Is the function $f(x,y)=\frac{x_1-y_1}{\|x-y\|}$ bounded?

Question

Consider $\mathbb R^3$ with the cartesian coordinates $x=(x_1,x_2,x_3)$. Let $K$ be compact of $\mathbb R^3$ and consider $f:\mathbb R^3\times K\to \mathbb R$ defined by: $$f(x,y)=\frac{x_1-y_1}{\|x-y\|}$$ It is clear that $|f|$ is bounded in $(\mathbb R^3-K)\times K$. But... is also $|f|$ bounded in its entire domain $\mathbb R^3\times K$?

Answer 1

Yes, for any vector $z$, $$ |z_1|^2 \leq \sum_k |z_k|^2, $$ so $|z_1|\leq \|z\|$. Therefore, $\frac{|z_1|}{\|z\|} \leq 1$ if $z≠0$. This works in particular with $z=x-y$. As a remark, if you are working with functions defined for each point, then the you have to remove the diagonal set $\{(x,y)\in\mathbb R^3×K:x=y\}$ from the domain.

Answer 2

Presumably $\|\;\|$ is the Euclidean norm on $\mathbb{R}^3$.

The function $f(\mathbf{x},\mathbf{y})=\frac{x_1-y_1}{\|\mathbf{x}-\mathbf{y}\|}$ is not defined in the set $\Delta:=\{(\mathbf{x},\mathbf{y})\in\mathbb{R}^3\times\mathbb{R}^3:\mathbf{x}=\mathbf{y}\}$. On the other hand $$|x_1-y_1|\leq \|\mathbf{x}-\mathbf{y}\|$$

Thus $$ |f(\mathbf{x},\mathbf{y})|\leq1,\qquad \mathbf{x}\neq\mathbf{y} $$ It is possible to define $f$ along $\Delta$ so that $f$ becomes bounded on $\mathbb{R}^3\times\mathbb{R}^3$