Find a function satisfies an integrals Can we find an Lebesgue integrable function $f(x)$ satisfies the following integrals 
$\int_0^1 x^3f(x)= 2$ and $\int_0^1 (f(x))^4 dx = 125$ 
This is a qualifying question (measure theory) 
Or in general $\int_0^1 x^n f(x)=c_1$ ,$\int_0^1 (f(x))^{n+1}dx= c_2$
I think that always true we can take step function and construct it by solving equations  right? 
 A: At least for odd $n$ (so that $n+1$ is even), one cannot satisfy the given conditions for arbitrary $c_1, c_2$ (that is, $c_1, c_2$ will have to satisfy a certain condition in order for the conditions to be satisfiable). To see this, note by Hölder's inequality that
\begin{align*}
|c_1| &= \Big| \int_0^1 x^n \, f(x) \, d x \Big| \\
& \leq \Big( \int_0^1 (f(x))^{n+1} \, d x \Big)^{1/(n+1)} \cdot \Big( \int_0^1 x^{n+1} \, dx \Big)^{n/(n+1)} \\
& = c_2^{1/(n+1)} \cdot (n+2)^{-n/(n+1)},
\end{align*}
and hence $c_2^{1/(n+1)} \geq |c_1| \cdot (n+2)^{n / (n+1)}$. Since $n+1$ is even, this is equivalent to $c_2 \geq c_1^{n+1} \cdot (n+2)^n$.

Conversely, let $f_0 (x) = x \cdot (n+2) \cdot c_1$, so that $\int_0^1 x^n \, f_0 (x) \, d x = (n + 2) \cdot c_1 \cdot \int_0^1 x^{n+1} \, dx = c_1$. Note that $\int_0^1 (f_0(x))^{n+1} \, d x = (n+2)^{n+1} \cdot c_1^{n+1} \cdot \int_0^1 x^{n+1}\, d x = (n+2)^n \cdot c_1^{n+1} =: a$.
Choose any function $g \not \equiv 0$ satisfying $\int_0^1 x^n \, g(x) \, dx = 0$, and consider the Ansatz $f(x) = f_t (x) : = f_0(x) + t \, g(x)$ for $t \in \Bbb{R}$. At least if $n$ is odd (so that $n+1$ is even), it is not hard to see $\Gamma(t) := \int_0^1 (f_t(x))^{n+1} \, d x \to \infty$ as $t \to \infty$. Furthermore, $\Gamma$ is continuous, so that (by the intermediate value theorem) if $c_2 \geq a = \Gamma(0)$, then there is $t = t(c_2) \geq 0$ such that $\Gamma(t) = c_2$. Since also $\int_0^1 x^n \, f_t(x) \, d x = c_1$ for all $t$, this implies that the conditions are satisfiable if $n$ is odd and $c_2 \geq (n+2)^n \cdot c_1^{n+1}$.

Overall, we have shown for odd $n$ that your conditions are satisfiable if and only if $c_2 \geq (n+2)^n \cdot c_1^{n+1}$.

In the concrete case that you gave, this is not satisfied, since $n=3$ is odd but $c_2 = 125 < 2000 = (3+2)^3 \cdot 2^4 = (n+2)^n \cdot c_1^{n+1}$.
