Let $f:\mathbb{R} \rightarrow [0,\infty)$ be a continuous probability density function on $\mathbb{R}$ such that \begin{equation} \int_{\mathbb{R}} |x| f(x)\, dx < \infty, \end{equation} and assume that $f$ has a strict global maximum $x_0$, that is $f(x) < f(x_0)$ for all $x \neq x_0$. For any fixed $q \in (0,1]$ consider the problem \begin{equation} \min_{y \in \mathbb{R}} \int_{\mathbb{R}}|y-x|^{q} f(x) \,dx, \end{equation}

and let $S_q$ be the set of all minimizers for this problem. My question is: does $S_q$ "shrink" to $x_0$ for $q \rightarrow 0$, in the sense that for every $\epsilon > 0$ there is some $\delta > 0$ such that $S_q \subset (x_0 - \epsilon, x_0 + \epsilon)$ for every $q \in (0, \delta)$?

I found this last property stated in a monograph about applied statistics without any proof, but I am not so convinced of its validity.

NOTE (1). If we put for a fixed $q \in (0,1]$: \begin{equation} F_q(y)=\int_{\mathbb{R}} |y-x|^q f(x) dx \quad (y \in \mathbb{R}), \end{equation} then $F_q$ is a continuous function and $\lim_{y \rightarrow - \infty} F_q(y) = \lim_{y \rightarrow \infty} F_q(y) = \infty$, so the minimization problem

\begin{equation} \min_{y \in \mathbb{R}} \int_{\mathbb{R}}|y-x|^{q} f(x) \,dx. \end{equation}

admits for sure some solution.

NOTE (2). For every $q > 0$ let $g_q:\mathbb{R} \rightarrow \mathbb{R}$ be the function: \begin{equation} g_q(x) = \begin{cases} 1 &\quad &\text{if} &\quad |x| > q,\\ 0 &\quad &\text{if} &\quad |x| \leq q, \end{cases} \end{equation} and let us put \begin{equation} G_q(y)=\int_{\mathbb{R}} g_q(y-x)f(x) dx. \end{equation} We have $G_q \leq 1$, $G_q(y) < 1$ for some $y \in \mathbb{R}$ and $\lim_{y \rightarrow - \infty} G_q(y) = \lim_{y \rightarrow \infty} G_q(y) = 1$, so that the problem \begin{equation} \min_{y \in \mathbb{R}} G_q(y) \end{equation} admits a non-empty compact set $T_q$ of minimizers. Moreover it is immediate to see arguing by contradiction that $T_q$ shrinks to $x_0$ as $q \rightarrow 0$ (in the precise sense explained above). This result might have suggested the conjecture about $S_q$ of our original problem.

notneed to be the mode even in the unimodal case. So I would bet on the claim being false. If nobody comes to a definitive conclusion in the next few days, I'll try to make an accurate computation. The suspicious case is when $f$ is essentially the characteristic function of $[-1,1]$ with a small upward bump near $1$. $\endgroup$isunimodal. Check it. $\endgroup$