• $f(v)=-dF(u)/du$
  • $f(1)=f(-1)=0$
  • $F(-1)=F(1)$
  • $g(v):=2(F(v)-F(-1))=-2\int_{-1}^v f(u)\, du$,
  • $\mu^2:=-f'(-1)=-f'(1)$

In particular, $g'(v)=-2f(v)$ and $g(1)=0$.

Now, it is claimed that by the definition of $g$ it is "quite standard" to get the estimate $$ f(v)g^{-1/2}(v)=\mu(1+O(1-v)), $$

And, to be honest, I have literally no idea how to see this!

  • $\begingroup$ That's not true as is, the leading order estimate of $g^{-1/2}$ will have a $\mu^{-1/2}$ in it that will spoil that result. Did you maybe want the expansion of $g$ to not involve $\mu$? $\endgroup$ – Ian Nov 8 '18 at 18:30
  • $\begingroup$ Sorry, I forgot to mention extra information which I now added. $\endgroup$ – Salamo Nov 8 '18 at 18:56
  • $\begingroup$ Still exactly the same issue as before. $\endgroup$ – Ian Nov 8 '18 at 19:40
  • $\begingroup$ You are right, I now added the whole context. Sorry for the chaos. $\endgroup$ – Salamo Nov 8 '18 at 19:40
  • $\begingroup$ It seems to be assuming $f'(1)$ and hence $g''(1)$ are nonzero, while $g'(1)=0$. Thus g scales like $(1-v)^2$. Multiply and divide by $(1+v)^{-1}$ to get to $(1-v)^{-1} f (g (1-v)^{-2})^{-1/2}$ and then estimate g. $\endgroup$ – Ian Nov 8 '18 at 19:54

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