Let $\displaystyle f:\mathbb{R}^{2} \to \mathbb{R}$ be defined as $\displaystyle f(x,y)=\text{max}\{|x|,|y|\}$. Prove that f is uniformly continuous

Printable View

- Jan 14th 2010, 04:02 AMChandru1uniform continuity
Let $\displaystyle f:\mathbb{R}^{2} \to \mathbb{R}$ be defined as $\displaystyle f(x,y)=\text{max}\{|x|,|y|\}$. Prove that f is uniformly continuous

- Jan 14th 2010, 07:29 AMHallsofIvy
- Jan 14th 2010, 08:00 AMBlackYou can also show that the function is Holder continuous on $\displaystyle \mathbb{R}^2$; that is,

$\displaystyle |f(\mathbf{y})-f(\mathbf{x})| \le C|\mathbf{y}-\mathbf{x}|^{\lambda}$ $\displaystyle \forall \mathbf{x}, \mathbf{y} \in \mathbb{R}^2,$

where $\displaystyle C$ and $\displaystyle \lambda$ are nonnegative real constants. If a function is Holder continuous, then it is uniformly continuous.