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.

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    $\begingroup$ 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 $f$? $\endgroup$ Sep 19, 2022 at 14:04
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    $\begingroup$ "Changes signs" and "sum of lengths of the intervals" implies $f$ is continuous. Is that a requirement? $\endgroup$ Sep 19, 2022 at 14:05
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    $\begingroup$ @ThomasAndrews what do you mean? The positive area will be compensated by the $x$ NOT in $A$, where the function will be $<0$ $\endgroup$
    – Amadeus
    Sep 19, 2022 at 14:13
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    $\begingroup$ @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)? $\endgroup$
    – Amadeus
    Sep 19, 2022 at 14:14
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    $\begingroup$ "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. $\endgroup$ Sep 19, 2022 at 14:38

1 Answer 1


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

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

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    $\begingroup$ Thank you, a similar approach had been used here- math.stackexchange.com/questions/3822324/… $\endgroup$
    – Amadeus
    Sep 26, 2022 at 6:06
  • $\begingroup$ 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 $\endgroup$
    – Amadeus
    Sep 26, 2022 at 6:09
  • $\begingroup$ @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. $\endgroup$ Sep 27, 2022 at 22:14
  • $\begingroup$ 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. $\endgroup$
    – Amadeus
    Sep 28, 2022 at 4:34
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    $\begingroup$ @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. $\endgroup$ Sep 28, 2022 at 8:20

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