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Thread: uniform continuity

  1. #1
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    uniform 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
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  2. #2
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    Quote Originally Posted by Chandru1 View Post
    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
    Okay, what is the definition of "uniform continuity"?
    Also, I would suggest dividing this into 4 cases: $\displaystyle x\ge 0$, $\displaystyle |y|\le |x|$; x< 0, |y|< |x|; $\displaystyle x\ge 0$, $\displaystyle |y|\ge |x|$, and x< 0, |y|< |x|.
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  3. #3
    Member Black's Avatar
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    You 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.


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