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?

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 $C[-1,1]$.
