Polar decompostion for the operator algebras I find that most of books discussing the polar decompostion at the W*-algebras, but not C*-algebras. I guess the rough reason is that the element of W*algebras has the well supported set, but I want to know more details.
1) What is the basic difference between the C*-algebras and W*-algebras such that the W*-algebras has the well supported set. 
2) If we insist to do the polar decompostion for a noninvertible element in the C*-algebras, what will happen?
 A: I'm not sure what you mean by the "well supported set", but in any case; C*-algebras are not closed under the polar decomposition of non-invertible elements. For this you have to go to the enveloping von Neumann (W*-) algebra.  This is essentially because von Neumann algebras are closed under the Borel functional calculus, whereas C*-algebras are only closed under the continuous functional calculus.  On the other hand, a (unital) C*-algebra is closed under the polar decomposition of its invertible elements.  This is easy to see using the continuous functional calculus.
For example, consider the identity map $f : [-1,1]\to [-1,1]$ as a member of the C*-algebra $C[-1,1]$.  It's easy to see that if $f = |f| v$ is the polar decomposition of $f$ (i.e. $v$ is a partial isometry), then $v$ must take the values $-1$ on $[-1,0)$ and $+1$ on $(0,1]$.  You can find such a $v$ in $L^\infty[-1,1]$ but not $C[-1,1]$.
A: This seems to be relevant for 2):
http://www.theta.ro/jot/archive/1987-017-002/1987-017-002-014.pdf
Since, this answer is two short to be submitted, I would just add redundantly as and answer to 1) that a C*-algebra doesn't have sufficiently many projections and that the proof that a von Neumann algebra does have enough is indeed the consequence of the double commutant theorem.
cf. also the algebraic proof of the polar decomposition Detail in the algebraic proof of the polar decomposition
