A Characterization of the Mode of a Distribution 
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}
(I) Can we find simple conditions on $f$ such that for some $\epsilon > 0$ and each given $q \in (0,\epsilon)$ there exists a unique solution to this problem?

Assume that for all $q$ small enough the problem has a unique solution, and put
  \begin{equation}
x_q= \arg \min_{y \in \mathbb{R}} \int_{\mathbb{R}}|y-x|^{q} f(x) \,dx.
\end{equation}
(II) Can we conclude that $x_q \rightarrow x_0$ for $q \rightarrow 0$?


I found this last property stated in a monograph about applied statistics without any proof.
Thank you very much in advance for your kind attention.
NOTE (1). For $q=1$ the solutions of our minimization problem are all the medians of the distributions defined by $f$: see Why does the median minimize $E[|X-c|]$. So a simple condition assuring that for $q=1$ our problem has a unique solution is that $f > 0$.
NOTE (2). Analogous questions can be asked about the midrange. Make the additional assumption that $f$ has compact support $S$ and put $a= \min S$ and $b = \max S$. If we define the probability measure 
\begin{equation}
\mu(A)=\int_{A} f(x)dx
\end{equation}
for every set $A$ in the Borel $\sigma$-algebra $\mathcal{B}(\mathbb{R})$, and we consider for every measurable function $g:\mathbb{R} \rightarrow \mathbb{R}$ the norm $||g||_{\infty}= \operatorname{ess} \sup |g|$ with respect to measure space $(\mathbb{R}, \mathcal{B}(\mathbb{R}), \mu)$, then the problem
\begin{equation}
\min_{y \in \mathbb{R}} || \mathbb{1} -y||_{\infty},
\end{equation}
where $\mathbb{1}(x)=x$ for all $x \in \mathbb{R}$, has as unique solution the midrange $x_{\infty}=(a+b)/2$.
Now we can ask:
(I') Are there simple conditions on $f$ such that for every $q$ greater then some fixed $M > 0$, there exists a unique solution to the problem
\begin{equation}
\min_{y \in \mathbb{R}} \int_{\mathbb{R}}|y-x|^{q} f(x) \,dx?
\end{equation}
(II') Assume that these conditions are verified, and put
\begin{equation}
x_q= \arg \min_{y \in \mathbb{R}} \int_{\mathbb{R}}|y-x|^{q} f(x) \,dx.
\end{equation}
Do we have $x_q \rightarrow x_{\infty}$ as $q \rightarrow \infty$?
 A: As for question (I), there is essentially no simple condition on $f$ assuring the uniqueness of the solution of the problem
\begin{equation}
\min_{y \in \mathbb{R}} \int_{\mathbb{R}}|y-x|^{q} f(x) \,dx.
\end{equation}
Th fact is that the function $x \mapsto |x|^q$ is not convex for $q \in (0,1)$ nor $f$ can be a convex function, since it is a probability density function. So we cannot say very much about the convolution integral
\begin{equation}
F_q(y) = \int_{\mathbb{R}}|y-x|^{q} f(x) \,dx
\end{equation}
The only very special case I could envisage in which the uniqueness is garanteed is the following one. Suppose that $f \in C^1(\mathbb{R})$, that it has compact support and that $f$ is symmetric around $x_0$. Without loss of generality we can take $x_0=0$ to simplify the notation. Then we have
\begin{equation}
F'_q(y) = \int_{\mathbb{R}}|x|^{q} f'(y-x) \,dx = \int_{\mathbb{R}}|y-x|^{q} f'(x) \,dx = \int_{0}^{\infty} -f'(x) [|x+y|^q - |x-y|^q ] dx.
\end{equation}
Now for any $x > 0$ we have $|x+y|^q - |x-y|^q > 0$ for $y >0$ and $|x+y|^q - |x-y|^q < 0$ for $y < 0$. So $F'_q(y) < 0$ for $y < 0$ and $F'_q(y) > 0$ for $y > 0$, and $x_0=0$ is the unique global minimum of $F_q$.
Anyway, essentially question (II) makes sense even if the uniqueness of the solution of our minimization problem is not assured. Indeed, if we call $S_q$ the set of all minimizers for the problem 
\begin{equation}
\min_{y \in \mathbb{R}} \int_{\mathbb{R}}|y-x|^{q} f(x) \,dx,
\end{equation}
then we can ask whether $S_q$ "shrinks" to $x_0$ when $q \rightarrow 0$, in the sense that for every $\epsilon > 0$ there exists $\delta > 0$ such that $S_q \subset (x_0 - \epsilon, x_0 + \epsilon)$ for every $q \in (0,\delta)$.
This last question has been answered in the negative by Maurice Fréchet in his work Les valeurs typiques d'ordre nul ou infini d'un nombre aléatoire: see the counterexample at pages 16-19. For further discussion and references on this subject see my post q-Means and the Mode of a Distribution.
