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how to prove the absolute value of any continuous function is continuous.
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Author Topic: how to prove the absolute value of any continuous function is continuous.  (Read 49619 times)
tcook10
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« on: November 08, 2009, 10:08:16 PM »

Please help me to understand how to prove The absolute value of any continuous function is continuous.
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« Reply #1 on: November 09, 2009, 09:41:44 AM »

I would go about it using the definition of absolute value, together with the fact that f(x) is continuous. Something like this:

Recall |f(x)| = f(x) if f(x) >= 0 or -f(x) if f(x) < 0

By assumption f(x) is continuous at c, i.e. |f(x) - f(c)| < epsi if |x - c| < delta

You need to show that | |f(x)| - |f(c)| | < epsi if |x - c| < delta

Now try three cases:

(1) f(c) > 0, which also implies that f(x) > 0 if x and c are close enough (why?) ... (use def of |f(x)|)
(2) f(c) < 0, which also implies that f(x) < 0 if x and c are close enough (why?) ... (use def of |f(x)| again but factor out a minus sign ...)
(3) f(c) = 0, which follows right away from f being continuous

Hope it helps, best of luck.
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« Reply #2 on: December 18, 2009, 02:20:32 PM »

Continuity of f(x) implies that |f(x) - f(c)| < epsilon whenever |x - c| < delta. Where "epsilon" and "delta" are positive real numbers.

Now,

| |f(x)| - |f(c)| | =< |f(x) - f(c)| for all x, which completes the proof.

i.e. | |f(x)| - |f(c)| | =< |f(x) - f(c)| < epsilon whenever |x - c| < delta, so |f(x)| is a continuous function.

Hope this helps.
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