The range of a argmin function I have the following function:
$$f(a)=\text{argmin}_{t\in\mathbb{R}} \left\{\sum_{i=1}^n\frac{1}{2}(1-a)(b_i-t)^2+a| b_i-t |\right\}\quad ,a\in[0,1]$$
Define $\bar{\beta}$ to be the mean of n observations $b_i$, and $\hat{\beta}$ the median.
Assume $\bar{\beta}\le \hat{\beta}$, how does one prove that
$$\bar{\beta}\le f(a)\le \hat{\beta} \quad \text{for all} \quad a\in[0,1]$$
with the lower inequality achieved for $a=0$ and upper for $a=1$
 A: When $a=0$ we have,
$$f(0) =\arg\min\{\sum\limits_{i=1}^n\frac{1}{2}(b_i-t)^2\}$$
Take the derivative of the objective function $h(t)$,
$$h'(t) = \sum\limits_{i=1}^nt-b_i$$
Find the critical point,
$$h'(t) = 0 \implies \sum\limits_{i=1}^n b_i = nt\implies t=\bar{\beta}$$
When $a=1$ we have,
$$f(1) = \arg\min\{\sum\limits_{i=1}^n|b_i-t|\}$$
Let $\textbf{t}$ be the vector of length $n$, $\textbf{b} = \begin{pmatrix}b_1\\\vdots\\ b_n\end{pmatrix}$, and $C=\{y\in\mathbb{R}^n:y_i=y_j\forall i,j\}$. Then we have,
$$f(1) = \arg\min\limits_{\textbf{t}\in\mathbb{R}^n}\|\textbf{b}-\textbf{t}\|_1+i_{C}(\textbf{t})\implies 0\in\partial \|\textbf{b}-f(1)\textbf{1}\|_1+N_C(f(1)\textbf{1})$$
The set $N_C(f(1)\textbf{1})$ is the set of vectors whose components sum to $0$. So $0=u+v$ for some $u\in \partial\|\textbf{b}-f(1)\textbf{1}\|_1$ and $v\in N_C(f(1)\textbf{1})$. For the vector $u$ we have,
$$u_i = \begin{cases}\mbox{sign}(f(1)-b_i) & b_i\neq f(1),\\ [-1,1] & b_i=f(1)\end{cases}$$
 and since $0=v+u$ we have that the components of $u$ must sum to $0$ as they do for $-v$. Without loss of generality choose $u_i=\mbox{sign}(f(1)-b_i)$ for all $i$. This gives,
$$\sum\limits_{i=1}^n u_i =0,\\
\implies \sum\limits_{i=1}^n\mbox{sign}(f(1)-b_i)=0$$
So, $f(1)$ must be the median $\hat{\beta}$ since we could just reorder the $b_i$ according to $\leq$ and find that half the $b_i$ are greater than or equal to $f(1)$ and half the $b_i$ are less than or equal to $f(1)$. This is kind of glossing over whether or not there are repeats in the observations, i.e. $b_i=b_j$ for some $i,j$ would mean we might have choose some $u_i$ to account for that instead.
