Let $b>a>0$ and $M>0$ be fixed. Let $F$ be the set of all functions $f:[a,b]\to[-M,M]$ such that $$\int_a^bf(x)\,\mathrm dx=0.$$Find$$\max_{f\in F}\int_a^b\frac{f(x)}x\,\mathrm dx.$$

I tried to use the CBS inequality for integrals for the functions $(f(x))^2$ and $\dfrac{1}{x^2}$, but when I tried to find the equality case, the function $f(x)=\dfrac{k}{x}$ I got did not satisfy $\displaystyle\int_a^bf(x)\,\mathrm dx=0$. So the maximum must be lower than what I found. But I do not know anything about the continuity of $f$ and I don't know how to continue.

  • $\begingroup$ I would guess that the maximum is reached when $f$ is the function equal to $-M$ on $[a, (a+b)/2]$ and equal to $M$ on $[(a+b)/2, b]$. $\endgroup$ Mar 12, 2019 at 17:13
  • $\begingroup$ Wouldn't it be the other way round @TheSilverDoe ? When you weight $f$ by $1/x$ the first half interval has higher weight and should be assigned $+M$, not $-M$, right? $\endgroup$ Mar 12, 2019 at 17:18
  • $\begingroup$ Yes, my mistake. $\endgroup$ Mar 12, 2019 at 20:09
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    $\begingroup$ Not sure how you applied CBS, but you need not have $f=k/x$. You can have $f=k_1/x$ for $x\in (a,c)$ and $f=-k_2/x$ for $x\in (c,b)$ in the maximal case, for suitable choices of $c,k_1,k_2$, for eg. As $f$ has no continuity restrictions, you can have stranger possibilities- where did you get this problem? $\endgroup$
    – Macavity
    Mar 13, 2019 at 1:03

2 Answers 2


$\def\d{\mathrm{d}}\def\peq{\mathrel{\phantom{=}}{}}$For any fixed $f \in F$, define$$ g(x) = \int_a^x f(u) \,\d u, \quad \forall x \in [a, b] $$ then $g' = f$ almost everywhere and $g(a) = g(b) = 0$. Since $|f(x)| \leqslant M$, then for any $a \leqslant x \leqslant \dfrac{a + b}{2}$,$$ g(x) = \int_a^x f(u) \,\d u \leqslant M (x - a), $$ and for any $\dfrac{a + b}{2} < x \leqslant b$,$$ g(x) = g(b) - \int_x^b f(u) \,\d u \leqslant M(b - x). $$ Therefore by integration by parts,$$ \int_a^b \frac{f(x)}{x} \,\d x = \int_a^b \frac{g'(x)}{x} \,\d x = \left. \frac{g(x)}{x} \right|_a^b + \int_a^b \frac{g(x)}{x^2} \,\d x = \int_a^b \frac{g(x)}{x^2} \,\d x\\ = \int_a^{\tfrac{a + b}{2}} \frac{g(x)}{x^2} \,\d x + \int_{\tfrac{a + b}{2}}^b \frac{g(x)}{x^2} \,\d x \leqslant M \int_a^{\tfrac{a + b}{2}} \frac{x - a}{x^2} \,\d x + M \int_{\tfrac{a + b}{2}}^b \frac{b - x}{x^2} \,\d x = M \ln\frac{(a + b)^2}{4ab}. $$

Note that the above inequality becomes equality for$$ f(x) = \begin{cases}M;& a \leqslant x \leqslant \dfrac{a + b}{2}\\-M;& \dfrac{a + b}{2} < x \leqslant b\end{cases}, $$ thus the maximum is $M \ln\dfrac{(a + b)^2}{4ab}$.

  • $\begingroup$ (+1) the final expression was rather unexpected for me $\endgroup$ Apr 25, 2019 at 8:35
  • $\begingroup$ As sanity checks this is (i) $\propto M$ and (ii) $a\leftrightarrow b$ symmetric, and (iii) it goes without saying, of course, that this maximum is positive by e.g. the AM-GM inequality, as it should be because I can change $f$ to $-f$. $\endgroup$
    – J.G.
    Apr 30, 2019 at 6:29

While this answer I cant claim is complete proof, I will share it, maybe others can complete it or find the ideas useful.

I would approach this using calculus of variations. However, if $f(x)$ must be continuous, I think the calculus of variations would dictate $f(x) \equiv 0$ (and the functional won't be a maximum). So, let us assume that $f$ is discontinuous at a point $ a < c < b$, such that: $$ f(x) = \begin{cases} f_1(x), & a \leq x \leq c \\ f_2(x), & c < x \leq b\end{cases}. $$

Our functional reads: $$ I_{f_1,f_2} = \int_a^b \dfrac{f(x)}{x} dx = \int_a^c \frac{f_1(x)}{x} dx + \int_c^b\frac{f_2(x)}{x} dx$$ and when we write the stationary (extremum) condition of $I$ with respect to independent variations of $f_1, f_2$ we get $\delta f_1 = \delta f_2 = 0$ (so they are constant functions). Let us assume now:

$$ f(x)/M = \begin{cases} k_1 , & a \leq x \leq c \\ -k_2, & c < x \leq b\end{cases}. $$

From the constraint $\int_a^b f(x) dx=0$, we get: $c = \dfrac{k_1 a + k_2 b}{k_1+k_2}$. We ask that $a < c < b$, so $0 < k_{1,2} \leq 1$ (because $\vert f(x) \vert \leq M$). Not all pairs $(k_1,k_2)$ need to be studied (see the scaling relation in the following, to keep in mind).

All that is left to do is to compute the functional and maximize it with respect to $k_1, k_2$. We have: $$I(k_1,k_2)/M = k_1 \ln \frac{c}{a} + k_2 \ln \frac{c}{b} = \ln \left[ \left(\frac{c}{a}\right)^{k_1} \left(\frac{c}{b}\right)^{k_2} \right],$$ with $c=c(k_1,k_2)$ given above.

Note that $I(\lambda k_1, \lambda k_2) = I^\lambda(k_1,k_2)$. This allows us to simplify the search for $k_1, k_2$ to obtain maximum value. Namely, we only could look for $0 < t < 1$ at $I(1,t)$, $I(t,1)$, and $I(1,1)$ to understand the behavior for all values $(k_1,k_2)$ (using the scaling relation above, we can scale the arguments down by the one that is the biggest). Turns out (I did plotting only) that the maximum is attained for $k_1=k_2=1$. This is a bit of a hand-waving proof, maybe somebody else can complete this point.

With the above in mind, the maximum value of the functional is the same as one obtained by Saad above: $$ \text{max } I_{f_1,f_2} = M \ln\dfrac{(a + b)^2}{4ab}.$$

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    $\begingroup$ Unless you have nonnegativity of $f(x)$ along with its continuity, you cannot conclude its integral being zero implies that it is almost always zero. For example, let $a=1, b=3$ and $f=x-2$. $\endgroup$
    – Sam
    May 18, 2020 at 6:28

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