This question mostly eluded me during the exam itself.

Problem: Suppose that $f: [-1, 1] \rightarrow \mathbb{R}$ continuously, and that

$$\begin{align} \text{(i)}\qquad &f(x) = \frac{2 - x^2}{2} f\left(\frac{x^2}{2-x^2}\right)\\ \text{(ii)}\qquad &f(0) = 1\\[6pt] \text{(iii)}\qquad &\lim_{x\to 1^-}\frac{f(x)}{\sqrt{1-x}}\ \text{exists} \end{align}$$

Determine the closed form of $f$, and prove that it is unique. ---

All that I managed to do was - using the eventual heuristic of "powers of $1 - x^2$ will cancel neatly in (iii) and work in (ii), so let's see if they work in (i)" - discover that

$$ f(x) = \sqrt{1 - x^2}$$

Now, this $f$ is an involution, so it suffices to prove that if $g$ is a function satisfying (i) to (iii), then $g(f(x)) = x$. This, however, was more than I could do.

How do you prove uniqueness? (I ran across one solution I did not understand; please, be gentle.)

  • 3
    $\begingroup$ Have you seen the discussion at artofproblemsolving.com/Forum/… ? $\endgroup$ – Gerry Myerson Dec 4 '12 at 4:56
  • $\begingroup$ @Gerry I had not. echinodermata's answer there is really clever. Thanks for the recommendation! $\endgroup$ – Chris Dec 4 '12 at 5:16
  • $\begingroup$ @user1296727: perhaps you can post a solution now, to keep this question from going unanswered? $\endgroup$ – user641 Mar 21 '13 at 23:05

Disclaimer: I am borrowing much from the Art of Problem Solving link alluded to in the comments.

Since you have guessed a solution, namely $x \mapsto \sqrt{1-x^2}$, so let us exploit it. Denote $s(x) := \sqrt{1-x^2}$, and note, as you did, that $s(s(x)) = x$. It is a good idea to express $f$ in an alternative way so that for $f(x) = s(x)$ things become very simple. This is vague, but a way to proceed is to write $f(x) = h(s(x))$ for some continuous $h:[0,1] \to \mathbb{R}$; since this is equivalent to saying that $h(y) = f(s(y))$, there is a 1 to 1 correspondence between possible $f$'s and $h$'s.

Let us see what the conditions say about $h$:

(i) $$h(y) = f(s(y)) = \frac{1 + y^2}{2} f\left(\frac{1-y^2}{1+y^2}\right) = \frac{1 + y^2}{2}h\left(s\left(\frac{1-y^2}{1+y^2}\right)\right) = \frac{1 + y^2}{2} h\left( \frac{2y}{1+y^2} \right)$$ (ii) $h(1) = 1$

(iii) $\lim_{y \to 0+} \frac{h(y)}{y}$ exists.

There are several ways to proceed at this stage. To keep things elegant, let us notice the striking resemblance between the expression $\frac{2y}{1+y^2}$ and the formula for $\tanh$ of doubled angle. In fact, if $y = \tanh \alpha$, then $\frac{2y}{1+y^2} = \tanh 2 \alpha$. Writing additionally $\frac{1+y^2}{2}$ as $\frac{y}{ \frac{2y}{1+y^2}} $ we conclude that (i) can be re-expressed in a nicer form: $$ h(\tanh \alpha) = \frac{\tanh \alpha}{\tanh 2 \alpha} h(\tanh 2\alpha)$$ Iterating this as many times as we like, we conclude: $$ h(\tanh \alpha) = \frac{\tanh \alpha}{\tanh 2 \alpha} \frac{\tanh 2 \alpha}{\tanh 4 \alpha} \dots \frac{\tanh 2^{k-1} \alpha}{\tanh 2^{k} \alpha} h(\tanh 2^k\alpha) = \tanh \alpha \frac{h(\tanh 2^k\alpha)}{\tanh 2^k \alpha}$$ Passing to the limit $k \to \infty$ (and remembering that $\tanh \beta \to 1 $ as $\beta \to \infty$ we conclude that: $$ h(\tanh \alpha) = \tanh \alpha$$ This means that $h(y) = y$. Translating this back to $f$ we conclude that: $$ f(x) = h(s(x)) = s(x) = \sqrt{1-x^2}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.