Stone-Weierstrass theorem (complex version)
, i.e. separates points
, i.e. contains all constant functions
If then , i.e. is a self-adjoint (http://planetmath.org/InvolutaryRing) subalgebra of
Then is dense in .
Proof: The proof follows easily from the real version of this theorem (see the parent entry (http://planetmath.org/StoneWeierstrassTheorem)).
Let be the set of the real parts of elements , i.e.
It is clear that contains (it is in fact equal) to the set of the imaginary parts of elements of . This can be seen just by multiplying any function by .
We can see that . In fact, and by condition 3 this element belongs to .
Moreover, is a subalgebra of . In fact, since is an algebra, the product of two elements , of gives an element of . But since is a real valued function, it must belong to . The same can be said about sums and products by real scalars.
Let us now see that separates points. Since separates points, for every in there is a function such that . But this implies that or , hence there is a function in that separates and .
Of course, contains the constant function .
Hence, we can apply the real version of the Stone-Weierstrass theorem to conclude that every real valued function in can be uniformly approximated by elements of .
Let us now see that is dense in . Let . By the previous observation, both and are the uniform limits of sequences and in . Hence,
Of course, the sequence is in . Hence, is dense in .
|Title||Stone-Weierstrass theorem (complex version)|
|Date of creation||2013-03-22 18:02:31|
|Last modified on||2013-03-22 18:02:31|
|Last modified by||asteroid (17536)|