# Absolute Value Function

• Feb 24th 2010, 07:43 PM
CrazyCat87
Absolute Value Function
How would you show that the absolute value function $f(x) := |x|$ is continuous at every point $c$ in $\mathbb{R}$
• Feb 24th 2010, 07:44 PM
Drexel28
Quote:

Originally Posted by CrazyCat87
How would you show that the absolute value function $f(x) := |x|$ is continuous at every point $c$ in $\mathbb{R}$

You've got to be kidding me. It's Lipschitz with Lipschitz constant one. $|f(x)-f(y)||\leqslant |x-y|$. So, in fact it's uniformly continuous on $\mathbb{R}$
• Feb 25th 2010, 03:49 AM
HallsofIvy
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, $\lim_{x\to a} f(x)$ exist, and $\lim_{x\to a} f(x)= f(a)$. 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 $\delta< a/2$ so that as long as $|x-a|<\delta$, x> 0 and $|f(x)- f(a)|= |x- a|< \epsilon$ will be true as long as $\delta$ is less than the smaller of a/2 and $\epsilon$.

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

Finally, if a= 0, |f(x)- f(a)|= | |x|- 0|= |x|. We can make that less than $\epsilon$ by taking $\delta= \epsilon$.
• Feb 25th 2010, 08:18 PM
CrazyCat87
Thank you HallsofIvy
• Feb 28th 2010, 05:27 AM
CrazyCat87
Quote:

Originally Posted by HallsofIvy
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, $\lim_{x\to a} f(x)$ exist, and $\lim_{x\to a} f(x)= f(a)$. 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 $\delta< a/2$ so that as long as $|x-a|<\delta$, x> 0 and $|f(x)- f(a)|= |x- a|< \epsilon$ will be true as long as $\delta$ is less than the smaller of a/2 and $\epsilon$.

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

Finally, if a= 0, |f(x)- f(a)|= | |x|- 0|= |x|. We can make that less than $\epsilon$ by taking $\delta= \epsilon$.

Why do you take $\delta
• Feb 28th 2010, 05:41 AM
CrazyCat87
Here's what I did, let me know if it's missing anything.

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

Similar approach for $x<0$....
• Feb 28th 2010, 08:22 AM
Plato
Quote:

Originally Posted by CrazyCat87
If $x>0$ then $f(x) = |x|=x$. As seen in some example, if $c\in\mathbb{R}$, then the limit of $f(x)$ as $x$ goes to $c$ equals $c$. Since $f(c) = c$, $f(x)$ is continuous at every point $c$.
Similar approach for $x<0$....

It is so easy. Learn the most useful inequality.
$\left( {\forall x~\&~y} \right)$ it is true that $\left| {\left| x \right| - \left| y \right|} \right| \leqslant \left| {x - y} \right|$.
Once we realize that, then we let $\delta = \varepsilon$.