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Does there exist a one-to-one function $f: \Bbb R \to \Bbb R $ such that $f(x^2) - (f(x))^2 \geq \frac 1 4\ \ \forall x \in \Bbb R$ ?

I've tested this with many one-to-one functions but the inequality does not hold, so I believe that there's no such function. Any ideas on how I can prove this?

My first move was to test this with $x = 0$ and $x = 1$ and got $\frac {-1}2 \leq f(0), f(1) \leq \frac 1 2$. Still, I don't know how to carry on.

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    $\begingroup$ Is there an additional assumption on $f$, e.g. is it continuous? $\endgroup$ – Asaf Karagila Nov 21 '14 at 8:58
  • $\begingroup$ @AsafKaragila Maybe the "one-to-one" assumption is enough to avoid pathological cases (that we would usually avoid using continuity). $\endgroup$ – Surb Nov 21 '14 at 9:04
  • $\begingroup$ No, the question does not mention anything about continuity. $\endgroup$ – Cranky Nov 21 '14 at 9:04
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    $\begingroup$ @Surb: Maybe. I don't know. $\endgroup$ – Asaf Karagila Nov 21 '14 at 9:05
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    $\begingroup$ I think $f(0)=f(1)=1/2$ $\endgroup$ – Empy2 Nov 21 '14 at 9:07
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For $x$ either $0$ or $1$, if we write $y = f(x)$, we get

$$ y - y^2 \geq \tfrac{1}{4}. $$

After completing the square, we find that

$$ 0 \geq (y - \tfrac{1}{2})^2, $$

so we must have $f(0) = f(1) = \tfrac{1}{2}$, a contradiction to the injectivity.


Here is a slightly different, yet trivial observation: Assume that $0 < a < 1 < b$ and $f : (a, b) \to \Bbb{R}$ satisfies

$$ f(x) \geq f(\sqrt{x})^{2} + \tfrac{1}{4} \quad \text{for } a < x < b. \tag{1} $$

(This makes sense since $\sqrt{x} \in (a, b)$ whenever $x \in (a, b)$.) First observation is that $f \geq \tfrac{1}{4}$ on $(a, b)$. In particular, $f$ is always positive.

Moreover, if we denote $g(t) = t^{2} + \frac{1}{4}$, then $g$ is increasing on $[0, \infty)$. So by iterating $(1)$ we have

$$ f(x) \geq g(f(x^{1/2})) \geq g^{2}(f(x^{1/2^{2}})) \geq \cdots \geq g^{n}(f(x^{1/2^{n}})) \geq g^{n}(0). $$

It is easy to check that $g^{n}(0) \to \frac{1}{2}$ as $n \to \infty$, so we have

$$ f(x) \geq \tfrac{1}{2}. $$

In particular, $f$ attains global minimum at $x = 1$. A consequence of this observation is that $f$ cannot be both injective and continuous.

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  • $\begingroup$ Thank you. I can't believe I solved that simple inequation wrongly. $\endgroup$ – Cranky Nov 21 '14 at 9:14
  • $\begingroup$ Would this still not be injective if we have the domain be restricted (i.e. not contain 0 or 1)? $\endgroup$ – Ryan Nov 21 '14 at 15:50

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