Define max(f,g) and min(f,g) as max(f,g)(x) = max {f(x),g(x)} and min(f,g)(x) = min {f(x),g(x)}. Show that?

Question

Define max(f,g) and min(f,g) as max(f,g)(x) = max {f(x),g(x)} and min(f,g)(x) = min {f(x),g(x)}. Show that max(f,g)=(1/2)*(f+g)+(1/2)*|f-g| and min(f,g)=(1/2)*(f+g)-(1/2)*|f-g|. (You can show this by first showing that for all m,n belonging to R, max{m,n}=(1/2)*(m+n)+(1/2)*|m-n|. Do this by considering the two cases m>=n and m

Answer 1

Well, the question pretty much outlines for you how to do it. Let m, n ∈ R. For the case m >= n, then max{m, n} = m and (1/2)(m+n) + (1/2)|m-n| = (1/2) ((m+n) + (m-n)) = m. For the case m < n, max{m, n} = n and (1/2)(m+n) + (1/2)|m-n| = (1/2) ((m+n) + (n-m)) = n. So in either case max{m,n}=(1/2)(m+n) + (1/2)|m-n|.
Note that it follows that min{m, n} = -max{-m, -n} = (1/2)(m+n) - (1/2)|m-n|.

So, for any x ∈ R, we have
max(f, g) (x) = (1/2) (f(x) + g(x)) + (1/2) |f(x) - g(x)|
= [(1/2) (f+g) + (1/2) |f-g|] (x).
Similarly min(f, g) (x) = [(1/2) (f+g) - (1/2) |f-g|] (x).
So max(f, g) = (1/2) (f+g) + (1/2) |f-g| and min(f, g) = (1/2) (f+g) - (1/2) |f-g|.

Answer 2

Define Max