a******y
发帖数: 2 | 1 Here is a rough provement.
Consider such a problem,
f(x) >= 0 and convex for x \in [a,b], a < 0 < b
f(x) = 0 otherwise.
\int_0^{-\infty} xf(x) dx = \int_0^{+\infty} xf(x)
S1 = \int_a^0 f(x) dx
S2 = \int_0^b f(x) dx
then S1 >= S2 * 4/5.
[PROVE]:
Let g(x) be the line passes (0, f(0)), (-2e, 0), where e > 0, and
S1 = \int_{-2e}^0 g(x) dx. Since f(x) convex, f(x) cross g(x) at most
once between -2e and 0, and
\int_0^{-\infty} xf(x) dx <= \int_0^{-2e} xg(x) dx.
It's not hard to show g(x) >= f(x) for | a******y
发帖数: 2 | 2 I do not have a very rigorous prove either. :)
Here are some explanations.
这个式子的几何意义我很清楚，可是如何从已知的条件里推出来呢？
- it seems it's not easy to prove 几何意义. :)
- f(x) is convex, that is f(tx + (1-t)y) >= tf(x) + (1-t)f(y)
for all x, y in (a,b), t in [0,1], It's not hard to prove
f(x) is continuous for x in (a,b).
- if |a| > |2e|, f(a) >= 0 > g(a), f(x) > g(x) by convex,
area under f(x) will be larger than area under g(x). impossible.
- if |a| = |2e|, f(a) >= 0 = g(a), it has to be f(x) = g(x), equali |
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