In article <b4be2fdf.0312191857.12869902@posting.google.com>,
qqquet@mindspring.com (Leroy Quet) wrote:
Let f be any real -> real integrable function which is strictly
monotonically increasing, f(0) = 0 and f(1) = 1, and
where integral{0 to 1} f(x) dx = 1/2.
Let g be the inverse of f, f(g(x)) = x for each x from 0 to 1.
What f(x)('s) has the minimum, and which the maximum, for
integral{0 to 1} f(x)*g(x) dx ??
Actually, the max is easy.
f(x) = x = g(x) works,
and gets the integral's value = to 1/3.
How do we know this is the maximum? Use the calculus of variations.
Let Df be a variation of f. To find the corresponding variation of g,
consider f(g(x)) = x. The variation of a composition of two functions
is given by D(f(g(x))) = (Df)(g(x)) + f'(g(x))Dg(x). Since D(f(g(x)))
is 0, we get that Dg(x) = -(Df)(g(x))/f'(g(x)).
Now to find a critical value of
|\1
| f(x) g(x) dx [1]
\| 0
we must have that the variation is 0:
0
|\1 |\1
= | Df(x) g(x) dx + | f(x) Dg(x) dx
\| 0 \| 0
|\1 |\1
= | Df(x) g(x) dx - | f(x) (Df)(g(x))/f'(g(x)) dx
\| 0 \| 0
|\1 |\1
= | Df(x) g(x) dx - | f(f(x)) Df(x) dx [2]
\| 0 \| 0
The last step follows from the substitution x -> f(x).
For [2] to be 0 for all variations of f, we must have g(x) = f(f(x)),
which means f(f(f(x))) = x. f is strictly monotonically increasing, so
if f(x) > x for any x then f(f(f(x))) > x; likewise for f(x) < x. Thus,
f(x) = x for all x in [0,1].
As for the minimum, the infimum of the integral is 0, but that cannot be
realized within the restrictions given. One family of functions which
tends to this limit is f_n(x) = x^n, for which [1] < 1/(n+1).
