# Uniformly Bounded (updating)

Suppose that $$a_1<1$$, $$a_1+a_2+a_3>1.$$ For $$x,y,z>0,$$

(1) define a fucntion $$H(x,y,z)=\frac{x^{\frac{1}{2}}\int_0^{\infty}\frac{1}{t^{a_1}~ (1+t)^{a_2+1}~ (1+t+z)^{a_3}}\exp\big\{-\frac{x}{1+t}-\frac{ y}{1+t+z}\big\}dt }{\int_0^{\infty}\frac{1}{t^{a_1}~ (1+t)^{a_2}~(1+t+z)^{a_3}}\exp\big\{-\frac{x}{1+t}-\frac{ y}{1+t+z}\big\}dt}.$$ Then $$H(x,y,z)$$ is uniformly bounded over $$x,y$$, i.e. there is a constant C, such that $$H(x,y)\le C.$$

(2) Furthermore, define $$L(y,z)=\frac{y^{\frac{1}{2}}\int_0^{\infty}\frac{1}{t^{a_1}~ (1+t)^{a_2}~ (1+t+z)^{a_3+1}}\exp\big\{-\frac{ y}{1+t+z}\big\}dt }{\int_0^{\infty}\frac{1}{t^{a_1}~ (1+t)^{a_2}~(1+t+z)^{a_3}}\exp\big\{-\frac{y}{1+t+z}\big\}dt}.$$

Then $$L(x,y)$$ is also uniformly bounded over $$x,y.$$

• What is the question? Jan 15 '19 at 20:02
• How to prove uniformly bounded?? Need help. Jan 15 '19 at 20:52
• There seems to be a typo. In the current version of the question, the numerator of your fraction is just $x^{1/2}$ times the denominator Jan 15 '19 at 21:00
• Thank you. Actually, I think it is right. The power of (1+t) is different. Jan 15 '19 at 21:08

We may assume that $$x>100(1+|a_2|+|a_3|)^2$$, for other $$x$$ simply $$C=10(1+|a_2|+|a_3|)$$ works perfectly.
Partition the integral in the numerator onto two parts: $$I_1$$ over $$(0,\sqrt{x}]$$ and $$I_2$$ over $$[\sqrt{x},\infty)$$. The second part does not exceed the denominator, since $$\sqrt{x}(1+t)^{-1}<1$$ for all $$t\geqslant \sqrt{x}$$. It suffices to estimate $$I_1\leqslant c I_2$$ for some fixed $$c>0$$ (depending only on $$a_1,a_2,a_3$$). For this it suffices to prove the pointwise estimate $$f(t)\leqslant 2^{a_1} f(2t)$$ for all $$t\in (0,\sqrt{x})$$ where $$f$$ is the integrated function. Indeed, integrating this over $$(0,\sqrt{x})$$ we get $$\int_0^{\sqrt{x}}f(t)dt\leqslant 2^{a_1}\int_0^{\sqrt{x}}f(2t)dt=2^{a_1-1}\int_0^{2\sqrt{x}} f(t)dt$$, thus $$I_1(1-2^{a_1-1})\leqslant 2^{a_1-1}I_2$$ as desired. We have $$\frac{f(t)}{2^{a_1}f(2t)}=\left(\frac{1+2t}{1+t}\right)^{a_2+1}\cdot \left(\frac{1+2t+z}{1+t+z}\right)^{a_3}\cdot \exp\left(-\frac{x}{1+2t}\left(\frac{1+2t}{1+t}-1\right)- \frac{z}{1+2t+z}\left(\frac{1+2t+z}{1+t+z}-1\right) \right).$$ We estimate $$\frac{1+2t+z}{1+t+z}\leqslant \frac{1+2t}{1+t}$$ and use the estimate $$1+s\leqslant e^s$$ for $$s:=\frac{1+2t}{1+t}-1$$. Also estimate $$-\frac{z}{1+2t+z}\left(\frac{1+2t+z}{1+t+z}-1\right)\leqslant 0$$ in the exponent. We get $$\frac{f(t)}{2^{a_1}f(2t)}\leqslant \exp\left(\left(1+|a_2|+|a_3|-\frac{x}{1+2t}\right)s\right)\leqslant 1,$$ since $$\frac{x}{1+2t}\geqslant \frac{x}{1+2\sqrt{x}}>\frac13\sqrt{x}>1+|a_2|+|a_3|$$ due to our assumptions.
• @xiaopai833 I think, everything works: we break the integral at the point $t_0$ for which $\sqrt{y}/(1+t_0+z)=100A$, where $A=|a_2|+|a_3|+1$, (if such a point does not exist, the ratio is clearly bounded.) On $(0,t_0]$ we estimate the ratio $\frac{f(t)}{2^{a_1}f(2t)}$ using the estimate $(1+2t)/(1+t)\leqslant 1+t\leqslant e^t$. And $tA\leqslant y(\frac1{1+t+z}-\frac1{1+2t+z})=\frac{ty}{(1+t+z)(1+2t+z)}$. Jan 16 '19 at 19:19
• I am afraid that it is no longer true if say $a_1=1/2,a_2=2,a_3=0$ (I say now about $L$), $b_1=1$, $b_2=100000000$. Then the integral in the denominator over $[0,1]$ looks to be much greater than all three integrals over $[0,1],[1,1+z]$ and $[1+z,\infty)$ in certain regime for $y$ and $1+z$. Within a bounded factor, you may significantly simplify these integrals (say, on $[0,1]$ you replace $1+t$ to 1, $1+t+z$ to $1+z$, $b_1y/(1+t+z)$ to $y/(2+z)$; on $[1,1+z]$ in the denominator you replace $1+t$ to $t$, $1+t+z$ to $z$, $e^{-b_2y/(1+t+z)}$ to $e^{-50y/(1+z)}$ etc.) Jan 17 '19 at 19:35