Let $v(x)=e^{-|x|^2}-e^{-1}$ for $x\in \mathbb{R}^n$ with $|x|\le 1$. For $x_0$ satisfying $|x_0|=1$ and $x$ with $|x|<1$, I'd like to show
$$\liminf_{x \to x_0} \frac{v(x)}{|x-x_0|}>0,$$ where the angle $x_0-x$ and $x_0$ is less than $\frac{\pi}{2}-\delta$ for some $\delta\in (0,1).$
Since $v(x)>0$ for $x\in \mathbb{R}^n$ with $|x|< 1$, it is clear that $\liminf_{x \to x_0} \frac{v(x)}{|x-x_0|}\ge 0.$ It seems that it is related to directional derivative.
Note that (i) $v$ is radial function.
(ii) $v(x_0)=0,$ and
(iii) for unit vector $w$ which is perpendicular to $x_0$, the directional derivative $D_w(x_0)=0.$
Intuitively, with the above information, it seems to be true that $\liminf_{x \to x_0} \frac{v(x)}{|x-x_0|}>0,$ where the angle $x_0-x$ and $x_0$ is less than $\frac{\pi}{2}-\delta$ for some $\delta\in (0,1),$ but I don'k know how to prove it with the definition of liminf rigorously
Please let me know if you have any idea or comment about my question. Thanks in advance!
Suppose $\{x[k]\}_{k=1}^{\infty}$ is a sequence of points in $\mathbb{R}^n$ such that $$x[k]=x_0+h[k]z[k] \quad \forall k \in \{1, 2, 3, ...\}$$ where $x_0 \in \mathbb{R}^n$ is a unit vector (i.e., $||x_0||=1$), and for each $k \in \{1, 2, 3, ...\}$ we have that $z[k] \in \mathbb{R}^n$ is a unit vector and $h[k]$ is a positive number. Notice that $$h[k] = ||x[k]-x_0|| \quad \forall k \in \{1, 2, 3, ...\}$$ Suppose that $$\lim_{k\rightarrow\infty} h[k]=0$$
Since $x_0$ and $z[k]$ are unit vectors, we have $x_0^Tz[k] = \cos(\theta[k])$ where $\theta[k]$ is the angle between vectors $x_0$ and $z[k]$. Suppose the angle is constrained so that there is an $\epsilon>0$ such that $x_0^Tz[k] \leq -\epsilon$ for all $k \in \{1, 2, 3, ...\}$. So \begin{align*} ||x[k]||^2 &= ||x_0+h[k]z[k]||^2 \\ &= ||x_0||^2 + 2h[k]x_0^Tz[k] + h[k]^2||z[k]||^2 \\ &= 1 + 2h[k](x_0^Tz[k]) + h[k]^2 \\ &\leq 1 -2\epsilon h[k] + h[k]^2 \end{align*} Then $$ \frac{e^{-||x[k]||^2}-e^{-1}}{||x[k]-x_0||} \geq \frac{e^{-(1-2\epsilon h[k]+h[k]^2)} - e^{-1}}{h[k]} $$ And you can take the $\lim\inf$ of both sides as $k\rightarrow\infty$ (the right-hand-side limit exists in the usual $\lim$ sense and indeed converges to a specific positive number).