Construct function with special decay

Question

Let $r,R>0$ such that $R>2r$. I want to construct a twice differentiable function $f=f_{r,R}$ so that \begin{align} f(x)= \begin{cases} 0, 0<|x|<r\\ 1, 2r<|x|<R\\ 0, |x|>2R \end{cases} \end{align} and \begin{align} |f'(r)|\leq \begin{cases} \frac{1}{r} \text{ if } r<|x|<2r\\ \frac{1}{R} \text{ if } |x|\geq 2r \end{cases} \end{align} and \begin{align} |f''(r)|\leq \begin{cases} \frac{1}{r^2} \text{ if } r<|x|<2r\\ \frac{1}{R^2} \text{ if } |x|\geq 2r. \end{cases} \end{align} On $(2r,R)$ the derivatives obviously fulfil the estimates. Now, e.g. on $(R,2R)$ I thought I might glue some $\log$ function, so take for example $f(x)=1$ on $(R,R+\varepsilon)$ and $f(x)=\log\left((2R_2-x)\frac{e}{R_2}\right)$ on $(R+\varepsilon,2R)$ for $\varepsilon >0$ small. At least all conditions in $R_2$ are fulfilled. Now the function should be zero in $(2R-\varepsilon, 2R)$. Any suggestions how I can modify the $\log$ part? The part $(r,2r)$ should then be similar.

Answer 1

If such a function $f$ existed on the reals, that would mean $f(R) = 1$ and $f(2R) = 0$. Since $f$ is twice differentiable, the first derivative of $f$ is continuous. We also know by continuity of the first derivative that $f'(R) = 0$ because $f$ is constant on $(2r,R)$.

Consider the first order Taylor expansion $P_1(x)$ of $f$ around $x = R$: we see that $$P_1(x) = 1 + 0(x-R) \implies P_1(2R) = 1.$$

Taylor's theorem (Lagrange form) for the remainder (https://en.wikipedia.org/wiki/Taylor%27s_theorem) applied on the interval $[R,2R]$ says that for $c = \sup_{x \in [R,2R]}|f''(x)|,$ $$|P_1(2R) - f(2R)| \leq \frac{c(2R-R)^2}{2!} \leq \frac{1}{R^2}R^2\frac{1}{2} = \frac{1}{2}.$$

This is bad since this means $f(2R) \geq \frac{1}{2}$. There are either some continuity assumptions that are missing, or you need to let your function decay over a greater distance.