Let $f: \mathbb{R} \to \mathbb{R}$,
$$f(x)=x^n+a_{n-1}x^{n-1}+\cdots + a_1x+a_0$$
where $n$ is even and $a_i\in \mathbb{R}$. It is then not difficult to show that $f$ is bounded below. This leads to the question:
- Does there exist a closed-form (not necessarily optimal) lower bound for such a polynomial in terms of its coefficients $a_i$?
- Can this closed-form be expressed as a (multi-variate) polynomial in the coefficients $a_i$?
I imagine that the answer to both is yes, but I have no idea how to prove this. I thought a bit about three cases below.
- $n=2$
For monic quadratics, a sufficient lower bound is $a_0 – \frac{a_1^2}{4}$. This satisfies (1), (2) and it is also optimal.
- $n=4$
In this case, we could differentiate to find the stationary points of $f$, and write the minimum of $f$ evaluated at these points in a closed-form (using the absolute value function). This satisfies (1) and is again optimal, but does not satisfy (2).
- $n \ge 6$
No immediate solutions, since polynomials of degree $\ge5$ have no closed-form solutions.
Suppose that $\{r_0,…r_k\}$ is the zero set of $f$, in ascending order. Then $f$ attains its minimum in $[r_0,r_k]$. (If the zero set is empty, then $0$ is a lower bound.) Thus, it may be possible to simplify this problem by using bounds on the locations of the roots of polynomials, if any are known.
Best Answer
If I’m not mistaken, let $S=1+\sum_{i=0}^{n-1}{|a_i|}$. Then $-S^n$ should be such a lower bound.
Indeed, if $|x| \geq S$, then $f(x) \geq S^n-\sum_{i=0}^{n-1}{S^{i}|a_{i}|}\geq S^n-\sum_{i=0}^{n-1}{|a_i|S^{n-1}} \geq S^{n-1} \geq 0$.
But if $|x| \leq S$, then $f(x) \geq -\sum_{i=0}^{n-1}{|a_i|S^i} \geq -S^{n-1}\sum_{i=0}^{n-1}{|a_i|} \geq -S^n$.