# How do I make this 'intuitive' argument into a rigorous proof?

This question has been asked multiple times on this website, but I was not able to find a proper proof that went along this line of reasoning so I'm asking it again

Find the maximum value of $$\int_0^1(f(x))^3 dx$$, if $$|f(x)|\leq 1$$ and $$\int_0^1 f(x)dx=0$$, where $$f$$ is a real valued function.

$$f(x)$$ clearly cannot have the same sign everywhere in $$(0,1)$$. Let $$a$$ be the sum of the lengths of all intervals on which $$f(x)\geq0$$, and $$A$$ the set of all $$x$$ for which $$f(x)\geq0$$. This contributes to all the positive area of in $$\int_0^1 f(x) dx$$. Now $$(f(x))^3\leq f(x)$$ for all $$x$$ in $$A$$ as $$f(x)<1$$ and $$(f(x))^3 \geq f(x)$$ for all $$x$$ for which $$f<0$$. For $$f(x)>0$$, we note that $$f(x)=1$$ is the only value for which the cubing doesn't decrease the value. For $$\int_0^1 f^3(x)$$ to be maximum, we thus need $$f(x)=1$$ for all $$x$$ in $$A$$ (of course, letting the $$+ve$$ values everywhere got to max will mean the $$-ve$$ area needs to adjust for it, ie the $$-ve$$ area will also increase. But, in the space of all possible curves that could have the same area but $$-ve$$, there will be one that distributes it perfectly uniformly, ie a rectangle. And since the area is uniformly smeared, its value at any point will be less, thus, after cubing it will contribute less than the $$+ve$$ area which remains unchanged).

Collecting all of the x in A together (since integral is just a sum we can rearrange intervals) we can break $$\int_0^1 f(x)$$ into $$\int_0^a f(x) + \int_a^1 f(x)$$. Let value of $$f(x)$$ when its $$<0 = -c$$, the given condition of $$\int_0^1 f(x)dx=0$$ will then set the value of $$c=\frac{a}{1-a}$$. Doing the same with $$(f(x))^3$$, we find the area as a function of $$a$$, namely $$a-c^3(1-a)$$. We just need to find maxima of this function for $$a$$ in $$(0,1)$$ now, which can be done with simple calculus, and it comes out to be $$1/4$$ at $$a=1/3$$, which is the correct answer (though its totally possible to do this as is, I think calculation is easier with substituting $$a=1-a$$).

However, clearly my argument for why $$f(x)=1$$ for $$x$$ in $$A$$ is lacking for a formal proof. How can it be made rigorous? Or perhaps it is even wrong generally?

Edit: calling $$b$$ the measure of the negative part, to account for the case of $$b$$<$$a$$ (since then setting $$f(x)=1$$ for $$x$$ in $$A$$ could never satisfy $$\int_0^1 f(x)dx=0$$) we will instead have to find the maximum of $$|f(x)|$$, since each such function $$f_i(x)$$ satisfying our conditions and having $$b$$<$$a$$ would have a corresponding $$-f_i(x)$$ also satisfying our conditions which has $$b$$>$$a$$. However even then we will just have to break $$|f(x)|$$ into two cases, and the same exact logic will be used just twice.

• I don't mean to be obnoxious, but presumably there is more the problem says about $f$ beyond it being a real-valued function. If not, $f$ might not even be (Riemann) integrable (e.g., take $f$ to be the restriction of the indicator function on the rationals to $[0, 1]$). So, what else do we know about $f$? Sep 19, 2022 at 14:04
• "Changes signs" and "sum of lengths of the intervals" implies $f$ is continuous. Is that a requirement? Sep 19, 2022 at 14:05
• @ThomasAndrews what do you mean? The positive area will be compensated by the $x$ NOT in $A$, where the function will be $<0$ Sep 19, 2022 at 14:13
• @ThomasAndrews " 'Changes signs' and "sum of lengths of the intervals" implies f is continuous. Is that a requirement?" I've edited the first line so that it doesn't seem to imply continuity, and for the second part, I didn't understand how that implies it is continuous. We're just collecting all $x$ for which f has the same sign. That should be possible even in a weird function like $f(x)=0$ when x is rational and $=1$ otherwise, right (of course there the length of the interval for $=0$ will be $=0$, I suppose)? Sep 19, 2022 at 14:14
• "One to one correspondence" is probably the wrong term. There is a $1-1$ correspondence between $[0,1/1000000)$ and $[1/1000000,1].$ It's more complicated than a $1-1$ correspondence. Sep 19, 2022 at 14:38

Your observation that $$f(x)^3 ≤ f(x)$$ when $$0 ≤ f(x) ≤ 1$$ is good. A similar observation extends this idea to the entire range:

For all $$y ≤ 1$$, we have $$y^3 ≤ y^3 + \frac14(1-y)(1 + 2y)^2 = \frac14 + \frac34y$$. Therefore,

$$\int_0^1 f(x)^3\,dx ≤ \frac14 + \frac34\int_0^1 f(x)\,dx = \frac14.$$

This is the maximum because it can be achieved by

$$f(x) = \begin{cases} -\frac12 & \text{if 0 ≤ x < \frac23,} \\ 1 & \text{if \frac23 ≤ x ≤ 1}. \end{cases}$$

• Thank you, a similar approach had been used here- math.stackexchange.com/questions/3822324/… Sep 26, 2022 at 6:06
• I wanted to specifically see a way to solve it with my approach since it also gives the right answer but is too informal, so I'll wait to see if I can get another answer before accepting Sep 26, 2022 at 6:09
• @Amadeus The biggest problem with the approach you’ve written, even before trying to formalize it, is at “letting the +ve values everywhere got to max will mean the −ve area needs to adjust for it”: you’ve given no argument that such an adjustment won’t result in a net decrease of $∫_0^1 f(x)^3\,dx$. The argument you need to fill this hole is very similar to this one, and once you see it, you also see that it solves the problem completely by itself. Sep 27, 2022 at 22:14
• The argument is essentially that $x^3<x$ in $(0,1)$. Let's say $a=1/2$ and $f(x)=1$ for all $x$ in it. Then $b=1/2$ and $c=-1$ to compensate, and thus even on cubing $f(x)$ the total area $=0$ as before. But now slowly pull the slider on the value of $a$ and start reducing it. The $+ve$ area to compensate for keeps decreasing, while the length of $b$ available to use for compensating keeps increasing. So $|c|$ will certainly decrease pretty fast. Sep 28, 2022 at 4:34
• @Amadeus The hole I’m referring to happens before you can assume $f(x) = 1$ for $x ∈ A$—you’ve still given no justification for that assumption. Sep 28, 2022 at 8:20