Infimum of an even degree polynomial

Question

I need to prove that every even degree polynomial function such that its higher-degree coefficient is positive reaches its infimum.I have it more or less proved, but there are some things that worry me. Let $p(x)=a_nx^n+...+a_1x+a_0$, being $n$ even and $a_n>0$:

1. As $\lim_{x\to-\infty} p'(x)=-\infty$ and $\lim_{x\to\infty}p'(x)=\infty$ and p'(x) is continuous; can I just say $p(x)$ has at least a relative minimum or do I have to prove something more?

2. How can I prove the lowest relative minimum of $p(x)$ is necessarily an absolute minimum? It seems obvious but I can´t think of a formal proof. I know both limits of p(x) in infinity are $\infty$, but I'm not able to apply this information to this concrete purpose.

Thank you.

Answer 1

1. I don't see why you want to prove the existence of a relative minimum. What you're after is an absolute minimum.
2. Since $\lim_{x\to\pm\infty}p(x)=+\infty$, there's a $M>0$ such that $|x|>M\implies p(x)>p(0)$. And since the interval $[-M,M]$ is closed and bounded, there's a $x_0\in[-M,M]$ such that$$\bigl(\forall x\in[M,M]\bigr):p(x_0)\leqslant p(x).$$So, $p$ reaches its absolute minimum at $x_0$, since outside $[-M,M]$ you have $p(x)\geqslant p(0)\geqslant p(x_0)$.

Answer 2

Step 1. Show that $$ \lim_{x\to\infty} p(x)=\lim_{x\to-\infty} p(x)=\infty. $$

Step 2. Hence, there exists an $M>0$, such that $$ |x|>M\quad\Longrightarrow\quad p(x)>p(0). $$ Step 3. Thus $$ \inf_{x\in\mathbb R}p(x)=\inf_{x\in[-M,m]}p(x) $$

Step 4. But $$ \inf_{x\in[-M,m]}p(x)=\min_{x\in [-M,M]}p(x), $$ since continuous functions in $[-M,M]$ assume their extrema.

Answer 3

You only need the second information, namely that $p(x)\rightarrow \infty$, when $\vert x \vert \rightarrow \infty$. Take $C>0$ such that $K=\{x\in \mathbb{R}\vert p(x) \le C\}$ is nonempty. $K$ is closed and it is bounded (because otherwise you would have a sequence $\vert x_n \vert \rightarrow \infty$, with $p(x_n)$ bounded by $C$, a contradiction), hence $K$ is compact and $p$, being continuous, achieves its minimum at some $x_0 \in K$. This automatically is a global minimum, as for any $x \notin K$ we have $p(x_0) \le C < p(x).$

Answer 4

.All you need to prove is that any such polynomial reaches its infimum. That is, there is a point $a \in \Bbb R$ such that $p(a) = \displaystyle\inf_{x \in \Bbb R} p(x)$.

First, we must prove that the infimum exists. Then, we must show that at some point $a$ the infimum is attained.

Now, what we do know is that $\lim_{x \to \infty} p(x) = \infty$ and $\lim_{x \to -\infty} p(x) = \infty$. In particular, there exists $R > 0$ such that $|x| > R \implies p(x) > 5000$, by the definition of the limit being infinite in both directions.

Next, note that $p(x)$ is continuous on $[-R,R]$, so $p(x)$ attains a minimum on $[-R,R]$, call it $m$. Note that $p(x) \geq \min\{5000,m\}$ for all $x \in \Bbb R$. In particular, the set $\{p(x) : x \in \mathbb R\}$ has a lower bound. By the greatest lower bound property, $\inf_{x \in \mathbb R} p(x)$ exists. Let's call it $L$.

By the limit property, there exists $R' > 0$ such that if $|x| > R'$ then $|p(x)| > L + 5000$. Let us focus on $p$ in the interval $[-R',R']$ now.

We claim that $\inf_{x \in [-R',R']} p(x) = L$. To see this, note that $L$ is the infimum of $p(x)$ over a larger interval, so $\geq$ follows. For the other direction, if $5000 > \epsilon > 0$, then there exists some $x \in \Bbb R$ such that $|f(x)| < L + \epsilon$. But then, from how $R'$ is defined, we know that such an $x$ can only lie in $[-R',R']$. So there is $x \in [-R',R']$ such that $f(x) > L + \epsilon$. Now, the inequality follows.

Finally, of course on compact sets ,infimums of continuous functions are attained, so there exists $a \in [-R',R']$ so that $p(a) = L = \inf_{x \in \Bbb R} p(x)$, completing the proof.

Key points of the proof:

• $\inf_{x \in \Bbb R} p(x)$ exists because we have $p(x)$ increasing to infinity uniformly on both sides, so we can cut $p(x)$ off after some time to get a compact set on which everything works out.

• Once the infimum is there, we cut off $p(x)$ at a point after which we know that the infimum won't surely be attained (hence distinguishing it from the first cutoff) and then use compactness.