# Polynomial function: infimum

1. Nov 18, 2012

### Felafel

1. The problem statement, all variables and given/known data

Given the function "P" defined by: P(x) := x^2n + a2n-1*x^2n-1 + ... + a1x + a0;
prove that there exists an x* in |R such that P(x*) = inf{P(x) : x belongs to | R}
Also, prove that:
|P(x*)| = inf{|P(x)| : x belongs to |R}

3. The attempt at a solution

As the function is the sum of continuous functions, it is contnuos too.
Then, I thought about the theorem according to which if we have a cont. function on a sequentially compact space, it has inf. and sup. therein.
But the space here is not sequentially compact.
Can I use this theorem all the same, by adding some restrictions, perhaps?

thanksss
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Nov 18, 2012

### tt2348

I'd say seperate your polynomial into it's even and odd powers.
IE P2 is all even powered x (including a0) and P1 is your odd, with some type of uniform bound on the coefficients. like |ak|<M
is this an analysis class? I assume so. we know P2>P1 at large x, and extremely small -x ( like -10^10 or whatever). your function goes to infinity on both sides. Since it is Continuous, think of how you could apply rolles theorem, squeeze theorem, IVT, and the fact that INF(aN+bN)>=Inf(aN)+Inf(bN)

3. Nov 24, 2012

### Felafel

thank you! i seem to have solved out the first question.
but how about the second part:
|P(x*)| = inf{|P(x)| : x belongs to |R}?
if I think about a parabola graphic with its vertex in, say, (0, -3), the vertex of the absolute value of the function (0, +3), is no more the infimum.
it would be possible if the infimum of this function were in the first or fourth quadrant, but i can't assume it, right?

4. Nov 24, 2012

### tt2348

I think assuming that the inf|P(x)|=/=inf(P(x)) in general is correct. unless there is a strict restrictions of P(x). you'll have two cases of x* that p(x*)=0 implies x* is in inf{|p(x)|}
or the inf is the same.