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Show that if $p(x)$ is a polynomial, $|p(x)|$ attains its minimum.

Attempt

Let $p(x) = a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$. Then if $|p(x)| = |a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0|$. If $x > 0,$ then $|p(x)| = |a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0| \leq |a_n|x^n+|a_{n-1}|x^{n-1}+\cdots+|a_1|x+|a_0|.$ Then we can take there derivative of $|p(x)|$ to get a bound to find critical points. I am not sure how to deal with the case of $x \leq 0$.

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  • $\begingroup$ First of all, $|p(x)|$ is not (usually) differentiable; secondly, just because $|a_n| x^n + ... + |a_0|$ attains a minimum doesn't imply that $p$ does. $\endgroup$
    – user296602
    Feb 22, 2016 at 2:09
  • $\begingroup$ @T.Bongers I didn't say it was differentiable and since $|a_n|x^n+\cdots+|a_0|$ attains a minimum implies $p$ does. I said the word bound. $\endgroup$
    – Puzzled417
    Feb 22, 2016 at 2:10
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    $\begingroup$ Show that $|p(x)|$ is continuous and that $\lim_{x\to \pm \infty} |p(x)| = \infty$ and deduce that the minimum is obtained in $[-R,R]$ for some $R$. Then use some known property of continuous functions and extemal points to finish it off. $\endgroup$
    – Winther
    Feb 22, 2016 at 2:11
  • $\begingroup$ @Puzzled417 You said "then we can take the derivative of $|p(x)|$," which is what I'm commenting about. And giving an upper bound on $|p|$ that attains a minimum is not sufficient. $\endgroup$
    – user296602
    Feb 22, 2016 at 2:13
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    $\begingroup$ See this math.stackexchange.com/questions/292714/… $\endgroup$
    – fleablood
    Feb 22, 2016 at 6:54

3 Answers 3

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observe that if the lead coefficient is $c_n x^n$ then for $|x|$ large we have $$ |p(x)| \geq |c_n/2| |x|^{n}. $$ So for some large $R$ the value on $|x|>R$ is greater than $|p(0)|.$

This means that the minimum can only be attained and approached on $|x| \leq R.$ So it's enough to work on the set $|x|\leq R.$ However, a continuous function on a compact set always attains its minimum, we are done.

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  • $\begingroup$ Why is $|p(x)| \geq (c_n/2) |x|^{n}$ true for large $x$? $\endgroup$
    – Puzzled417
    Feb 22, 2016 at 3:11
  • $\begingroup$ @Puzzled417 Write $$|p(x)| = |c_nx^n + c_{n-1}x^{n-1} \ldots + c_0| = |c_n x^n|\cdot\left|1 + \frac{c_{n-1}}{c_n x} + \ldots + \frac{c_0}{c_n x^n}\right|$$ When $|x|\to\infty$ the last factor above goes to $1$. Pick $R$ so large that each of the terms $\frac{c_{n-i}}{c_n x^i}$ is less than $\frac{1}{2n}$ in absolute value when $x = R$ then we have $|p(x)| > |c_n x^n|/2$ for all $|x|>R$. $\endgroup$
    – Winther
    Feb 22, 2016 at 3:40
  • $\begingroup$ @Winther What is the point of saying So for some large $R$ the value on $|x| > R$ is greater than $|p(0)|$ and why is it true? $\endgroup$ Feb 22, 2016 at 3:49
  • $\begingroup$ @user19405892 The point is that since $|p(x)| > |p(0)|$ for $|x|>R$ the minimum cannot be attained for $|x|>R$. You can pick for example $1$ instead of $0$ if you want. It's true since the inequality Mark has given here shows that $\lim_{x\to\pm \infty} |p(x)| = \infty$. $\endgroup$
    – Winther
    Feb 22, 2016 at 3:59
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Hint: Consider $q(x)=\mid p(x)\mid^2$ it is a polynomial of even degree, its derivative has an odd degree. This implies $lim_{x\rightarrow -\infty}q'(x)=-\infty$ $lim_{x\rightarrow +\infty}q'(x)=+\infty$. So there exists $a<0, b>0, a<b$ such that $q'(x)<0, x<a$ and $q'(x)>0, x>b$. Thus $p^2(x)$ is a decreasing function on $(-\infty,a]$ and an increasing function on $[b,+\infty)$. The minimum of the restriction of $p^2$ to $[a,b]$ exists and its square root is the minimum of $\mid p(x)\mid$.

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  • $\begingroup$ there is no need for $a<0<b$. $\endgroup$
    – user251257
    Feb 22, 2016 at 3:32
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Does $|p(x)|$ always attain $p(x)$'s minimum?

Try using a $p(x)$ such that $a_0<0$ and $a_0$ happens to be its minimum [example: $p(x)=x^2-1$]. Then $|p(x)|$ will never attain $a_0$, but $|a_0|$.

Maybe I'm misunderstanding the question. Try to word it better.

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  • $\begingroup$ It says: "show that $|p(x)|$ obtains it's minimum" and not that $|p(x)|$'s minimum is the same as that of $p(x)$. What this mean is that we need to show that there exist a $x_*\in\mathbb{R}$ such that $|p(x_*)|$ is the minimum value of $|p(x)|$. $\endgroup$
    – Winther
    Feb 22, 2016 at 2:45
  • $\begingroup$ @Winther How can't a function obtain its own minimum? Isn't that the definition of a minimum is the smallest value the function reaches? $\endgroup$
    – Puzzled417
    Feb 22, 2016 at 3:00
  • $\begingroup$ @Puzzled417 It's your question so you should clarify if I'm mistaken here (and I will delete my comments), but the standard interpretation of the phrase "$|p(x)|$ attains it's minimum" is what I explained above: there exist a value $x_*\in\mathbb{R}$ such that $|p(x_*)|$ is the minimum value of $|p(x)|$. $\endgroup$
    – Winther
    Feb 22, 2016 at 3:05
  • $\begingroup$ @Winther Oh I am sorry, I see what you mean. Of course if the function were not bounded it wouldn't so we need only check that. $\endgroup$
    – Puzzled417
    Feb 22, 2016 at 3:06
  • $\begingroup$ @Puzzled417 And also if we are not talking about polynomials then it does not need to be true. For example $f(x) = e^{x}$ does not attain it's minimum value $0$ for any $x\in\mathbb{R}$. The key property that makes it true for polynomials is continuity + the fact that $|p(x)| \to \infty$ as $|x|\to \infty$. $\endgroup$
    – Winther
    Feb 22, 2016 at 3:08

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