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Thread: Absolute Value Function

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    Absolute Value Function

    How would you show that the absolute value function $\displaystyle f(x) := |x|$ is continuous at every point $\displaystyle c$ in $\displaystyle \mathbb{R}$
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    Quote Originally Posted by CrazyCat87 View Post
    How would you show that the absolute value function $\displaystyle f(x) := |x|$ is continuous at every point $\displaystyle c$ in $\displaystyle \mathbb{R}$
    You've got to be kidding me. It's Lipschitz with Lipschitz constant one. $\displaystyle |f(x)-f(y)||\leqslant |x-y|$. So, in fact it's uniformly continuous on $\displaystyle \mathbb{R}$
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    Well, if he is asking this, he probably doesn't know about "Lipschitz"- or "uniformly continuous"!

    CrazyCat87, The definition of f(x) continuous at x= a is that f(a) exist, $\displaystyle \lim_{x\to a} f(x)$ exist, and $\displaystyle \lim_{x\to a} f(x)= f(a)$. Typically, showing that the two are equal requires showing they exist so that is enough.

    Here, if x> 0, f(x)= |x|= x. If a> 0, we can take $\displaystyle \delta< a/2$ so that as long as $\displaystyle |x-a|<\delta$, x> 0 and $\displaystyle |f(x)- f(a)|= |x- a|< \epsilon$ will be true as long as $\displaystyle \delta$ is less than the smaller of a/2 and $\displaystyle \epsilon$.

    If x< 0, f(x)= |x|= -x. If a< 0, we can take $\displaystyle \delta< -a/2$ so that as long as $\displaystyle |x-a|< \delta$, x< 0 and $\displaystyle |f(x)- f(a)|= |-x-(-a)|= |x-a|< \epsilon$ will be true as long as $\displaystyle \delta$ is less than the smaller of -a/2 and $\displaystyle \epsilon$.

    Finally, if a= 0, |f(x)- f(a)|= | |x|- 0|= |x|. We can make that less than $\displaystyle \epsilon$ by taking $\displaystyle \delta= \epsilon$.
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    Thank you HallsofIvy
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    Quote Originally Posted by HallsofIvy View Post
    Well, if he is asking this, he probably doesn't know about "Lipschitz"- or "uniformly continuous"!

    CrazyCat87, The definition of f(x) continuous at x= a is that f(a) exist, $\displaystyle \lim_{x\to a} f(x)$ exist, and $\displaystyle \lim_{x\to a} f(x)= f(a)$. Typically, showing that the two are equal requires showing they exist so that is enough.

    Here, if x> 0, f(x)= |x|= x. If a> 0, we can take $\displaystyle \delta< a/2$ so that as long as $\displaystyle |x-a|<\delta$, x> 0 and $\displaystyle |f(x)- f(a)|= |x- a|< \epsilon$ will be true as long as $\displaystyle \delta$ is less than the smaller of a/2 and $\displaystyle \epsilon$.

    If x< 0, f(x)= |x|= -x. If a< 0, we can take $\displaystyle \delta< -a/2$ so that as long as $\displaystyle |x-a|< \delta$, x< 0 and $\displaystyle |f(x)- f(a)|= |-x-(-a)|= |x-a|< \epsilon$ will be true as long as $\displaystyle \delta$ is less than the smaller of -a/2 and $\displaystyle \epsilon$.

    Finally, if a= 0, |f(x)- f(a)|= | |x|- 0|= |x|. We can make that less than $\displaystyle \epsilon$ by taking $\displaystyle \delta= \epsilon$.
    Why do you take $\displaystyle \delta<a/2$
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    Here's what I did, let me know if it's missing anything.

    If $\displaystyle x>0$ then $\displaystyle f(x) = |x|=x$. As seen in some example, if $\displaystyle c\in\mathbb{R}$, then the limit of $\displaystyle f(x)$ as $\displaystyle x$ goes to $\displaystyle c$ equals $\displaystyle c$. Since $\displaystyle f(c) = c$, $\displaystyle f(x)$ is continuous at every point $\displaystyle c$.

    Similar approach for $\displaystyle x<0$....
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    Quote Originally Posted by CrazyCat87 View Post
    If $\displaystyle x>0$ then $\displaystyle f(x) = |x|=x$. As seen in some example, if $\displaystyle c\in\mathbb{R}$, then the limit of $\displaystyle f(x)$ as $\displaystyle x$ goes to $\displaystyle c$ equals $\displaystyle c$. Since $\displaystyle f(c) = c$, $\displaystyle f(x)$ is continuous at every point $\displaystyle c$.
    Similar approach for $\displaystyle x<0$....
    It is so easy. Learn the most useful inequality.
    $\displaystyle \left( {\forall x~\&~y} \right)$ it is true that $\displaystyle \left| {\left| x \right| - \left| y \right|} \right| \leqslant \left| {x - y} \right|$.
    Once we realize that, then we let $\displaystyle \delta = \varepsilon $.
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