Landau asymptotics for a product

Suppose

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

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!

• 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 $\mu$? – Ian Nov 8 at 18:30
• Sorry, I forgot to mention extra information which I now added. – Salamo Nov 8 at 18:56
• Still exactly the same issue as before. – Ian Nov 8 at 19:40
• You are right, I now added the whole context. Sorry for the chaos. – Salamo Nov 8 at 19:40
• 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. – Ian Nov 8 at 19:54