Taylor expansion at infinity and optimal value at infinity of a function. Given a function $f(x)$ that we need to minimize on the hold space, i.e., $$\mbox{minimize} \;\; f(x):\;\;\; \mbox{subject to }\; x\in X.$$ 
Suppose this function is bounded, i.e., $|f(x)|\leq \gamma$ for all $x\in X$ for some $\gamma$. I would like to find the optimal value of $f$ at infinity. i.e., among all directions we can choose to tend to infinity, which directions make the value of $f$ is minimal. The value of $f$ when $\|x\|\to +\infty$ in such a direction is called the optimal value at infinity. 
Now consider the function $f(x)=\sum_{i=1}^pw_i \|x-a_i\|$ with $\sum_{i=1}^p w_i=0$. To find the optimal value of $f$ at infinity, I utilize the Taylor expansion at a point $x$ with $\|x\|$ is very large, we call $x$ is at infinity. 
Define $\varphi(x)=\|x\|$. The gradient and Hessian of $\varphi$ at every $x\neq 0$ are given by 
$$\nabla \varphi(x) = \frac{x}{\|x\|} \;\; \mbox{ and }\; \; \nabla^2\varphi(x) = \frac{1}{\|x\|^3}\left(\|x\|^2I - xx^{T}\right).$$
Using three terms of Taylor expansion of $\varphi$ we can approximate 
$$\varphi(x-a)\approx \varphi(x)- \langle \nabla\varphi(x), a\rangle + \frac{1}{2}\langle \nabla^2\varphi(x)a, a\rangle.$$
This means
$$\|x-a\| \approx \|x\| - \langle \frac{x}{\|x\|}, a \rangle + \frac{1}{2} \langle \dfrac{\|x\|^2I - xx^{T}}{\|x\|^3}a, a\rangle $$
$$\; \; \; \approx  \|x\| -\langle \frac{x}{\|x\|}, a \rangle + \frac{1}{2}(\dfrac{\|a\|^2}{\|x\|} - \dfrac{|\langle x, a\rangle|^2}{\|x\|^3})$$
Since $|\langle x, a\rangle|^2 \leq \|x\|^2 \|a\|^2$ by Cauchy-Schwarz inequality, when $\|x\|$ is large, the last two terms can be ignored and we have 
$$\|x-a\|  \approx  \|x\| -\bigg\langle \frac{x}{\|x\|}, a\bigg \rangle.$$
Thus, the value of $f(x)$ for a large $\|x\|$ is approximately: 
$$\sum_{i=1}^p w_i \|x\| - \langle \frac{x}{\|x\|}, \sum_{i=1}^p w_ia_i\rangle $$
Since $\sum_{i=1}^p w_i=0$, the solution at infinity is obtained by maximizing   $\langle \frac{x}{\|x\|}, \sum_{i=1}^p w_ia_i\rangle$, the Cauchy-Schwarz give us
$$\langle \frac{x}{\|x\|}, \sum_{i=1}^p w_ia_i\rangle \leq \|\sum_{i=1}^p w_ia_i\|.$$
The equality holds if $x=t\sum_{i=1}^p w_ia_i, \; t>0$. And I can conclude that the optimal value of $f$ at infinity is $-\|\sum_{i=1}^p w_ia_i\|$ when $x$ tends to infinity along with the direction $v=\sum_{i=1}^p w_ia_i$.
I wonder whether my argument as above is correct or not? The main key is that: In Taylor expansion, $f(x+h)=\ldots$, $x$ is often at finite, $h$ is near $0$ to ensure that $x+h$ is belongs to a small enough neighborhood of $x$. But my point $x$ is at infinity and the points $a_i$ is at finite? 
That was my great trouble? Please help me to know this. Thanks in advance.
 A: Assuming $\frac{|a_i|}{|x|}\le\epsilon$ and since $|x-a_i|^2=|x|^2-2x\cdot a_i+|a_i|^2$, we have that
$$
|x-a_i|=|x|\left(1-\frac{x\cdot a_i}{|x|^2}+O\left(\epsilon^2\right)\right)\tag{1}
$$
If we set $\bar{w}=\sum\limits_{i=1}^p|w_i|$ and $\bar{a}=\sum\limits_{i=1}^pw_ia_i$, then $(1)$ yields
$$
\begin{align}
f(x)
&=\sum_{i=1}^pw_i|x-a_i|\\
&=-\frac{x}{|x|}\cdot\bar{a}+|x|\bar{w}\,O\left(\epsilon^2\right)\tag{2}
\end{align}
$$
Thus, the optimal value of $f$ is $-|\bar{a}|$ achieved in the direction of $\bar{a}$.
A: A Taylor series is difficult and may be overkill. There is a simpler method that I see using the reverse triangle inequality:
$$
        |\,\|x\|-\|y\|\,| \le \|x-y\|
$$
For example,
$$
     |\,\|x-a\|-\|x-b\|\,| \le \|(x-a)-(x-b)\|=\|b-a\|.
$$
Because your weights add up to 0, you should be able to regroup with some effort. For example,
$$
\begin{align}
   |\,\|x-a\|+\|x-b\|-2\|x-c\|\,| & \le |\,\|x-a\|-\|x-c\|\,|+|\,\|x-b\|-\|x-c\|\,| \\
      & \le \|a-c\|+\|b-c\|.
\end{align}
$$
I don't think I've said anything wrong, but I am prone to idiotic mistakes. The trick is to rearrange the bits until you can do what is suggested in this last step.
