Prove application is a contraction

[burritoloco]

Homework Statement

We have the application h: (C[0,1], R) --> (C[0,1], R), defined by
h(f(x)) = $$\[\int_0^x cos(f(t))/2\, dt \]$$

Prove that h is a contraction.

Homework Equations

Need to prove there exists k, 0<k<1 s.t.

sup_(x in [0,1]) $$\left|\int_0^x (cos(f(t)) - cos(g(t)))/2\, dt \right|\]$$
<= K sup_(x in [0,1]) |f(x) - g(x)|

for all f, g in domain.

The Attempt at a Solution

I tried saying that the cosine function is smaller or equal to 1, but the LHS became smaller or equal to 1, in which case I can't derive the RHS relationship. Not sure what to do now...
Thanks for your help! (sorry about my latex code)

 

[burritoloco]

btw there should be absolute symbols outside the integral

 

[micromass]

Hi burritoloco!

A personal question: do you speak French? It's just a guess

I guess you already did the following step?

$$\int_0^x{(\cos(f(t))-\cos(g(t)))/2dt}\leq x|(\cos(f(t))-\cos(g(t)))/2|$$

Now, what happens if you apply the sum-to-product formulas? (or Simpson formula)

 

[burritoloco]

Hello and thanks for your help! I don't speak french, but I do speak spanish :) However my prof is french, very good prof.

Using that inequality (which I had forgotten :P) then using the sum-to-product formulas, and finally using |sin x| <= |x|, I got LHS <= |f(x) - g(x)|/2 as needed.

Thanks again :).