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$\textbf{Question:}$ 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?

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  • $\begingroup$ You lost me going from $\frac{E_0}{E_k} \geq \frac{E_{n-k}}{E_n}$ to $x^{-k} \geq x^{n-2k}$. $\endgroup$
    – aschepler
    Apr 21, 2015 at 1:21
  • $\begingroup$ Ya this part is flawed $\endgroup$ Apr 21, 2015 at 1:25
  • $\begingroup$ My reasoning, which I know isn't entirely correct, is that the ratio would really only depend on the ratio of the integrals of the powers of $x$. $\endgroup$
    – user7090
    Apr 21, 2015 at 1:27

2 Answers 2

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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,

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

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  • $\begingroup$ Thought it was hölder's - just couldn't find the p and q. Thanks! Anyway this question is from a graduate qualifying exam. $\endgroup$
    – Ilham
    Apr 21, 2015 at 13:28
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    $\begingroup$ @Ilham, You are welcome. And thanks for letting me know where the question comes from. :) $\endgroup$
    – ki3i
    Apr 21, 2015 at 17:59
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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.

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