If $u\in L^1(0,1)$ is nonnegative and $E_n = \int_0^1 x^n u(x) \, dx$, prove $E_{n-k} E_k \leq E_0 E_n$. $\textbf{Question:}$ Let $ u \in L^1(0,1)$ be a nonnegative function. Define $$E_n := \int_0^1 x^n u(x) dx$$ Prove the following inequality, $\forall n \ge 0$, and $\forall k \in [0,n]$, we have $$ E_{n-k} E_k \le E_0 E_n$$
$\textbf{My Attempt:}$ We have, $$
E_0 := \int_0^1 u(x) dx$$
$$E_1= \int_0^1 x u(x) dx$$
$$E_2 = \int_0^1 x^2 u(x) dx$$
Since $x \in (0,1)$, we have $$E_0 \ge E_1 \ge E_2 \dots$$
Thus $$ E_n E_0 \ge E_n E_1 \ge E_n E_2 \ge \dots $$
To show that $E_0 E_n \ge E_k E_{n-k}$, we must show that $$\frac{E_0}{E_k} \ge \frac{E_{n-k}}{E_n}$$
This is equivalent to show that $x^{-k} \ge x^{n-2k}$ which is true as lons as $-k \le n-2k$, which is whenever $k \ge n$. 
Is the above proof correct?
 A: Holder's Inequality:
Assume, wlog, that $k\geqslant n-k$.
$$
\begin{align}
\Big(\int_0^1 x^{n-k}u~\text dx\Big)\Big(\int_0^1x^k u~\text dx\Big) &= \Big(\int_0^1 (x^k)^{\frac{n-k}{k}}u~\text dx\Big)\Big(\int_0^1x^k u~\text dx\Big)\\ 
&\leqslant \Big(\int_0^1 u~\text dx\Big)^{2-\frac{n}{k}}\Big(\int_0^1 x^ku~\text dx\Big)^{\frac{n-k}{k}}\Big(\int_0^1x^k u~\text dx\Big) \\
& =  \Big(\int_0^1 u~\text dx\Big)^{2-\frac{n}{k}}\Big(\int_0^1 (x^n)^{\frac{k}{n}}u~\text dx\Big)^{\frac{n}{k}} \\
& \leqslant \Big(\int_0^1 u~\text dx\Big)^{2-\frac{n}{k}}\Big(\int_0^1 x^nu~\text dx\Big)\Big(\int_0^1 u~\text dx\Big)^{(1-\frac{k}{n})\frac{n}{k}} \\
& = \Big(\int_0^1 u~\text dx\Big)\Big(\int_0^1 x^nu~\text dx\Big)\,.
\end{align}
$$
Therefore,

$$\Big(\int_0^1 x^{n-k}u~\text dx\Big)\Big(\int_0^1x^k u~\text dx\Big)\leqslant  \Big(\int_0^1 u~\text dx\Big)\Big(\int_0^1 x^nu~\text dx\Big)$$

A: Recall Hölder's inequality for $f$, $g$ non-negative and $\alpha\in[0,1]$:
$$
\int f^\alpha g^{1-\alpha}\le\left(\int f\right)^\alpha\left(\int g\right)^{1-\alpha}\tag{*}
$$
To relate $E_k$ to $E_n$ and $E_0$ using (*), the obvious thing to try is to find $\alpha$ such that
$
\int_0^1 x^ku=\int_0^1 (x^nu)^\alpha (u)^{1-\alpha}$.
The only choice is $\alpha=k/n$. Hölder then gives
$$
E_k
\le E_n^{k/n} E_0^{1-k/n}.\tag1
$$
Apply (1) with $k$ replaced by $n-k$ to obtain
$$
E_{n-k}\le E_n^{1-k/n}E_0^{k/n}.\tag2
$$
Multiplying (1) and (2) gives the result.
