# uniform continuity

• January 14th 2010, 04:02 AM
Chandru1
uniform continuity
Let $f:\mathbb{R}^{2} \to \mathbb{R}$ be defined as $f(x,y)=\text{max}\{|x|,|y|\}$. Prove that f is uniformly continuous
• January 14th 2010, 07:29 AM
HallsofIvy
Quote:

Originally Posted by Chandru1
Let $f:\mathbb{R}^{2} \to \mathbb{R}$ be defined as $f(x,y)=\text{max}\{|x|,|y|\}$. Prove that f is uniformly continuous

Okay, what is the definition of "uniform continuity"?
Also, I would suggest dividing this into 4 cases: $x\ge 0$, $|y|\le |x|$; x< 0, |y|< |x|; $x\ge 0$, $|y|\ge |x|$, and x< 0, |y|< |x|.
• January 14th 2010, 08:00 AM
Black
You can also show that the function is Holder continuous on $\mathbb{R}^2$; that is,

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

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