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May 11, 2022 at 21:08 history edited YCor CC BY-SA 4.0
formatting, added tag
May 11, 2022 at 7:16 history edited MathArt CC BY-SA 4.0
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May 8, 2022 at 12:21 answer added Christophe Leuridan timeline score: 0
May 7, 2022 at 18:10 vote accept MathArt
May 7, 2022 at 16:55 answer added Christian Remling timeline score: 3
May 7, 2022 at 15:52 comment added Andrea Marino 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 $D_s \frac{ K(f(s)) }{K(s) }\ge 0$.
May 7, 2022 at 15:08 comment added username @AnthonyQuas not concave increasing..
May 7, 2022 at 14:43 comment added Anthony Quas 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?
May 7, 2022 at 14:43 comment added Andrea Marino 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.
May 7, 2022 at 14:18 comment added Andrea Marino 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?
May 7, 2022 at 13:53 history edited Martin Sleziak CC BY-SA 4.0
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May 7, 2022 at 12:33 history asked MathArt CC BY-SA 4.0