NetNews Usenet Archive 1992 #30/NN_1992_30.iso/spool/sci/math/17235

Internet Message Format | 1992-12-21 | 1.9 KB

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From: [e-mail] (Kanad Chakraborty)
Newsgroups: sci.math
Subject: Re: Integral Puzzle (SPOILER)
Message-ID: <[e-mail]>
Date: 20 Dec 92 17:25:00 GMT
References: <[e-mail]> <[e-mail]>
Sender: [e-mail] (Mr. News)
Organization: University of Michigan EECS Dept., Ann Arbor
Lines: 43

In article <[e-mail]> [e-mail] (Chris Long) writes:
>In article <[e-mail]>, Yuzuru Hiraga writes:
>
>> Let f be any function such that:
>>     * for 0<=x<=1; 0<=f(x)<= a
>
>>       /1
>>     * |  f(x)dx = A
>>       /0
>> where a and A are given constants (obviously, A<=a).
>
>> What are the maximum and minimum values that
>
>>     /1       2
>>     |  {f(x)}  dx
>>     /0
>
>> can take?
>
>Let I = \int_0^1 f(x)^2 dx; the fact that I >= A^2 follows immediately
>from the Cauchy-Schwarz inequality.  For the other (more interesting)
>direction, consider that if f(x) is non-negative, then so is a-f(x),
>and so (f(x)+a-f(x))^2 >= f(x)^2+(a-f(x))^2, hence \int_0^1 a^2 dx >=
>\int_0^1 f(x)^2 dx + \int_0^1 (a-f(x))^2 dx ==> a^2 >= I + a^2 - 2aA
>+ I ==> I <= aA.  To finish, such functions exist.  For the first case,
>consider f(x)=A.  For the second, consider f(x)=a for 0 <= x <= A/a,
>f(x)=0 elsewhere.
>--
>Chris Long, 265 Old York Rd., Bridgewater, NJ  08807-2618
>
>"S.B., with an I.Q. of 161, failed to complete his course of study,
>running away instead with his professor's wife."
>H. J. Eysenck, _Know Your Own I.Q._

The fact that I must be <= aA can be deduced more easily as follows :

f(x)^2 = f(x)*f(x) <= a * f(x), hence :

     /1       2       /1
     |  {f(x)}  dx <= | a * f(x) = aA
     /0               /0

Kanad