If $f(x) = \max\left|2\sin y-x\right|,$ Then $\min.$ value of $f(x)$ 
If $f(x) = \max\left|2\sin y-x\right|\;,$ Where $y\in \mathbb{R}\;,$ Then $\min.$ value of $f(x)$

$\bf{My\; try}$ We know that $-2 \leq 2\sin y\leq 2$. Now I did not Understand How Can I open That
help me, Thanks 
 A: $$f(x) = \sup_{y \in \mathbb{R}}\left|2\sin y-x\right|$$
Method-$1$
Define a function, $g_{x}(y)=\left|2\sin y-x\right|, \text{ where } y \in \mathbb{R}$. So, $f(x) = \sup_{y \in \mathbb{R}}g_{x}(y)$.
Now, $-1\leq \sin(x) \leq 1$. 
So, this optimization problem of $g_x(y)$ can be posed as another equivalent optimization problem. That is,
$$f(x) = \sup_{a \in [-2,2]} h_{x}(a)$$
where,
$$h_{x}(a) = |a-x|, \text{ where } a \in \mathbb{R}.$$
$$\implies f(x) = \max_{t \in [(-2-x),(2-x)]} |t|.$$
(From the attainability argument we can replace $sup$ by $max$.)
$$ \implies f(x) =\begin{cases} 2-x, & x \leq 0 \\ x+2, & x > 0 \end{cases}$$
Method-$2$
Credit to @A.G.'s comment.
$$\left|2\sin y-x\right| \leq \left|2\sin y\right|+\left|-x\right|$$
Equality holds when, $sgn(sin{y}) = -sgn(x)$. 
$$\implies \sup_{y \in \mathbb{R}}\left|2\sin y-x\right| \leq 2 + \sup_{y \in \mathbb{R}}\left|x\right|.$$
Because, $\sup(A+B) \leq \sup{A}+\sup{B}$, but here equality will hold. A detailed discussion can be found here- link.
So, we can say that-
$$f(x) = 2+|x|, x \in \mathbb{R}$$  
So, then it is clear that,
$$\inf_{x \in \mathbb{R}} f(x) = 2, \text{ when } x=0$$ 
(Again one can replace $inf$ by $min$ from the same argument.)
A: Since $2 \sin y$ ranges continuously from $-2$ to $2$, $f(x)$ is the maximum of the absolute value function on the interval from $-2 -x$ to $2 - x$. As $x$ varies this interval can be any interval of length 4, and this maximum absolute value is minimized when this interval is exactly $[-2,2]$, in other words when $x = 0$. So the minimum of $f(x)$ would be $2$.
