every orthonormal basis is a parseval frame. but what about the converse in the finite dimensional case?  

Let's say $H$ is a n-dimensional Hilbert space and $a\_1,..,a\_n$ a parseval frame. then, of course, $a\_1,..,a\_n$ is a basis, since every frame spans the whole space (or maybe the closure in the infinite dimensional case).  

But is it also an orthonormal basis then? I mean it satisfies Parsevals identity by definition. Does anybody know how to prove or contradict that?


Greatings  

Roman



When is a Parseval frame an orthonormal basis?
Yes, because a Parseval frame has the property that for any $i$, 
$$a\_i = \sum\_{j=1}^n \langle a\_i, a\_j \rangle a\_j$$
and since the $a\_i$ form a basis, the coefficients of $a\_i$ on both sides must be the same, which implies that all the $a\_i$ are orthogonal to one another, and also that $||a\_i||^2 = 1$.


(If you are starting from the definition $||x||^2 = \sum |\langle x, e\_i \rangle |^2$ for all $x$, it is a standard fact that this is equivalent to $x = \sum \langle x, e\_i\rangle e\_i$ for all $x$.)


