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Let f, g be functions$: \mathbb{R} \mapsto \mathbb{R}$ with $f(x)=x^2+ax+b$ with $a,b \in \mathbb{R}$, such that $(f\circ g)=(g\circ f)$.

If the equation $g(x)=x$ has precisely one solution for all $x \in \mathbb{R}$, prove that $(a-1)^2=4b$.

epimorphic
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3 Answers3

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From $f \circ g = g \circ f$ it follows that:

$$ g^2(x) + ag(x) + b = g(x^2 + ax+b)$$

Now we plug into the above equation $x^*$ which is the unique solution of $g(x^*)=x^*$ and we get:

$$g^2(x^*) + ag(x^*) + b = g(x^{*2}+ ax^*+b)$$

and then:

$$ x^{*2} + ax^* + b = g(x^{*2} + ax^* + b) $$

But then $\tilde{x} = x^{*2} + ax^* + b$ also solves the above equation, i.e. $g(\tilde{x})=\tilde{x}$. By uniqueness of the solution we get $\tilde{x} = x^*$. Plugging in again into the definition of $\tilde{x}$ we see that:

$$ x^{*2} + (a-1)x^* + b = 0 $$

Thus the quadratic equation $x^2+(a-1)x+b=0$ has at least one solution, namely $x^*$.

EDIT: Before I claimed that in fact it only has one solution. But as Karl Kronenfeld pointed out in his answer, this is false (in fact the original question is false).

Nevertheless the above steps are still correct and because the quadratic has at least one solution, it follows that $(a-1)^2 \geq 4b$ (but equality does not necessarily hold).

air
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    Yes, $f(x)=x$ may have multiple solutions--just have $g$ map all of them to $x^{\ast}$. – Karl Kroningfeld Jul 08 '15 at 22:11
  • Gah, I was just taking a walk when I realized that I had screwed up and wanted to build a counterexample , but you beat me to it. What is the etiquette in such cases, do I delete my answer? – air Jul 08 '15 at 22:12
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    I don't think you can delete accepted answers. An edit explaining the error is typical, though I am no expert on MSE etiquette. – Karl Kroningfeld Jul 08 '15 at 22:14
  • Thank you, I updated my post accordingly. – air Jul 08 '15 at 22:19
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    Yes, +1 for the statement and proof of the (correct) inequality. – Karl Kroningfeld Jul 08 '15 at 22:20
  • Well, you need to have precisely one solution in the quadratic equation, or else there could be two solutions which contradicts the hypothesis of the original question, that $g(x)=x$ has strictly one solution for all $x \in R$. –  Jul 09 '15 at 18:51
  • No you don't, look at Karl Kronenfeld's counterexample. – air Jul 09 '15 at 18:52
  • OK, I accept the counterexample. But what is the error in considering that the quadratic has only one solution? Why can it have two (since the solution has to be unique)? –  Jul 09 '15 at 18:59
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    The key point is that the solution to $g(x)=x$ is unique, while the quadratic equation is actually $f(x)=x$. My arguments above show that from $g(x^)=x^$ it follows that $f(x^)=x^$. The problem is that this implication does not go into the other direction: So assume there exists another $x^{} \neq x^*$ with $f(x^{})=x^{}$. It then does not follow that $g(x^{}) = x^{}$. What actually follows (can be shown) is simply that $g(x^{}) = x^{}$ OR $g(x^{}) = x^{*}$. So for a counterexample you just need $f$ with 2 solutions to $f(x)=x$ and $g$ which maps both to the first solution. – air Jul 09 '15 at 19:13
  • I would like to add the following to the question and would like to be given a counterexample with the following taken into consideration:

    $a\neq 0 \text{ and } b\neq 0 \text{ and } g(x)\neq c, c\in \mathbb{R} \text{ and } g(x)\neq x^n, n\in \mathbb{Z}^*$

     –  Jul 13 '15 at 15:15
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This is false. Define $f$ and $g$ by $$f(x)=x^2+ax \qquad\text{and} \qquad g(x)=0.$$

We have, for all $x$, $f(g(x))=f(0)=0$ and $g(f(x))=0$. Thus, the condition $f\circ g=g\circ f$ holds. Moreover, $g(x)=x$ if and only if $x=0$. So, $g(x)=x$ has a unique solution.

Yet, the coefficient $a$ is arbitrary, so $(a-1)^2$ can certainly take nonzero values, contradicting the claim in the OP.

Karl Kroningfeld
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  • @air I would like to add the following to the question and would like to be given a counterexample with the following taken into consideration:

    $a\neq 0 \text{ and } b\neq 0 \text{ and } g(x)\neq c, c\in \mathbb{R} \text{ and } g(x)\neq x^n, n\in \mathbb{Z}^*$

     –  Jul 13 '15 at 14:58
  • @Jason The a,b nonzero condition should have a negligible effect. I suspect a counterexample still exists, even given those conditions on $g$. – Karl Kroningfeld Jul 13 '15 at 18:53
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From the equallity, we have

$g^2(x) + ag(x) + b = g(x^2 + ax + b)$

If we have a solution $c$ for $g(x) = x$, so:

$c^2 + ac + b = g(c^2 + ac + b)$

But from the unicity of the solution of $g(x) = x$, we have

$c^2 + ac + b = c \implies c^2 + (a-1)c + b = 0$

If $\Delta < 0$, then there no exist such $c$. If $\Delta > 0$, then we would have other solution different from $c$ to $g(x) = x$. So, $\Delta = (a-1)^2 - 4b = 0$

João Dos Reis
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