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**The Setup**

Let $f:\mathbb{R} \to\mathbb{R}$ be a smooth function with support in the

interval $[-R,R]$ and satisfying $\int f = 0$. By manipulating some integrals, I found the surprising inequality

$$

\int\int f(x)f(y) |x-y| \,dxdy \leq 0.

$$

My questions are

- Is this inequality true? Is my derivation below correct?
- Is there a different reason why this is true?
- Are there other nontrivial functions $g(x,y)$ for which

$$

\int \int f(x)f(y)g(x,y)\,dxdy \leq 0?

$$

What properties should I expect of these functions $g(x,y)$?

**The Derivation**

- Inequality involving rearrangement: $ \sum_{i=1}^n |x_i - y_{\sigma(i)}| \ge \sum_{i=1}^n |x_i - y_i|. $
- Inequality.$\sqrt{\frac{2a}{b+c}}+\sqrt{\frac{2b}{c+a}}+\sqrt{\frac{2c}{a+b}} \leq 3$
- Prove independence of a pairwise independent subsequence of independent events
- Inequality $\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq a + b+ c$ when $abc = 1$
- Prove the inequality $n! \geq 2^n$ by induction
- Prove that if $a,b,$ and $c$ are positive real numbers, then $\frac{a^3}{a^3+2b^3}+\frac{b^3}{b^3+2c^3}+\frac{c^3}{c^3+2a^3} \geq 1.$

Let $H:\mathbb{R}\to\mathbb{R}$ be the Heaviside function, so $H(x) = 1$

for $x>0$, and $H(x) = 0$ for $x\leq 0$. I was interested in the convolution $H\ast f(x)$ defined by

$$

H\ast f(x) = \int f(y) H(x-y)\,dy = \int_{-\infty}^x f(y)\,dy.

$$

Notice that since $f$ has support in $[-R,R]$ and $\int f = 0$,

$H\ast f$ also has support in $[-R,R]$.

Now we’re prepared to start the derivation of the inequality, beginning

with the simple observation

$$

0 \leq \int |H\ast f(x)|^2\,dx = \int_{-R}^R |H\ast f(x)|^2\,dx.

$$

First expand the square and the convolution, and then

rearrange the order of the integrals:

\begin{align*}

\int_{-R}^R |H\ast f(x)|^2\,dx

&= \int_{-R}^R \left(\int_{-R}^R f(y)H(x-y)\,dy\right) \left(\int_{-R}^R f(z)H(x-z)\,dz\right) \,dx \\

&= \int_{-R}^R\int_{-R}^R f(y)f(z) \left(\int_{-R}^R H(x-y)H(x-z)\,dx\right)\,dydz

\end{align*}

The integrand $H(x-y)H(x-z)$ is $1$ when both $x>y$ and $x>z$, and $0$ otherwise. Thus the integral comes out to $\min\{R-y,R-z\}$. We plug this

back into the integral, and use again the fact that $\int f = 0$:

\begin{align*}

\int_{-R}^R\int_{-R}^R f(y)f(z) \min\{R-y,R-z\} \,dydz

&= \int_{-R}^R \int_{-R}^R f(y)f(z)(\min\{-y,-z\} – R)\,dydz

\\&= \int_{-R}^R\int_{-R}^R f(y)f(z)\min\{-y,-z\}\,dydz.

\end{align*}

Now split up the domain according to which of $-y$ or $-z$ is smaller:

\begin{align*}

\int_{-R}^R\int_{-R}^R f(y)f(z)\min\{-y,-z\}\,dydz

&= -\int_{y=-R}^R f(y)\left(y\int_{z=-R}^y f(z)\,dz + \int_{z=y}^R zf(z)\,dz\right)\,dy.

\end{align*}

Since $\int_{-R}^R f(z)\,dz = 0$, $\int_{-R}^yf(z)\,dz = -\int_y^R f(z)\,dz$, so we can combine the integrals to conclude that

\begin{align*}

\int |H\ast f(x)|^2\,dx &= \int_{-R}^R f(y) \int_y^R f(z) (y-z)\,dz\,dy\\

&= – \int_{-R}^R f(y) \int_y^R f(z) |y-z|\,dz\,dy

\end{align*}

To get to the integral above, swap the names of the dummy variables and

then swap the order of the integrals:

\begin{align*}

\int |H\ast f(x)|^2\,dx &= -\int_{-R}^R f(z) \int_z^R f(y) |z-y|\,dy\,dz\\

&= -\int_{-R}^R f(y) \int_{-R}^y f(z) |z-y|\,dy\,dz.

\end{align*}

In conclusion, adding both of these formulas together, we obtain

$$

0\leq 2\int |H\ast f(x)|^2\,dx = -\int\int f(y)f(z)|y-z|\,dy\,dz.

$$

The derivation is quite long so there’s a good chance I’ve made a mistake.

- Show that if $f(x) > 0$ for all $x \in $, then $\int_{a}^b f(x) dx > 0$
- Why should we get rid of indefinite integration?
- Yet another multivariable integral over a simplex
- Equal integral but only one of them converges absolutely .
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- Proof of the inequality $(x+y)^n\leq 2^{n-1}(x^n+y^n)$
- $\frac{a+b-2c}{b+c} + \frac{b+c-2a}{c+a} + \frac{c+a-2b}{a+b} \geq 0$
- Algebraic way to change limits of integration of a double integral
- Is this closed-form of $\int_0^1 \operatorname{Li}_3^2(x)\,dx$ correct?
- $S\subseteq T$ implies $\inf T\leq\inf S\leq\sup S\leq \sup T$

You don’t need smoothness: indeed, integrability is probably enough. Perhaps the “most illuminating” proof is to use Fourier series: let $f \in L^2[-R,R]$, then

$$ \tilde{f}(k) = \frac{1}{2R}\int_{-R}^R e^{i\pi kx/R}f(x) \, dx $$

all exist, and Parseval’s identity says that

$$ C\sum_{k=-\infty}^{\infty} \overline{\tilde{f}(k)} \tilde{g}(k) = \int_{-R}^R \overline{f(x)}g(x) \, dx, $$

where $C$ is an uninteresting normalisation constant (most likely $2R$).

Also, we have the convolution theorem, so taking $g=f \star h $, where $h(x)=-\lvert x \rvert$, we find that

$$ \tilde{g}(k) = \tilde{f}(k) \tilde{h}(k), $$

and a calculation shows that

$$ \tilde{h}(k) = \begin{cases} -R/2 & k=0 \\ \frac{(1-(-1)^k) R}{k^2 \pi^2} & k \neq 0 \end{cases}, $$

and in particular, all of these but $\tilde{h}(0)$ are nonnegative. Also, $\tilde{f}(0)=\int f = 0$. So therefore we have

$$ -\iint f(x) f(y) \lvert x -y \rvert \, dx \, dy = 2R \sum_{k=-\infty}^{\infty} \lvert \tilde{f}(k) \rvert^2 \tilde{h}(k) = 2R\sum_{k \neq 0} \frac{(1-(-1)^k)R}{k^2\pi^2} \lvert \tilde{f}(k) \rvert^2 \geq 0 $$

More generally, it is apparent that if $g(x,y)=h(x-y)$, you require that all the Fourier coefficients of $h$ be nonnegative save for $k=0$. Beyond that, it’s considerably more difficult (and in fact I am writing a paper that includes some such functions at the moment).

You can also generalise this to infinite intervals, if you have $\int \lvert x\rvert f(x) \, dx < \infty$, by a similar argument using the Fourier transform.

Setting $F(x)=\int_{-R}^xf(y)\mathrm dy=H\star f(x)$ and $G(x)=\int_{-R}^xyf(y)\mathrm dy$, we have $F'(x)=f(x)$, $G'(x)=xf(x)$,

$F(-R)=F(R)=0$. Proceed with integration

$$\begin{split}A&=\iint f(x)f(y)|x-y|\,\mathrm dy\\

&=\int_{-R}^Rf(x)\left[x\int_{-R}^xf(y)\mathrm dy-\int_{-R}^xyf(y)\mathrm dy-x\int_x^Rf(y)\mathrm dy+\int_x^Ryf(y)\mathrm dy\right]\,\mathrm dx\\

&=\int_{-R}^Rf(x)\left(xF(x)-G(x)+xF(x)+G(R)-G(x)\right)\mathrm dx\\

&=\big[F(x)\left(2xF(x)-2G(x)+G(R)\right)\big]_{-R}^R-\int_{-R}^RF(x)\left(2xf(x)+2F(x)-2xf(x)\right)\mathrm dx\\

&=-2\int_{-R}^RF(x)^2\mathrm dx\le0.

\end{split}$$

We have use integration by parts and $\int_x^Rf(y)\mathrm dy=F(R)-\int_{-R}^xf(y)\mathrm dy=-F(x)$.

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