Can I find a bump function $\psi$ such that $\nabla\log\psi$ vanishes too?

Question

Consider a bump function supported in the ball of radius $1$, that is $\psi:\mathbb R^n\to\mathbb R$ such that

1. $\ \psi(x)>0$ for $|x|<1$

2. $\ \psi(x)=0$ for $|x|\geq 1$

3. $\ \psi\in C^\infty$.

Is it possible to find such a function $\psi$ that satisfies also one of the following conditions? For all $i,j=1,\dots,n$

4. $\ \displaystyle \frac{\partial_{x_i}\psi(x)}{\psi(x)}\to 0\ $, $\ \displaystyle\frac{\partial_{x_i}\partial_{x_j}\psi(x)}{\psi(x)}\to0\ $ as $|x|\to1\ $

5. $\ \displaystyle\lim\frac{\partial_{x_i}\psi(x)}{\psi(x)}\in\mathbb R\ $, $\ \displaystyle\lim\frac{\partial_{x_i}\partial_{x_j}\psi(x)}{\psi(x)}\in\mathbb R\ $ as $|x|\to1\ $.

Intuitively this would mean that both $\psi$ and its derivatives vanish approaching the boundary of the ball, but the derivatives vanish faster than $\psi$ (or at least not slower).

A typical example of function satisfying conditions 1.-3. is given by $$ \psi_0(x) = \begin{cases} e^{-\frac{1}{1-|x|^2}} &\textrm{ if }|x|<1 \\[2pt] 0 &\textrm{ if }|x|\geq1\end{cases} $$ but 4.,5. are clearly not satisfied by $\psi_0$ since $$ \frac{\nabla\psi_0(x)}{\psi_0(x)} \,=\, \frac{2x}{(1-|x|^2)^2}\,\xrightarrow[|x|\to1]{}\,\infty \,.$$

Answer 1

Elaborating the comment by Wojowu: If we take a look at $n=1$ and $\psi\in C^\infty_{\text c}(\mathbb R)$ is a function satisfying conditions 1., 2. and 3. of your question, then for every $x\in]-1,1[$, we have, by smoothness of $\ln\psi$ on $]-1,1[$ and the fact that $\frac{\psi'}{\psi}$ is continuous on every $[-K,K]$ for $K\in]0,1[$,

$$\ln\psi(x)=\ln\psi(0)+\int_0^x\frac{\psi'(s)}{\psi(s)}\,\mathrm ds.$$

Now, if we had $$\lim_{s\to 1}\frac{\psi'(s)}{\psi(s)}=r$$ for any real number $r\in\mathbb R$, then the function $$s\mapsto\frac{\psi'(s)}{\psi(s)}$$ would be well-defined and continuous on $[0,1]$, and therefore we would have $$\lim_{x\to1}\ln\psi(x)=\ln\psi(0)+\int_0^1\frac{\psi'(s)}{\psi(s)}\,\mathrm ds\in\mathbb R,$$ which is impossible since $\lim_{x\to1}\psi(x)=0$.

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Similarly, for a natural number $n>1$, any $\phi\in C_{\text c}^\infty(\mathbb R^n)$ with your properties 1., 2. and 3. defines a function

$$\psi:\mathbb R\to\mathbb R, t\mapsto\phi(t,0,0,\dots, 0)$$ and we can proceed our argumentation as above.