I wanted to prove the following statement: in case of a polynomial of an even degree, there exists some $y\in\mathbb{R}$ for which there is NO $x\in\mathbb{R}$ such that $f(x)=y$.

The proof I initially came up with was as following. I thought of splitting the proof into two cases: with positive and negative leading coefficient. Then, we can can prove that the polynomial with a positive leading coefficient attains a global minimum, and hence our statement follows for every $y<y_{\min}$. Similarly, we can prove that each polynomial with a negative leading coefficient attains a global maximum, and hence the statement follows for every $y>y_{\max}$.

However, is there an alternative (or perhaps a simpler) way to prove this statement?

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    $\begingroup$ I can imagine a proof that any polynomial which yields all $y \in \mathbb{R}$ must be of odd degree, but this is only a slight variation on the approach you've outlined. $\endgroup$ – hardmath Feb 27 '13 at 19:08
  • $\begingroup$ Your argument looks pretty simple to me. I suppose you can say that without loss of generality the leading coefficient is positive as otherwise you could apply the proof to $- f(x)$. $\endgroup$ – JavaMan Feb 27 '13 at 19:14
  • $\begingroup$ @hardmath - well, I think the proof of such a statement would be a bit easier (we could resort to intermediate value theorem), but I don't see how we could move from that to the statement above. If you see how this could be done, it would be great if you could share your thoughts :) $\endgroup$ – Johnny Westerling Feb 27 '13 at 19:17
  • $\begingroup$ @JavaMan - it just seemed to me that such a statement shouldn't require the hassle involved with proving that a function attains an extremum (derivatives, etc) $\endgroup$ – Johnny Westerling Feb 27 '13 at 19:19
  • $\begingroup$ @JohnnyWesterling I think that's overkill showing it attains its extrema, all you really needed is boundedness. $\endgroup$ – muzzlator Feb 27 '13 at 19:22

I will answer this formally since it hasn't been (rigorously) answered yet, in my opinion. This is likely a duplicate, but I couldn't find a duplicate.

If we're considering polynomials of degree $0$, this is trivial, since such polynomials are constant. If the degree is $\geq 2$, then note

$$p(x) \stackrel{\text{def}}{=}\sum_{1 \leq i \leq 2n} a_ix^i = a_{2n}x^{2n}\left(1 + \sum_{1 \leq i \leq 2n-1} \frac{a_i}{a_{2n}}\frac{1}{x^{2n-i}}\right) $$

The sum inside the brackets clearly goes to $0$ as $|x| \to \infty$, and hence $$\lim_{x \to ±\infty} p(x) = \text{sgn}(a_{2n}) \cdot \infty$$

This implies there's an $M>0$ such that $p(x) > 0$ $\text{XOR}$ $p(x) < 0$ for all $x \in \mathbb{R} \setminus [-M,M]$. Hence, were $p$ surjective onto $\mathbb{R}$, we would need one of $(-\infty, 0]$ or $[0, +\infty)$ to be subsets of the image of $[-M, M]$ under $p$.

However, this leads to a contradiction since the continuous image of a (finite) closed interval is a (finite) closed interval.


It is possible to totally avoid the notions of continuity and limits in a proof of the claim (and therefore make the claim applicable for polynomials with coefficients in arbitrary ordered rings):

First assume that $p$ is not constant and that the leading coefficient of $p(X)$ is $1$, i.e., $$p(X) = X^{2n}+\sum_{k=0}^{2n-1}a_kX^k $$ with $n\ge1$. Let $M=2n\max\{\,|a_k|\mid 0\le k<2n\,\}$. Then for $|x|\ge M':= \max\{M,1\}$, we have $$\begin{align}p(x)&\ge x^{2n}-\sum_{k=0}^{2n-1}|a_kx^k|\\&\ge x^{2n}-\sum_{k=0}^{2n-1}\frac M{2n}|x^k|\\&\ge x^{2n}-\sum_{k=0}^{2n-1}\frac 1{2n}|x^{k+1}|\\&\ge x^{2n}-\sum_{k=0}^{2n-1}\frac 1{2n}x^{2n}\\&=0\end{align}$$ On the other hand, for $|x|<M'$, we have $$p(x)\ge -\sum_{k=0}^{2n-1}|a_kx^k|\ge -\sum_{k=0}^{2n-1}|a_k|M'^k.$$ Hence $p$ is bounded from below.

If the leading constant of the non-constant polynomial $p(X)$ is $a(\ne 0)$, then the leading term of $\frac 1ap(X)$ is $1$ and so $\frac 1ap(X)$ is bounded from below. It follows that $p(X)$ is bounded from below if $a>0$ and bounded from above if $a<0$. (And of course the case of constant polynomials is trivial).


Here's an approach (though I haven't rigorously checked its validity), which is a bit more 'complicated' than your approach.

Suppose $f$ is any even polynomial and we assume that it is surjective on $\mathbb{R}$, seeking a (handwavey) contradiction.

By our assumption, $f$ always has a root in $\mathbb{R}$, and in fact it must come in pairs (as a result of the complex conjugate root theorem), hence there must be at least 2 roots.

By looking at roots of $f(x)-y$ we can deduce that for any $y \in \mathbb{R}$, there are at least two values of $x$ for which $f(x)=y$.

Now I believe we can then find a contradiction, though I haven't rigorously validated the following direction:

Without loss of generality, suppose $$\lim_{x \to {\infty^{+}}}f(x) = \lim_{x \to {\infty^{-}}}f(x) = \infty^+$$. (We may assume this as $f$ is an even polynomial)

Then perhaps we could somehow prove that this means there must be some $x=a \in \mathbb{R}$ such that $\lim_{x \to a} f(x) = \infty^-$ because every element in the image is hit at least twice. Hence, if this is true, we have found the contradiction we've been looking for, as the limit to a finite point of a real valued polynomial is finite.

As I've said, I missed many details, not as nearly as simple as your original approach, but I've included it for comparison.


A polynomial is continuous and so bounded on any bounded interval.
$$\lim_{x\rightarrow \pm \infty} f(x) = \sigma \cdot \infty$$

where $\sigma$ is the sign of the leading coefficient and so $f(x)$ is bounded above or below accordingly.


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