# Absolute value anxiety

• October 2nd 2010, 10:08 AM
wilbursmith
Absolute value anxiety
Hi!

I got stuck in an assignment in real analysis.

I have $\mid\mid f(x)-g(x)\mid-\mid f(x_0)-g(x_0)\mid\mid$ and I would like it to be less than or equal to

$\mid f(x)-f(x_0)\mid+\mid g(x)-g(x_0)\mid$

Can I do that, and if I can then why? Im assuming it has something to do with triangle inequality but...??

I hated absolute values back in the days and now it comes back biting me in my a##
• October 2nd 2010, 10:23 AM
Plato
I think that you need to provide more context for the question.
More detail about $f~\&~g$.
More about what you are proving.
• October 2nd 2010, 11:21 AM
wilbursmith
Quote:

Originally Posted by Plato
I think that you need to provide more context for the question.
More detail about $f~\&~g$.
More about what you are proving.

Sure, the question is

"Let f and g be continuous functions on R. Show that the functions max(f(x), g(x)) and min(f(x), g(x)) are continuous."

So far I have defined h(x)=max(f(x),g(x)) and with the epsilon delta theorem reached the point I described

$\mid\mid f(x)-g(x)\mid-\mid f(x_0)-g(x_0)\mid\mid$

Now if I get

$\mid f(x)-f(x_0)\mid+\mid g(x)-g(x_0)\mid$

then I´d almost be done.
• October 2nd 2010, 11:35 AM
Plato
With that information we know that both $\left| {f - g} \right|\;\& \,\left| {f + g} \right|$ are continuous functions.

Notice that $\max \left\{ {f,g} \right\} = \dfrac{{f + g + \left| {f - g} \right|}}{2}\;\& \,\min \left\{ {f,g} \right\} = \dfrac{{f + g - \left| {f - g} \right|}}{2}$
• October 2nd 2010, 12:12 PM
emakarov
Quote:

Originally Posted by wilbursmith
I have $\mid\mid f(x)-g(x)\mid-\mid f(x_0)-g(x_0)\mid\mid$ and I would like it to be less than or equal to

$\mid f(x)-f(x_0)\mid+\mid g(x)-g(x_0)\mid$

Consider four numbers a, a', b, b'. Assume that |a - b| >= |a' - b'|. Then

||a - b| - |a' - b'|| <= |a - a'| + |b - b'| iff
|a - b| - |a' - b'| <= |a - a'| + |b - b'| iff
|a - b| <= |a - a'| + |b - b'| + |a' - b'| iff
|a - b| <= |a - a'| + |a' - b'| + |b' - b|

Now, |a - b| = |a - a' + a' - b' + b' - b| <= |a - a'| + |a' - b'| + |b' - b|.

If |a - b| < |a' - b'|, then do a similar thing since ||a - b| - |a' - b'|| = ||a' - b'| - |a - b||.