Maximizing $\int_0^1 f(x) \, {\rm d} x$ given $\int_0^1 x f(x) \, {\rm d} x = 0$

Define $F:[0, 1] \to \Bbb R$ as $$ F(x) = \int_0^x t f(t) \, dt = -\int_x^1 t f(t) \, dt \, . $$ The idea is to determine a sharp upper bound for $F(x)$ and then express $\int_0^1 f(x) \, dx$ in terms of $F$ to get a sharp upper bound for that integral.

From $-1 \le f(x) \le 1$ we get $$ |F(x)| \le \int_0^x t |f(t)| \, dt \le \frac 12 x^2 $$ and $$ |F(x)| \le \int_x^1 t |f(t)| \, dt \le \frac 12 (1-x^2) \, , $$ so that $$ \tag{$*$} |F(x)| \le \frac 12 \min(x^2, 1-x^2) $$ for all $ \in [0, 1]$.

The function $F$ is absolutely continuous with $F'(x) = x f(x)$ almost everywhere, therefore we can do integration by parts: For $0 < \epsilon < 1/\sqrt 2$ is $$ \int_\epsilon^1 f(x) \,dx = \int_\epsilon^1 \frac{F'(x)}{x} \,dx = - \frac{F(\epsilon)}{\epsilon} + \int_\epsilon^1 \frac{F(x)}{x^2} \, dx \\ \underset{(*)}{\le} \frac 12 \epsilon + \int_\epsilon^{1/\sqrt 2} \frac 12 \,dx + \int_{1/\sqrt 2}^1 \frac{1-x^2}{2x^2} \, dx \, . $$ Taking the limit $\epsilon \to 0$ gives $$ \int_0^1 f(x) \, dx \le \int_0^{1/\sqrt 2} \frac 12 \,dx + \int_{1/\sqrt 2}^1 \frac{1-x^2}{2x^2} \, dx = \boxed{\sqrt 2 - 1} \, . $$

The bound is sharp. Equality holds if $f(x) = 1$ for almost all $x \in [0, 1/\sqrt 2)$ and $f(x) = -1$ for almost all $x \in (1/\sqrt 2, 1]$.

I'm sure someone can convert this argument into a more mathematical one (currently it's essentially a manipulation of means)

Let $g(x) = f(x)+1$ for convenience. Then maximising $\int_0^1 f(x)dx = \int_0^1 (g(x) -1)dx$ is the same as maximising $\int_0^1 g(x)dx$.

Consider a rod of length 1 unit and mass $m$ units. Let $g(x)$ be the mass per unit length at distance $x$ from the left end. The given conditions imply:

1. mass per unit length cannot exceed $2$
2. $0=\int_0^1 xf(x)dx = \int_0^1 x(g(x)-1)dx \implies \int_0^1 xg(x) = 0.5$, i.e., the centre of mass of the rod lies at $0.5/m$.

Note that the centre of mass must have a minimum $x$ co-ordinate of $\frac 12 \left(\frac m2 \right)$ (half of mass divided by its maximum length density), where it achieves the minimum if all the mass is concentrated towards the left.

This gives $$0.5/m \ge m/4 \iff m \le \sqrt 2$$ This gives that the maximum value of $\int_0^1 f(x)dx$ is $\sqrt2 - 1$.
Corresponding to the conditions of the rod having all it's mass towards the left as much as possible, we also have the value of $f$ as in the OP.

This is Just a user's excellent argument, only written a bit differently.

Let $s = 1/\sqrt 2$ and $$ f(x) = \begin{cases} 1 & 0 \le x \le s \\ -1 & s < x \le 1 \end{cases} $$ be the suspected extremal function. It satisfies $\int_0^1 xf(x) \, dx = 0$ and $\int_0^1 f(x) \, dx = \sqrt 2 - 1$.

Any function $g:[0, 1] \to [-1, 1]$ satisfies $$ (s-x) \cdot (f(x)-g(x)) \ge 0 $$ for all $x \in [0, 1]$. If $g$ is also integrable with $\int_0^1 xg(x) \, dx = 0$ then $$ 0 \le \int_0^1 (s-x) \cdot (f(x)-g(x)) \, dx = s \left(\int_0^1 f(x) \,dx - \int_0^1 g(x) \, dx \right) $$ and therefore $\int_0^1 g(x) \, dx \le \int_0^1 f(x) \, dx = \sqrt 2 -1$. Equality holds if and only if $g(x) = f(x)$ for almost $x \in [0, 1]$.

Here is an intuitive geometric explanation:

If $\int_0^1 f<\int_0^1 g$, we have $A<B$, but then with the density $x$ (or any other increasing density $\delta$), the total mass of area $A$ should be even smaller than $B$, hence they cannot be balanced.

Now we write down a rigorous proof based on the above observation for the natural generalization:

Let $\delta(x)$ be an arbitrary density function that's positive almost everywhere and increasing over $[a, b]$, then there exists a unique $s\in [a,b]$, such $\int_a^s \delta(x)dx=\int_s^b \delta(x)dx$. We have

$$\max \int_a^b f(x)dx \text{ subject to } \begin{cases} \int_a^bf(x)\delta(x)=0 \\ |f(x)|\le 1 , \forall x\in [a,b] \end{cases}$$

is achieved by $$f(x)=\begin{cases} 1 & a\le x \le s \\ -1 & s<x\le b \end{cases}$$

Note that since $|f|$ is bounded, we no longer have to assume $f$ is integrable, but only measurable. Also, we can even further generalize by replacing $dx$ with another measure, and $\delta(x)$ is the Radon–Nikodym derivative of two measures.

Suppose there exists admissible $g$ such that $F[g]>F[f]$, let $h:=f-g$, we have $$F[h]<0, \int_a^b h(x)\delta(x)=0, \text{ and } \begin{cases} h(x)\ge 0 & a\le x \le s \\ h(x)\le 0 & s<x\le b \end{cases}$$

Now we have a contradiction: $$\begin{align}\int_a^b h(x)\delta(x) &= \int_a^s |h|\delta(x) - \int_s^b |h|\delta(x) \\ &\le \delta(s)\int_a^s|h|-\delta(s)\int_s^b|h| \\ &=\delta(s)F[h]<0\end{align}$$

Remark: I started with the observation that the constraint domain is actually convex, and the functional $F[f]:=\int_a^b f(x)dx$ to be optimized is linear/convex, then drew the above image to prove $f(x)$ achieves a local maximum (for the $L^\infty$ norm, or whatever norm that's convenient). But the solution turned out to be simple and doesn't depend on any result from convex optimization or norm, especially the modified version given by Martin R in the comment.

We prove the existence of a maximum for your functional, and we show that the function you found is indeed the only point of maximum.

Existence of a maximum.

For each measurable $f\colon [0,1]\rightarrow [-1,1]$ we can associate the Lipschitz function $F\colon [0,1]\rightarrow [-1,1]$ given by

$$F(x):=\int_0^x f(t) \ dt$$

Note that

1. $F(0)=0$,
2. $|F(x)-F(y)|\leq |x-y|$,

Moreover every function $F$ satisfying $1-2$ is associated to an unique function $f\in L^\infty$ given by $f=F'$, with $f(x)\in [-1,1]$ almost everywhere. One can see that

$$\int_0^1 f(t) \ dt=F(1),$$

$$\int_0^1 t f(t) \ dt = \int_0^1 t F'(t) \ dt = F(1) - \int_0^1 F(t) \ dt.$$

Consider the condition

3. $F(1) - \int_0^1 F(t) \ dt =0$.

Let $\mathcal{F}$ be the family of functions that satisfies $1-2-3$. By Ascoli-Arzelá Theorem this is a compact subset of functions in $C^0[0,1]$. So the continuous linear functional

$$F\in C^0[0,1]\mapsto F(1)$$

has points of maximum in $\mathcal{F}$. Since $F\in \mathcal{F}\implies -F \in \mathcal{F}$, it is easy to see that the maximum value is no negative.

Note that $F_0$ is a point of maximum of this problem if and only if $f_0=F_0'$ is a point of maximum of your original problem.

Finding the points of maximum: there is only one point of maximum.

Let $F_0$, with $f_0=F_0'$, be a point of maximum.

Claim 1. We claim that $|f_0|=1$ almost everywhere.

Otherwise there is $\epsilon\in (0,1]$ such that $$m(\Omega_\epsilon)> 0,$$ where $$\Omega_\epsilon:=\{x\in [0,1]\colon f_0(x)\in [-1+\epsilon,1-\epsilon] \}.$$

Then there is $c\in (0,1)$ such that

$$\int_0^c t 1_{\Omega_\epsilon}(t) \ dt = \int_c^1 t 1_{\Omega_\epsilon}(t) \ dt$$

Define

$$h = \frac{\epsilon}{2} \left( 1_{\Omega_\epsilon \cap [0,c]}- 1_{\Omega_\epsilon \cap [c,1]}\right).$$

Then

$$\int_0^1 t h(t) dt =0,$$

$$\int_0^1 h(t) dt > 0,$$

$g(x):=f_0(x)+h(x)\in[-1,1]$ and

$$\int_0^1 g(t) dt > \int_0^1 f(t) dt,$$

that contradicts the maximality of $f_0$. This proves the claim.

Claim 2. There is $c\in (0,1)$ such that

$$\{x\in [0,1]\colon f_0(x)=1\}=[0,c]$$

and

$$\{x\in [0,1]\colon f_0(x)=-1\}=[c,1]$$

up to subsets of zero Lebesgue measure.

Indeed, suppose this does not occur. Then there are two disjoint subsets $A$ and $B$ with positive Lebesgue measure and $d\in (0,1)$ such that

$A\subset [0,d]$, $B\subset [d,1]$, $f_0(x)=-1$ on $A$, and $f_0(x)=1$ on $B$. Reducing $A$ and $B$ if necessary we can assume that

$$\int t 1_A(t) \ dt = \int t 1_B(t) \ dt > 0$$.

Note that $m(A) > m(B)$. Then

$$h = \frac{1}{2} \left(1_A- 1_B\right)$$

satisfies

$$\int_0^1 t h(t) \ dt =0,$$

$$\int_0^1 h(t) \ dt > 0,$$

$g(x) = f_0(x)+ h(x)\in [-1,1]$ and we have

$$\int_0^1 g(t) \ dt > \int_0^1 f_0(t) \ dt,$$

that contradicts the maximality of $f_0$.

This concludes the proof of the second claim. Now it is easy to show that the only possible choice for $c$ in the second claim is $c=1/\sqrt{2}.$ So there is only one point of maximum, and the maximum value is $\sqrt{2}-1$.

Final Remark. The argument seems to be fairly general. For instance we could replace the condition

$$\int_0^1 t f(t) \ dt=0$$

by the condition

$$\int_0^1 \alpha(t) f(t) \ dt=0,$$

where $\alpha\colon [0,1]\rightarrow \mathbb{R}$ is continuous, positive and strictly monotone increasing. In this case

$$\int_0^1 \alpha(t) f(t) \ dt = F(1)\alpha(1)- F(0)\alpha(0)- \int_0^1 F(t) \ d\alpha,$$

where the last integral is a Riemann–Stieltjes integral.

Then we can prove the existence of maximum and Claim 1 and Claim 2 using analogous arguments. The uniqueness and nature of the point of maximum follows, that is, the unique point of maximum is

$$f_c= 1_{[0,c]}- 1_{(c,1]},$$

where $c$ is the only $c\in (0,1)$ satisfying

$$\int \alpha(t)f_c(t) \ dt =0.$$

$$ \begin{array}{ll} \underset{f : [0,1] \to [-1,1]}{\text{maximize}} & \displaystyle\int_0^1 f(x) \, {\rm d} x \\ \text{subject to} & \displaystyle\int_0^1 x f(x) \, {\rm d} x = 0 \end{array} $$

Introducing a Lagrange multiplier $\mu$,

$$ \int_0^1 f(x) \, {\rm d} x + \mu \int_0^1 x f(x) \, {\rm d} x = \int_0^1 (1 + \mu x) f(x) \, {\rm d} x $$

Since $f (x) \in [-1,1]$, maximizing the integrand pointwise, we obtain the bang-bang solution$^\color{magenta}\star$

$$ f(x) = \begin{cases} +1 & \text{if} & 1 + \mu x > 0 \\ -1 & \text{if} & 1 + \mu x < 0 \\ \text{any value} \in [-1,1] & \text{if} & 1 + \mu x = 0 \end{cases} $$

with at most one switch, at $x = -\frac{1}{\mu}$, if $x_{\star} := -\frac{1}{\mu} \in [0,1]$, i.e., if $\mu < -1$. Whoever happens to be acquainted with optimal control is (almost) surely acquainted with such solutions. There are two cases to consider:

• If $\mu \geq -1$, there is no switch. From the equality constraint, $$0 = \int_0^1 x f(x) \, {\rm d} x = \int_0^1 x \, {\rm d} x = \frac12 \neq 0$$ Thus, the switchless case is an impossibility.

• If $\mu < -1$, there is a single switch at $x_{\star} := -\dfrac{1}{\mu} \in [0,1]$. From the equality constraint, $$0 = \int_0^1 x f(x) \, {\rm d} x = \int_0^{x_\star} x \, {\rm d} x - \int_{x_\star}^1 x \, {\rm d} x = x_\star^2 - \frac12 $$ Thus, the single switch occurs at $\color{blue}{x_\star := \dfrac{1}{\sqrt{2}}}$. Also, $\mu = -\sqrt{2}$.

Thus, the maximum is

$$ \int_0^1 f(x) \, {\rm d} x = \int_0^{x_\star} {\rm d} x - \int_{x_\star}^1 {\rm d} x = 2 x_\star - 1 = \frac{2}{\sqrt{2}} - 1 = \color{blue}{\sqrt{2} - 1} $$

and the maximizing bang-bang solution is

$$ f(x) = \begin{cases} +1 & \text{if} & x \in [0, x_\star) \\ \,\,\,\xi & \text{if} & x = x_\star \\ -1 & \text{if} & x \in (x_\star, 1] \end{cases} $$

where $\xi \in [-1,1]$ is arbitrary.

_{$\color{magenta}\star$ Lawrence C. Evans, An Introduction to Mathematical Optimal Control Theory, UC Berkeley.}

In the following, we use $$ \mu\left(T\right) = \int_T 1 dx $$ for a measurable set $T\subseteq [0,1].$

In order to reduce typing effort, let $s=\frac{\sqrt{2}}{2}.$ Let $$ f(x) = \begin{cases} \phantom{-}1 & \text{if} & x\in[0,s] \\ -1 & \text{if} & x\in(s,1] \end{cases} $$ that is, the function proposed by the OP.

Now let $g$ be a function that deviates from $f$ in $[0,s]$ in a way that affects the integrals. That means, there is a set $S_1\subset [0,s]$ such that $g(x)\leq 1-\varepsilon_1$ with $\varepsilon_1 >0$ for $x\in S_1,$ and $S_1$ has positive measure $\int_{S_1}1dx.$

From $\int_0^1 xg(x)dx =0$ it follows $$ \int_0^s xg(x)dx = - \int_s^1 xg(x)dx $$ which means that there also must be a measurable deviation between $g$ and $f$ in $(s,1].$

There is a set $S_2\subset (s,1]$ such that $g(x)\geq -1+\varepsilon_2$ with $\varepsilon_2 >0$ for $x\in S_2,$ and $S_2$ has positive measure $\int_{S_2}1dx.$

Now we select subsets $T_1\subseteq S_1$ and $T_2\subseteq S_2$ with $$ \int_{T_1}xdx = \int_{T_2}xdx =: d > 0. $$ As all elements in $T_1$ are smaller than or equal to $s$, we have $$ d = \int_{T_1}xdx \leq \int_{s-\mu(T_1)}^s xdx = \mu(T_1)s-\frac{1}{2}\mu^2(T_1) $$ As all elements in $T_2$ are greater than or equal to $s$, we have $$ d = \int_{T_2}xdx \geq \int_s^{s+\mu(T_2)} xdx = \mu(T_2)s+\frac{1}{2}\mu^2(T_2) $$ From that, it follows $$ \mu(T_1) - \mu(T_2) \geq \frac{\mu^2(T_1)+\mu^2(T_2)}{2s} =: m $$ with $m >0.$

We set $\varepsilon =\min(\varepsilon_1,\,\varepsilon_2).$

We create a new function as follows: $$ h(x)= \begin{cases} g(x)+\varepsilon & \text{if} & x\in T_1 \\ g(x)-\varepsilon & \text{if} & x\in T_2 \\ g(x) & \text{if} & x\in [0,1]\setminus (T_1 \cup T_2) \end{cases} $$ Due to the choice of $\varepsilon$, the image of this function is still in $[-1,1].$

Then $$ \int_0^1 h(x) dx= \int_0^1 g(x) dx+ \int_{T_1} \varepsilon dx -\int_{T_2} \varepsilon dx \\ \geq \int_0^1 g(x) dx + \varepsilon m $$ which is really greater than $\int_0^1 g(x)dx$, and $$ \int_0^1 xh(x) dx= \int_0^1 xg(x) dx+ \int_{T_1} x\varepsilon dx -\int_{T_2} x\varepsilon dx \\ = \int_0^1 xg(x) dx + \varepsilon d- \varepsilon d = \int_0^1 xg(x) dx = 0 $$ This shows: Whenever we have a function $g$ that deviates from $f$ (in a measurable way), we can construct a function $h$ that yields a better value of the objective function. Therefore, $f$ must be optimal.

Regarding this answer as an experimental verification...

After discretizing, this problem can be handled via linear programming. Consider the discretized version

$$ \min_{a_k}\sum_{k=0}^{k=N}a_k, \ \ \text{s.t.}\ \ \cases{\sum_{k=0}^{k=N}\frac kN a_k = 0\\ -1\le a_k \le 1},\ k = 0,\cdots,N $$

The formulation is straightforward with MATHEMATICA:

n = 200;
m0 = Table[k/n, {k, 0, n}];
m[k_] := Table[If[j == k, 1, 0], {j, 0, n}];
Flatten[Table[{m[k], -m[k]}, {k, 0, n}], 1];
c = Table[-1, {k, 0, n}];
M = Flatten[Table[{m[k], -m[k]}, {k, 0, n}], 1];
b = Flatten[Table[{1, 1}, {k, 0, n}], 1];
A = Table[a[k], {k, 0, n}];
constr = Join[{A . m0 == 0}, Thread[M . A + b >= 0]];
sol = Maximize[{Total[A]/n, constr}, A]
ListLinePlot[Transpose[Join[{m0}, {A /. sol[[2]]}]]]

We reformulate the problem a little bit by doing a change of variables $y=x^2$ in the integration dummy variable and then we make the problem symmetric around the origin. Then we end up with the following problem: "among the integrable functions $f:[-1,1]\to [-1,1]$ for which $\int_{-1}^{+1}dy\,f(y)=0$, maximize $\int_{-1}^{+1}dy\,|y|^{-1/2}f(y)$."

We now reformulate the problem once more to redirect the range of $f$ to the interval $[0,1]$ in stead of $[-1,1]$. We do this by doing the substitution $f_{new}=\frac{f_{old}+1}{2}$. The problem then becomes... "among the integrable functions $f:[-1,1]\to [0,1]$ for which $\int_{-1}^{+1}dy\,f(y)=1$, maximize $\int_{-1}^{+1}dy\,|y|^{-1/2}f(y)$."

From the Hardy-Littlewood inequality, we have $$\int_{-1}^{+1}dy\,|y|^{-1/2}f(y)\leq \int_{-1}^{+1}dy\,\{|.|^{-1/2}\}^*(y)f^*(y)=\int_{-1}^{+1}dy\,|y|^{-1/2}f^*(y)$$ so we know the maximum is to be found among the symmetrically non-increasing functions. Assuming $f:[-1,1]\to[0,1]$ is indeed non-increasing, one can easily test its performance against that of $\chi_{[-\frac{1}{2},\frac{1}{2}]}$ and check that the latter candidate "wins".

Remark: the use of Hardy-Littlewood is not pedantic showing off of math skills. Rearrangement lies at the core of this problem (I think) and is a worthy subject of study, so better not re-invent the wheel with some "more elementary" approach.

Although the existing answers are pretty clever and mathematically correct, if I were a professor teaching a physics course and I gave out this problem, I would expect (hope?) the students would attack it using the Euler-Lagrange equations. (That is, use the more general technique of the calculus of variations.)

Surely the professor wants to guide you into a working understanding of the use of Lagrangians, and has provided this somewhat contrived problem as a gentle way of letting you see how this works.