\frac BA}$. Larger condition numbers correspond to larger round off errors and iterative algorithms are less stable.  If your frame is Parseval then $A=B=1$ and the condition number is $1$, the best possible.
It may seem at first glance that, since $\Psi \Phi^* = I$ for dual frame pairs, the condition number is $1$ for first representing your data against any frame and then using its canonical dual to reconstruct your data but this is not the case.
Any dual frame will have optimal frame bounds $\frac 1B \le \frac 1A$ and condition number $\sqrt{\frac{1/A}{1/B}}=\sqrt{\frac BA}$, so that the condition number for first representing your data against one frame then reconstructing with a dual is $\frac BA$.
If your frame is not Parseval but tight, $A = B \ne 1$, then your condition number is still $1$ and your canonical dual is just $\frac{1}{B^2}$ times your frame.

Nota bene: Any values $0 < A \le B < \infty$ such that $$A \le \| \Phi^*x \|^2 \le B, \quad \forall \|x\| =  1$$ are frame bounds for the frame $\Phi$.  But just having any two frame bounds doesn't tell you the condition number, for that you need the optimum ones.  If you happen to have a lower bound equal to an upper bound (as you state in your question), then you know they must be the optimal bounds.
