How can I estimating $$ \arctan(\lVert x-y\rVert), $$ to below (where $(x,y)\in\Omega\times\Omega, x\neq y$, $\Omega\subset\mathbb{R}^n, n>1$ bounded domain)?
Can you give me a hint please?
Someone said me to use
$$ \lvert x\rvert\ll 1\Rightarrow \arctan(x)\approx x $$ but I do not see how that can help.
On
For some $c,d >0$ you have $\min(d,c|t|) \le \arctan |t|$.
For example, $\min( \sqrt{\frac{3}{8}}, \frac{1}{2} |t| ) \le \arctan |t|$.
To see the latter, suppose $ x \ge 0$ and let $ p(x) = x - \frac{1}{3} x^3$. Consider $\phi(x) = \arctan x - p(x)$. We have $\phi'(x) = { x^4 \over x^2+1} \ge 0$, and $\phi(0) = 0$, hence $\arctan x \ge p(x)$ for all $x \ge 0$. Furthermore, $p$ is concave and has a maximum at $x_0=\sqrt{3 \over 2 }$, hence we have $p(x) \ge {x\over x_0} p(x_0) = \frac{1}{2} x$, for $x \in [0,x_0]$. Since $\arctan$ is increasing, we have $\arctan x \ge p(x_0) = \sqrt{\frac{3}{8}}$ for all $x \ge x_0$. It follows that $\arctan x \ge \min(\sqrt{\frac{3}{8}}, \frac{1}{2} x)$ for all $x \ge 0$.
It's hard to say without knowing what you need the estimate for, but one simple bound is the one suggested by the following picture:
Pick some point $P = (x_0,\arctan(x_0))$ on the graph of arctan. Then, since arctan is concave on $[0,\infty)$ (as you can verify by computing the second derivative), it lies above the secant from the origin to $P$, as shown; also, since arctan is increasing, it lies above the horizontal line extending to the right from $P$, as shown. This yields the lower bound $$ \arctan(t) \ge \begin{cases} t\arctan(x_0)/x_0 &\text{if $0\le t\le x_0$,}\\ \arctan(x_0) &\text{if $x_0\le t$.} \end{cases} $$ Or, more compactly, $$ \arctan(t) \ge \min(\arctan(x_0), t\arctan(x_0)/x_0) \quad\text{if $t\ge 0$.} $$ You can choose $x_0$ to suit your purposes; if you take $x_0=1$, for example, you get $$ \arctan(t) \ge \min(\tfrac\pi4, \tfrac\pi4 t) \quad\text{if $t\ge 0$.} $$