For a = (a_0, a_1, ..., a_n) element R^n+1, a_n notequalto 0 let P_a (x) = a_n x

Question

For a = (a_0, a_1, ..., a_n) element R^n+1, a_n notequalto 0 let P_a (x) = a_n x^n + ... + a_1x + a_0. Suppose that for a^0 = (a^0 _0, a^0 _1, ..., a^0 _n), a^0 _n notequalto 0 the polynomial P_a^0 (x) has n distinct real roots. Prove that there exist epsilon > 0 and C^infinity smooth functions lambda_1, ..., lambda_n: B^n+1 (a^0, epsilon) rightarrow R such that for any a element B^n+1(a^0, element), lambda_1, ..., lambda_ n(a) are distinct roots of the polynomial P_a (x). In other words, prove that in a small neighborhood of a^0, roots of the polynomial P_a depend smoothly on the coefficients a_0, a_1, ..., a_n.

Explanation / Answer

Here is a version of continuity of the roots.
Consider the monic complex polynomial f(z)=zn+c1zn1+...+cnC[z]

and factor it as

f(z)=(za1)...(zan)     (akC)

where the roots ak are arranged in some order, and of course needn't be distinct.
Then for every >0, there exists >0 such that every polynomial g(z)= zn+c1zn1+...+cnC[z] satisfying |dkck|< (k=1,...,n) can be written

g(z)=(zb1)...(zbn)    (bkC)

with |bkak|<(k=1,...,n).

A more geometric version is to consider the Viète map v:CnCn

sending, in the notation above, (a1,...,an) to (c1,...,cn) (identified with zn+c1zn1+...+cn=(za1)...(zan)).
It is a polynomial map (and so certainly continuous!) since ck=(1)ksk(a1,...,an), where sk is the k-th symmetric polynomial in n variables.
There is an obvious action of the symmetric group Sn on Cn and the theorem of continuity of the roots states that the Viète map descends to a homeomorphism w:Cn/SnCn. It is trivial (by the definition of quotient topology) that w is a bijective continuous mapping, but continuity of the inverse is the difficult part.
The difficulty is concentrated at those points (c1,...,cn) corresponding to polynomials zn+c1zn1+...+cn having multiple roots.

Here is a version of continuity of the roots.
Consider the monic complex polynomial f(z)=zn+c1zn1+...+cnC[z]

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