Fix \(0\le a\le1\). Determine all nonnegative continuous functions \(f\) on \([0,1]\) satisfying \[ \int_0^1 f(x)\,dx=1, \qquad \int_0^1 x f(x)\,dx=a, \qquad \int_0^1 x^2f(x)\,dx=a^2. \]
Solution
1. Any function satisfying the three displayed integral conditions would satisfy\[ \int_0^1 (x-a)^2f(x)\,dx=0. \]
Expand the square and use the three hypotheses: \[ \begin{aligned} \int_0^1 (x-a)^2f(x)\,dx &= \int_0^1x^2f(x)\,dx -2a\int_0^1xf(x)\,dx +a^2\int_0^1f(x)\,dx\\ &= a^2-2a^2+a^2\\ &=0. \end{aligned} \]
2. If \(f\) satisfied the hypotheses, then\[ (x-a)^2f(x)=0 \]for every \(x\in[0,1]\).
The function \[ g(x)\coloneqq(x-a)^2f(x) \] is continuous and nonnegative on \([0,1]\). By step 1 its integral is \(0\). If \(g(x_0)>0\) at some point, continuity would make \(g\) positive on a nontrivial interval relative to \([0,1]\), forcing its integral to be positive. Hence \(g\equiv0\).
3. Any such \(f\) would be identically zero.
For every \(x\neq a\), step 2 and \((x-a)^2>0\) give \[ f(x)=0. \] Thus \(f\) vanishes on \([0,1]\setminus\{a\}\). Since \(f\) is continuous, taking points \(x\neq a\) tending to \(a\) from within \([0,1]\) gives \[ f(a)=0. \] This also covers the endpoint cases \(a=0\) and \(a=1\) by one-sided convergence. Therefore \(f\equiv0\).
4. No nonnegative continuous function satisfies all three conditions.
Step 3 would imply \[ \int_0^1f(x)\,dx=0, \] contradicting the required value \(1\).
5. Q.E.D.
The set of functions requested in the problem is empty by step 4.