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"If f(x) is continuous and f(a) and f(b) are of opposite signs then there exist at least one or an odd number of roots between a and b." Is it true for polynomial equations only or any continuous function?

user463280
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If $f$ is continuous function, then there exists at least a root between $f(a)$ and $f(b)$. This is true by intermediate value theorem.

However, for a general function, it is possible that the zero set can be uncountable.

For example let

$$f(x) = \begin{cases} x-1 &, x \ge 1 \\ 0 & , x \in (0,1) \\ x+1 & , x \le -1\\ \end{cases}$$

Siong Thye Goh
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At least one: yes. An odd number: This is not so clear. Of course for polynomials this is true, but we have to count roots "with multiplicity". And as far as I can see if $f$ is just continuous and $f(c)=0$ there's no reasonable definition of the multiplicity.

Example. Let $f(x)=|x|(x-1)$, $a=-1$, $b=1$. Then $f$ is continuous, $f(a)$ and $f(b)$ have different signs, but $f$ has exactly $2$ zeroes between $a$ and $b$.

Similarly for $f(x)=|x|^\alpha(x-1)$ for any $\alpha>0$; if we want to say there are an odd number of zeroes between $-1$ and $2$ we have to somehow define the multiplicity of a zero in such a way that for every $\alpha>0$ the function $|x|^\alpha$ has a zero of even order at the origin.

in fact, come to think of it, not only is there no standard definition of the order of a zero of a continuous function, it's easy to see that it's impossible to give such a definition, assuming two natural conditions:

Define $\Bbb N = \{0,1,2\dots\}$.

Prop There does not exist a function $O:C(\Bbb R)\to\Bbb N$ such that (i) $O(f)>0$ if $f(0)=0$ and (ii) $O(fg)=O(f)+O(g)$.

Proof. If $O$ is such a function then for every postitive integer $n$ we have $$O(|x|)=nO(|x|^{1/n})\ge n.$$

David C. Ullrich
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  • How at most one? – user463280 Sep 17 '18 at 14:54
  • Typo - I meant at least one. – David C. Ullrich Sep 17 '18 at 14:55
  • Thanks.Have u any idea about the situation when f(a) and f(b) are of same signs? – user463280 Sep 17 '18 at 15:00
  • Obviously in that case there need not be any zero at all. Consider $f=1$, for example.. – David C. Ullrich Sep 17 '18 at 15:10
  • I don't see why there is any need to count multiplicities here, the number of different $x_i$ such that $f(x_i) = 0$ in that interval is always odd (as long as it is a finite number, of course). – Dirk Sep 18 '18 at 11:35
  • @DirkLiebhold No, we need to count multiplicity to ensure the number is odd. Let $f(x) = x^2(x-1)$, $a=-1$, $b=2$. Then $f(a)<0$, $f(b)>0$, but the number of zeroes without counting multiplicity between $a$ and $b$ is $2$. – David C. Ullrich Sep 18 '18 at 12:48
  • @DavidC.Ullrich Yes, you are right of course, thanks. I think I forgot the case of the graph just touching the $x$ axis without crossing, my bad, sorry and thanks for pointing that out. – Dirk Sep 19 '18 at 11:32