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The text from Robust Estimation of Location Parameter (1964), Huber, P.:

It will be assumed thad $\rho$ is continuous convex real-valued function of a real variable $t$. tending to $+\infty$, as $t > \rightarrow \pm \infty$.

Let $x_1, x_2, ..., x_n$ be independent identically distributed random variables with common distribution function $F$. Let $\left[T_n(x)\right]$ bet the set of all those $\xi$ for which $Q(\xi) = \sum_{i=1}^n \rho(x_i-\xi) $ reaches it's minimum $Q_{\inf}$. Obviously $\left[T_n(x)\right]$ is invariant under translations: $\left[T_n(x+c)\right] = \left[T_n(x)\right]+c$. By $T_n(x)$ we shall denote any representation of the set valued function $(x_1,x_2,...,x_n) \rightarrow \left[T_n(x)\right]$ by a single valued function $(x_1,x_2,...,x_n) \rightarrow T_n(x) \xi \left[T_n(x)\right]$, e.g., $T_n(x) - $ midpoint of $\left[T_n(x)\right]$.

Why it's obvious that $\left[T_n(x)\right]$ is invariant under translation? How does $T_n(x)\xi \left[T_n(x)\right]$ translates to midpoint of $\left[T_n(x)\right]$, and if the set is not ordered, how the middle point be of any use?

Lemma 1. $Q(\xi)$ is a convex function of $\xi$, and $\left[T_n(x)\right]$ is convex, non-empty and compact. If $\rho$ is strictly convex, then $\left[T_n(x)\right]$ is reduced to a single point.

PROOF. Strict convexity of Q follows immediately from strict convexity of $\rho$. The sets $\{\xi| Q(\xi) \le Q_{\inf} + m^{-1}\}$ form a decreasing sequency of non empty convex compact sets as $m\to\infty$ , hence their intersection $\left[T_n(x)\right]$ is non-empty convex compact. If $Q$ is convex, and $\xi', \xi''$ are two distinct points from $\left[T_n(x)\right]$, then we would have $Q_{\inf} = Q(\frac{1}{2}\xi' + \frac{1}{2}\xi'') < \frac{1}{2}Q(\xi') + \frac{1}{2}Q(\xi'') = Q_{\inf}$, which is a contradiction.

I don't get the last expression $Q_{\inf} = Q(\frac{1}{2}\xi') + Q(\frac{1}{2}\xi'')$. If for example $Q(\xi') =Q(\xi'') = 0$, does it follow that $Q(\frac{1}{2}\xi') + Q(\frac{1}{2}\xi'')=0$? Why $Q(\frac{1}{2}\xi') + Q(\frac{1}{2}\xi'')$ is less than $ Q(\frac{1}{2}\xi' + \frac{1}{2}\xi'')$?

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