0
$\begingroup$

Question: let $a_{i},(i=1,2,\cdots,n)$ be real number,show that $$\sum_{i,j=1}^{n}\dfrac{a_{i}a_{j}}{\max{(i,j)}}\le 4\sum_{i=1}^{n}a^2_{i}\tag{1}$$

Hilbert's Inequality:

For any real numbers $a_1,a_2\cdots,a_n$ the following inequality holds: $$\sum_{i=1}^n\sum_{j=1}^n\frac{a_ia_j}{i+j}\leq\pi\sum_{i=1}^na_i^2$$.

Proof: I was reading a proof of this inequality where first they applied Cauchy Schwarz to get $$(\sum_{i=1}^n\sum_{j=1}^n\frac{a_ia_j}{i+j})^2\leq(\sum_{i=1}^n\sum_{j=1}^n\frac{\sqrt{i}a_i^2}{\sqrt{j}(i+j)})(\sum_{i=1}^n\sum_{j=1}^n\frac{\sqrt{j}a_j^2}{\sqrt{i}(i+j)})$$.

Then they stated that it suffices to prove $$\sum_{n=1}^{\infty}\frac{\sqrt{m}}{(m+n)\sqrt{n}}\leq\pi$$ for any positive integer $m$. and $$\displaystyle\sum_{n=1}^\infty\,\frac{\sqrt{m}}{(m+n)\sqrt{n}}<\int_0^\infty\,\frac{\sqrt{m}}{(m+x)\sqrt{x}}\,\text{d}x=\Bigg.\Bigg(2\arctan\left(\sqrt{\frac{x}{m}}\right)\Bigg)\Bigg|_{x=0}^{x=\infty}=\pi$$

But for $(1)$ I think use integral to solve it,can you help?

$\endgroup$
  • $\begingroup$ I don't understand your question. Is your problem finding a primitive function for the integral? $\endgroup$ – Steven Van Geluwe Apr 29 '18 at 15:25
3
$\begingroup$

The approach for the proof of your version is exactly the same, the constant $4$ is optimal and it comes from $$\int_{0}^{+\infty}\frac{dz}{\max(1,z)\sqrt{z}}=4.$$

$\endgroup$
2
$\begingroup$

A different proof: It's not hard to show that it's sufficient to prove the continuous version $$\int_0^\infty\int_0^\infty\frac{f(x)f(y)}{\max(x,y)}dxdy\le 4||f||_2^2$$for $f\in L^2((0,\infty))$. Assume $f\ge0$.

Note first that if $\lambda>0$ then $$\left(\int_0^\infty(f(r^2\lambda^2))^2r\,dr\right)^{1/2}=\frac1{\sqrt 2\lambda}||f||_2\quad(*).$$

Take the integral above, apply a slight change of variables, convert to polar coordinates, apply Cauchy-Schwarz and (*) on each ray: $$\begin{aligned}\int_0^\infty\int_0^\infty\frac{f(x)f(y)}{\max(x,y)}dxdy &=4\int_0^\infty\int_0^\infty\frac{xy}{\max(x^2,y^2)}f(x^2)f(y^2)dxdy \\&=4\int_0^{\pi/4}\int_0^\infty\frac{\cos(\theta)\sin(\theta)}{\max(\cos^2(\theta),\sin^2(\theta))}f(r^2\cos^2(\theta))f(r^2\sin^2(\theta))r\,drd\theta \\&\le 2||f||_2^2\int_0^{\pi/2}\frac{d\theta}{\max(\cos^2(\theta),\sin^2(\theta))} \\&=4||f||_2^2\int_0^{\pi/4}\frac{d\theta}{\cos^2(\theta)} \\&=4||f||_2^2.\end{aligned}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.