# How to prove the inequality including Bessel functions?

As well-known the modified Bessel functions of orders zero and one of the first kind, $$I_0(x)$$ and $$I_1(x)$$ are positive and increasing functions over $$x\in(0, \infty)$$, while the modified Bessel functions of orders zero and one of the second kind, $$K_0(x)$$ and $$K_1(x)$$ are positive and decreasing functions over $$x\in(0, \infty)$$.

How can I prove the following inequality with $$\alpha>1$$ and $$k>0$$?

$$I_0(k\alpha)K_1[k(\alpha-1)]-K_0(k\alpha)I_1[k(\alpha-1)]>0.$$

In other words, how to prove $$I_0(x)K_1(y)>K_0(x)I_1(y)$$ with $$x>y>0$$?

I believe the inequality is true as can be observed by plotting. Thank you very much!

• Inequation or inequality? Jun 19, 2022 at 14:11
• @PeterMortensen: Indeed, "inequality". Jun 19, 2022 at 15:06

Introduce the function $$f(x,t)=I_0(x+t)K_1(x)-K_0(x+t)I_1(x).$$ We want to show that $$f(x,t)>0$$ for all $$x>0,t\geq 0$$.

Consider first the case $$f(x,0)$$. We can prove that $$f(x,0)>0$$ by writing $$f(x,0)=(I_0(x)-I_1(x))K_1(x)+I_1(x)(K_1(x)-K_0(x))$$ and using the monotonicity properties (see here: https://dlmf.nist.gov/10.37) $$I_0(x)>I_1(x)\mbox{ and }K_1(x)>K_0(x)\mbox{ for all }x>0.$$

Next, by a direct computation, we have $$\frac{\partial}{\partial t}f(x,t) =I_1( t + x) K_1(x) + I_1(x)K_1(t + x)$$ which is manifestly positive for $$x>0,t\geq 0$$. Hence $$f(x,t)=f(x,0)+\int_0^t (I_1( \tau+ x) K_1(x) + I_1(x)K_1(\tau + x))\mathrm d \tau>0.$$

• thank you for your answer! Is there a typo in line 2 of the link: $0<\nu<\infty$ should be changed to $0\le\nu<\infty$? Jun 20, 2022 at 9:11

In other words, you want to prove that, for the worst case, $$f(x)=I_0(x)\, K_1(x)-I_1(x)\, K_0(x)~>~0$$

It is simple if you consider small and large values of $$x$$.

Using their asymptotic values,

$$f(x)=\frac{1}{x}+O\left(x\right)$$ $$f(x)=\frac{1}{2 x^2}+O\left(\frac{1}{x^4}\right)$$

If you want to go further, use Hankel's asymptotic values which will give $$f(x) = \frac{1}{2 x^2}+\frac{3}{16 x^4}-\frac{45}{32768 x^6}+\cdots$$ which is extremely good as soon as $$x>\frac 32$$.

• thank you for your answer! Should your function $f$ have two independent variables, that is, $f(x,y)$ with $x>y>0$. Btw, could you please give the asymptotic form of $I_0$, $K_0$, $I_1$ and $K_1$, respectively, since I found it is not that clear in the link. Thank you again! Jun 19, 2022 at 7:31
• @user95273. I used Hankel's formulations which are on the Wikipedia page Jun 20, 2022 at 1:32