Proofs with continuity and absolute values

[mscbuck]

Homework Statement

-F is a continuous function on [0,1], so let ||f|| be the maximum value of |f| on [0,1]

a. Prove that for any number c we have ||cf|| = |c|$$\ast$$||f||

b. Prove that ||f + g|| $$\leq$$ ||f|| + ||g||.

c. Prove that ||h - f|| $$\leq$$ ||h - g|| + ||g - f||

Homework Equations

Based on the statement, we know f is continuous so we know a limit exists for f(x), f(a) exists, and that the limit of f(x) as x-->a = f(a).

The Attempt at a Solution

I think I have solved part a.). My attempt begins by saying that since |cf| (x) = |c| $$\ast$$ f(x), so if y is where |f| will reach its maximum value. Therefore:

f|y| = ||f|| , so therefore:

|c|$$\ast$$|f(y)| $$\geq$$ |cf(x)|

|cf(y)| = ||cf||

Is that enough for part a? I am stuck somewhat at parts b and c. My instructor said this is an exercise in placing the absolute value signs and parentheses perfectly at just the right times, but I"m afraid I'm not seeing those times! Some first steps to take would be greatly appreciated!

Thanks

 

[statdad]

You do know that [tex] |f| [/itex] is continous, so there is some point $$y \in [0,1]$$ that satisfies

$$|f(y)| = \max_{x \in [0,1]} |f(x)|$$

This should help with `a'.

For the others, think along these lines to start

a) For any numbers (or function values) $$|a + b| \le |a| = |b|$$

b) It is also true that $$|a - b| \le |a - c| + |c - b|$$.