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Let $0<f(x)<1$ be a smooth periodic function defined on $[0,2\pi]$. Is it possible to prove the following inequality,

$ \overline{f^{-3}}\cdot \overline{f}\ge\overline{f^{-2}}$

where $\overline{f}:=\frac{1}{2\pi}\int_0^{2\pi}f\,\mathrm{d}x$, or even a more general case,

$ \overline{f^{-n}}\cdot \overline{f}\ge\overline{f^{-n+1}}$ for $n\ge3\in\mathbb{Z}$

Do I need more information on $f$ ? Thanks.

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3 Answers 3

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The standard trick is putting distinct variables in the product of averages and symmetrizing: \begin{align*} \overline{f^{-n}} \cdot \overline{f} &= \frac1{2\pi}\int_{x=0}^{2\pi} f^{-n}(x) \mathrm{d}x \cdot \frac1{2\pi}\int_{y=0}^{2\pi} f(y) \mathrm{d}y = \\ &= \frac1{8\pi^2} \int_{x=0}^{2\pi} \int_{y=0}^{2\pi} \Big( f^{-n}(x) f(y) + f(x) f^{-n}(y) \Big) \mathrm{d}x \mathrm{d}y \end{align*} and similarly $$ \overline{f^{-n+1}} = \frac1{8\pi^2} \int_{x=0}^{2\pi} \int_{y=0}^{2\pi} \Big( f^{-n+1}(x) + f^{-n+1}(y) \Big) \mathrm{d}x \mathrm{d}y. $$ For the comparison of the two, you only need to verify $$ a^{-n}b + ab^{-n} - a^{-n+1}+b^{-n+1} = a^{-n}b^{-n} (a-b)(a^n-b^n) \ge 0 $$ for $a=f(x)>0$ and $b=f(y)>0$.

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Let $g, h \in C[0,L]$ be any continuous functions defined on interval $[0,L]$.
For any $t \in \mathbb{R}$, consider following integral:

$$\mathcal{I}(t) \stackrel{def}{=} \frac1L \int_0^L g(x) e^{th(x)} dx$$

Recall for any $0 < |x| < 1$, we have $\displaystyle\;\left|\frac{e^x - 1}{x} - 1\right| \le |x|$.

Let $M = \max( 1, \| h\|_\infty )$ where $\| h \|_\infty = \sup\limits_{x\in [0,L]} |h(x)|$. For any $0 < |\delta| < \frac{1}{M}$, we have

$$\begin{align} \left|\frac{\mathcal{I}(t + \delta) - \mathcal{I}(t)}{\delta} - \frac{1}{L}\int_0^L g(x)h(x)e^{th(x)} dx\right| & \le \frac{1}{L}\int_0^L |g(x)|e^{th(x)}\left|\frac{e^{\delta h(x)} - 1}{\delta} - h(x)\right| dx\\ & \le \frac{\delta}{L} \int_0^L |g(x)| h(x)^2 e^{th(x)} dx\end{align} $$ Since $\delta$ can be arbitrary small, this implies as a function of $t$, $\mathcal{I}(t)$ is differentiable and

$$\mathcal{I}'(t) = \frac{1}{L}\int_0^L g(x)h(x) e^{th(x)} dx$$

Repeat apply this to the integral

$$\mathcal{J}(t) \stackrel{def}{=} \frac{1}{2\pi}\int_0^{2\pi} f(x)^t dx$$

We find $\mathcal{J}(t)$ is a smooth function in $t$ and

$$\frac{d^k}{dt^k} \mathcal{J}(t) = \frac{1}{2\pi}\int_0^{2\pi} (\log f(x))^k f(x)^t dx\quad\text{ for } k = 1, 2, 3 \ldots$$

Let $\mu = \frac{\mathcal{J}'(t)}{\mathcal{J}(t)}$, it is easy to deduce

$$\frac{d^2}{dt^2} \log\mathcal{J}(t) = \frac{\mathcal{J}''(t)}{\mathcal{J}(t)} - \left(\frac{\mathcal{J}'(t)}{\mathcal{J}(t)}\right)^2 = \frac{1}{2\pi\mathcal{J}(t)}\int_0^{2\pi}\left( \log f(x) - \mu \right)^2 f(x)^t dt \ge 0$$ This implies $\log\mathcal{J}(t)$ is a convex function in $t$. As a result,

$$\log\mathcal{J}(s-1) + \log\mathcal{J}(s+1) \ge 2\log\mathcal{J}(s)\quad\text{ for all } s$$ Summing this for $s = -n+1, -n+2, \ldots, 0$ and cancelling terms on both sides, we obtain

$$\log\mathcal{J}(-n) + \log\mathcal{J}(1) \ge \mathcal{J}(-n+1) + \log\mathcal{J}(0) \iff \mathcal{J}(-n)\mathcal{J}(1) \ge \mathcal{J}(-n+1)\mathcal{J}(0)\\ \iff \overline{f^{-n}}\cdot \overline{f} \ge \overline{f^{-n+1}}\cdot\overline{f^0} = \overline{f^{-n+1}}$$

Update

To extend this to non-integer $n$, arrange the two set of numbers $\{ -n, 1 \}$ and $\{ -n+1, 0 \}$ into non-increasing lists: $(x_1, x_2) = ( 1, -n )$ and $(y_1, y_2) = ( 0, -n+1 )$.

Since they satisfy $\displaystyle\; \begin{cases} x_1 \ge y_1\\ x_1 + x_2 = y_1 + y_2, \end{cases}$ the finite sequence $(x_k)$ majorizes $(y_k)$.

By Karamata's inequality, we have $$\sum_k\log\mathcal{J}(x_k) \ge \sum_k\log\mathcal{J}(y_k) \implies \log\mathcal{J}(-n) + \log\mathcal{J}(1) \ge \mathcal{J}(-n+1) + \log\mathcal{J}(0) $$ and the assertion for non-integer $n \ge 1$ follows.

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Suppose $\mu$ is a positive measure on $X$ with $\mu(X) = 1.$ Let $f:X\to [a,b]$ be measurable, where $a>0.$ Then for any $n\in \mathbb N,$

$$\tag 1 \smallint f\cdot \smallint f^{n-1} \le \smallint f^n.$$

(The notation $\smallint $ means $\smallint_X \, d\mu.$) Proof: Holder shows

$$\smallint f \le (\smallint f^n)^{1/n},\,\,\, \smallint f^{n-1} \le (\smallint f^n)^{(n-1)/n}.$$

The result in $(1)$ follows. From $(1)$ we get

$$\tag 2 \smallint f^{n-1} \le \smallint f^{n}\cdot\frac{1}{\smallint f} \le \smallint f^{n}\cdot \smallint \frac{1}{f}.$$

The second inequality in $(2)$ follows from Jensen, using $g(\smallint f)\le \smallint g\circ f,$ with $g(x) = 1/x.$

In your problem, we are in the setting $X=[0,2\pi], d\mu = dx/(2\pi),$ where $dx$ is Lebesgue measure. To finish, simply apply $(2)$ to the function $1/f.$ The result falls right out.

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  • $\begingroup$ Edited a little to make it clearer. (The range of $f$ contained in $[a,b]$ prevents any integrals from being infinite, avoiding $\infty/\infty$ problems and the like.) $\endgroup$
    – zhw.
    Commented Jan 28, 2016 at 18:15

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