first, we need to show that P is linear.

this means showing that for two functions f and g, and two real numbers a and b:

P(af + bg) = aP(f) + bP(g).

to do this, we'll show that P(af+bg)(x) = [aP(f) + bP(g)](x), for all x.

this is mostly a matter of straight-forward calculation:

P(af+bg)(x) = (1/2)[(af+bg)(x) + (af+bg)(-x)] = (1/2)[(af)(x) + (bg)(x) + (af)(-x) + (bg)(-x)]

= (1/2)[a(f(x)) + b(g(x)) + a(f(-x)) + b(g(-x))] = (1/2)[a(f(x) + f(-x)) + b(g(x) + g(-x))]

= (a/2)(f(x) + f(-x)) + (b/2)(g(x) + g(-x)) = a[(1/2)(f(x) + f(-x))] + b[(1/2)(g(x) + g(-x))] = a(P(f(x)) + b(P(g(x)) = (aP(f))(x) + (bP(g))(x) = (aP(f) + bP(g))(x).

this shows P is linear.

next we need to show that P^{2}(f) = P(P(f)) = P(f). to do this we will show that:

P(P(f))(x) = P(f)(x), for all x.

P(P(f))(x) = (1/2)[P(f)(x) + P(f)(-x)] = (1/2)[(1/2)(f(x) + f(-x)) + (1/2)(f(-x) + f(-(-x))] = (1/4)(f(x) + f(-x) + f(-x) + f(x)) = (1/4)(2(f(x)) + 2(f(-x))) = (1/2)(f(x) + f(-x)) = P(f)(x).