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This question is probably trivial for those with decent training in math. Unfortunately, I have very limited training, and I got it over forty years ago and have forgotten a lot. I have tried getting this problem answered on other sites, but those sites seem to be designed to help students in high school or early in college math.

My math training ended after 2 semesters of calculus, 1 semester of abstract algebra, 1 semester of probability and statistics, and 1 semester of numerical methods. Based on what I have found in wikipedia, I suspect my problem is a very simple one in the calculus of variations. Unfortunately, I know nothing about that field and had trouble understanding even the notation in the article.

I want to determine whether the definite integral over a defined interval of ANY function that meets certain minimum constraints has a maximum value, whether that value is caclulable if it exists, and, if so, what that maximum is. If the constraints specified are insufficient, then I need to determine what additional constraints to impose. Ideally, I'd like to understand not merely what the answer is, but why that is the answer.

Given

(1) $0 < a < b$.

(2) $f(x)$ is continuous at every real.

(3) $f(x)$ can be integrated over any closed real interval.

(4) The derivative of $f(x)$ may or may not exist at every real.

(5) If $0 \leq x < y$, then $f(x) \geq f(x + y)$.

(6) $f(x) \geq g(x) = 1 /(a + b)$ if $0 \leq x \leq a$.

(7) $0 < f(x) \leq g(x) = [(b - x)/(b^2 - a^2)]$ if $a \leq x < b$.

(8) $f(x) = g(x) = 0$ if $b \leq x$.

(9) $\int_0^b f(x)dx = \int_0^b g(x)dx = 1/2 $.

(10) Set $j(x) = xf(x)$.

(11) Set $k(x) = xg(x)$.

Questions: (a) Are the constraints on $f(x)$ sufficient to set a maximum bound on $\int_0^b j(x)dx$?

(b) If the bound exists, is it calculable?

(c) If it is calculable, what is it?

(d) If the answers to a, b, and c are negative, what are the minimum additional constraints that must be imposed on $f(x)$?

I have a hand-waving argument that says the maximum of $\int_0^b j(x)dx$ over all possible $j$ equals $\int_0^b k(x)dx$ (with $j$ and $k$ as above). If that is right (and I am not sure that it is), what is the argument that supports it?

I hope my question is specific enough. I apologize in advance for its probable triviality, but the high school sites do not handle this sort of question.

Jeff Morrow

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  • $\begingroup$ I edited the question so it is easier to read mathematically. If I messed up any of the conditions when I changed the typesetting, please say so. $\endgroup$
    – KConrad
    Commented Mar 21, 2011 at 21:04
  • $\begingroup$ First, thank you for the type setting. That certainly is easier to read. Second, I believe that you have re-stated the problem correctly except "all possible k" may be confusing because k(x) = xg(x), and g(x) is fully defined so only one possible k(x) exists. $\endgroup$
    – Jeff Morrow
    Commented Mar 21, 2011 at 22:00
  • $\begingroup$ I do not know how to get the neat integrals, but the integral of g(x) from 0 to a = [a / (b + a)]. The integral of g(x) from a to b = (1/2)(b^2 - 2ab + a^2) / (b^2 - a^2) = (b - a) / 2(b + a). So the integral from 0 to b = [(2a + b - a) / 2(b + a)] = [(b + a) / 2(b + a)] = (1/2). If I have erred in integrating, I am ashamed because it is essential that the integral = 1/2. $\endgroup$
    – Jeff Morrow
    Commented Mar 21, 2011 at 22:16
  • $\begingroup$ $$ 0 \geq \int_0^b (x-a) (f(x) - g(x)) dx = \int_0^b x (f(x) - g(x)) dx = \int_0^b x f(x) dx - \int_0^b x g(x) dx $$ See answer below by Charles Matthews. $\endgroup$
    – Will Jagy
    Commented Mar 21, 2011 at 22:46

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I think this isn't too bad, and your guess in below point (d) is justified. I doubt it depends on the detailed formulae.

Suppose we concentrate on g - f, which is constrained to have integral 0? And by a change of origin to have a graph that is non-positive on the negative real axis, and non-negative on the positive real axis. The main issue is that multiplying it by the independent variable (x in the current notation) you will get a non-negative function, so a non-negative integral. I believe this is what you want.

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  • $\begingroup$ Yes I think you have given an elegantly simple answer. It is a little abbreviated so I am going to go work it out. If I have a question, I'll be back. If have none, I'll make sure to give you your reputation points. In any case, thank you for pointing out a direction. I suspected I was going to need a function that combined f and h so I reserved the letter g (and likewise L between j and k). Of course when we shifted to showing functions with lower-case letters L became unusable. BUT THANKS A LOT. $\endgroup$
    – Jeff Morrow
    Commented Mar 21, 2011 at 22:45
  • $\begingroup$ Yes, that is exactly what I wanted. Thank you very much. I am also thankful to KConrad for the reformatting and to both him and Will Jagy for taking the time to look at a problem posed by the mathematically ignorant. I appreciate all the help. $\endgroup$
    – Jeff Morrow
    Commented Mar 21, 2011 at 23:22
  • $\begingroup$ I typeset the answer, using the symbols of KConrad's edit, in a comment above. As Charles says, most of your conditions have no bearing on the answer. $\endgroup$
    – Will Jagy
    Commented Mar 21, 2011 at 23:43
  • $\begingroup$ This answer is great because it does even more than I asked for. It shows that the maximum value of the integral is attained only if f = g over [0, b]. I get the point that it is the general relationship of f to g and the equality of the integrals that drives the result, not the nature of g itself, but the specification of g grew out of the overall problem I had posed to myself. Anway thanks again to all. $\endgroup$
    – Jeff Morrow
    Commented Mar 22, 2011 at 13:45

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