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Let $f:[0,1]\rightarrow[0,\infty)$ be an increasing function, $a \in (0,1)$, and $\displaystyle \int_0^1 f(x) \ dx =1 $. What are the maxima of $$i)\int_0^a f(x)^2 \ dx$$ $$ii)\int_0^a xf(x)^2 \ dx$$

Some clues?

EDIT: maxima need to be expressed as $g(a)$, $a \in (0,1)$

Damn_crazy_girl
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    Could you please be more precise! are you looking for the maximum of the function $a \mapsto \int_0^a\ldots$? – HorizonsMaths Jul 17 '13 at 09:48
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    Both of $(i)$ and $(ii)$ are increasing functions so the maxima in the domain $a\in (0,1)$ will be $\lim_{a\rightarrow 1-}\int_{0}^a f(x)^2 dx$ and $\lim_{a\rightarrow 1-}\int_{0}^a xf(x)^2 dx$ respectively. – Samrat Mukhopadhyay Jul 17 '13 at 09:50
  • a is fixed number here, so the maximum will depend on a. – leshik Jul 17 '13 at 09:54
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    Some clarification would indeed help. As functions in $a$, neither $i)$ nor $ii)$ have maxima, only suprema (see Samrat's comment). If you're asking for the maximal values the terms can assume varying over functions $f$ with the stated constraints, the question becomes more interesting. – m_l Jul 17 '13 at 09:57
  • @Mercy Yes. I try to find the precise maxima when $a$ runs in $(0,1)$. – Damn_crazy_girl Jul 17 '13 at 10:03
  • @leshik are you saying that the question asks for maximizing the functionals $(i)$ and $(ii)$? Then, of course the answer is completely different. – Samrat Mukhopadhyay Jul 17 '13 at 10:05
  • If the maxima is to be expressed as $g(a)$ you should probably reformulate your question because there is no way you can get it by maximizing over $a \in (0,1)$. It seems to me that you want to maximize over $f:[0,1] \to [0,\infty)$, then again it might be difficult (even impossible) to prove that the maximizer is an increasing function. – HorizonsMaths Jul 17 '13 at 10:42
  • Just note that, $ \int_0^a f(x)^2 \ dx \leq \int_0^1 f(x)^2 \ dx $. – Mhenni Benghorbal Jul 17 '13 at 10:59
  • @MhenniBenghorbal the core idea is to express the maximum as a function of $a, \forall a \in (0,1)$. – Damn_crazy_girl Jul 17 '13 at 11:06
  • Is this a homework question or some problem in a book? If that is the case, please give some context. Is the question stated exactly as you posted it? – m_l Jul 17 '13 at 11:09
  • @m_l neither homework (who would ask for such questions?), nor a problem in a book. It comes from a friend of mine. – Damn_crazy_girl Jul 17 '13 at 11:18
  • @Damn_crazy_girl I'd like to find out who asks for such questions. This could be a (mildly strange) homework assignment in variational calculus, or an even stranger one in first semester calculus. Basically, I'm still confused about whether or not $f$ is fixed. I assume you want to find the maxima for varying $f$ and $a$ satisfying the given constraints, correct? – m_l Jul 17 '13 at 13:04
  • @m_l since I didn't mention a certain function, all functions family satisfying the given constraints must be taken into account. I repet, the maxima must be expressed as a function in variable $a$. Honestly, I doubt one may see it in first semester calculus. It's been here for 3 hours and no answer. The instructors I know didn't solve it either. The question is meant for the best of the best. – Damn_crazy_girl Jul 17 '13 at 13:23
  • @Damn_crazy_girl Yes, but that may be in part due to confusion as to what is being asked here. There was, by the way, an answer, but it has been deleted. I'm not sure why, it seemed correct enough: for each fixed $a \in (0,1), c > 0$ you can give a function $f_{a,c}$ such that $\int_0^a f_{a,c}(x)^2 \operatorname{d}x = c$, so there is no maximum. The same holds for case $ii)$. – m_l Jul 17 '13 at 13:30
  • @m_l What I mean above is that you can choose any $f$ from that family functions, and then there must be a way to express the maxima as a function in $a$. I don't know what you understand from all I say. – Damn_crazy_girl Jul 17 '13 at 13:41
  • @Damn_crazy_girl I'll try again. Are you asking for a function $g : (0,1) \rightarrow \mathbb{R} : a \mapsto \operatorname{max}_{f \in C} \int_0^a f(x)^2 dx$ where $C := { f : [0,1] \rightarrow [0,\infty) ~|~ f \text{ increasing }, \int_0^1 f(x) dx = 1 }$? If not, please explain what kind of maximum you are asking for. – m_l Jul 17 '13 at 13:56
  • @m_l Exactly. This is my point. – Damn_crazy_girl Jul 17 '13 at 14:00
  • @Damn_crazy_girl If you agree with the last comment made by m_1 then you definitely have to reformulate your question. In one of your comments you claim that $f$ is arbitrary and at the same time the statement of the problem still says that $f$ is fixed!. – HorizonsMaths Jul 17 '13 at 16:00

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