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Can we find a counterexample to the following assertion?

Assume that $f:[-1,1]\to [-1,1]$ an odd function of class $C^3$, and assume thaht $f$ is a concave increasing diffeomorphism of $[0,1]$ onto itself. Then my examples say that $$\frac{1-f(x)^2}{1-x^2}\le f'(x),\; x\in (0,1).$$

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  • $\begingroup$ I am not fresh enough to produce an effective calculation, but what about trying to minimize the functional $f'(x) (1-x^2) -(1-f(x) ^2) $ over the space of concave increasing diffeomorphisms? $\endgroup$
    – Andrea Marino
    Commented May 7, 2022 at 14:18
  • $\begingroup$ This approach shows that a minimiser (if it exists) $f$ satisfies $f(x) = \frac{-u'(x) (1-x^2) -u(x) ^2}{2u(x) }$ in all $x$ such tha $u(x)\neq 0$, for all $u$ such that $f+u \in \textrm{YourSpace}$. Suppose now that by contradiction $f$ is $\neq 0,1$ in some point (so in an interval $(a, b) $). Consider the $u(x) $ that is constantly $\epsilon$ in a subinterval $(a', b') $ and $0$ outside $(a, b) $. You obtain that $f(x) \equiv \pm \epsilon/2 $ for all $\epsilon$ small enough. Thus the only candidates $f$ are the constant func. 0,1 which does not belong to your space. Thus there is no min. $\endgroup$
    – Andrea Marino
    Commented May 7, 2022 at 14:43
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    $\begingroup$ Isn’t your inequality false for the piecewise linear function through $(0,0)$, $(\frac 12,\frac 34)$, $(1,0)$ and hence for anything smooth nearby? $\endgroup$
    – Anthony Quas
    Commented May 7, 2022 at 14:43
  • $\begingroup$ @AnthonyQuas not concave increasing.. $\endgroup$
    – username
    Commented May 7, 2022 at 15:08
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    $\begingroup$ If you write it as $\frac{f'}{1-f^2} - \frac{1}{1-t^2}$, you can rewrite the espression as $D_s \log \frac{ K(f(s)) }{K(s) }\ge 0$, where $K(x) =\frac{1+x}{1-x}$. This is equivalent to $D_s \frac{ K(f(s)) }{K(s) }\ge 0$. $\endgroup$
    – Andrea Marino
    Commented May 7, 2022 at 15:52

2 Answers 2

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A counterexample is provided by any function that equals $f(x)=1+m(x-1)$ near $x=1$, with $0<m<1$. (Maybe this is in fact just restating Anthony's comment, with a typo corrected?)

What is actually true is the trivial observation that (by the mean value theorem) $$ \frac{1-f^2(x)}{1-x^2}=f'(c) \frac{1+f(x)}{1+x} \le f'(x)\frac{1+f(x)}{1+x} , $$ and it seems this is as far as we can go in general. (In my example, we have equality here, and $(1+f)/(1+x)>1$, since $f>x$.)

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  • $\begingroup$ This is the counterexample: $$\begin{array}{ll} f(x)=\{ & \begin{array}{ll} \frac{5 x}{2}-2 x^2 & 0 < x <\frac{1}{2} \\ 1+\frac{1}{2} (-1+x) & \frac{1}{2} \leq x \leq 1 \\ 0 & \text{True} \end{array} \end{array}$$ $\endgroup$
    – MathArt
    Commented May 7, 2022 at 18:10
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I complete the reformulation given by Andrea Marino and give another counterexample.

First, the inequality of the beginning can be written $$\forall x \in (0,1), \quad \frac{f'(x)}{1-f^2(x)} \ge \frac{1}{1-f^2(x)}$$ and means that the function $x \mapsto \arg\tanh(f(x))-\arg\tanh(x)$ is non-decreasing on $(0,1)$, or equivalently (composing with $\tanh$) that the function $$g : x \mapsto \frac{f(x)-x}{1-xf(x)}$$ is non-decreasing on $(0,1)$.

Now, let $$f(x) := \frac{1}{2} [x+1-(1-x)^3].$$ The function $f$ thus defined satisfies the assumptions. Let us compute the corresponding function $g$. \begin{eqnarray*} g(x) &=& \frac{[x+1-(1-x)^3]-2x}{2-x[x+1-(1-x)^3]} \\ &=& \frac{1-x-(1-x)^3}{2-x-x^2+x(1-x)^3} \\ &=& \frac{1-(1-x)^2}{2+x+x(1-x)^2} \\ &=& \frac{2x-x^2}{2+2x-2x^2+x^3} \\ \end{eqnarray*} The quantities $u(x):=2x-x^2$ and $v(x):=2+2x-2x^2+x^3$ are positive on $[0,1]$ (since $x \ge x^2$ on $[0,1]$), and $u'(1)=2-2=0$ whereas $v'(1)=2-4+3=1$. Hence $g$ is decreasing at the neighborhood of $1$.

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