uniform continuity

**Chandru1** · January 14th 2010, 04:02 AM

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

**HallsofIvy** · January 14th 2010, 07:29 AM

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|.

**Black** · January 14th 2010, 08:00 AM

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.

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