let $\mu$ be a finite positive borel measure on $\mathbb{R}$ and let $\mathbb{H}$ denote the upper half plane $\{(x,y) \in \mathbb{R}^2: y > 0\}$. consider the functions

$$f(x,y,t)=\frac{1+t^2}{(t-x)^2+y^2},\quad F(x,y)=\int_\mathbb{R}f(x,y,t)\,d\mu(t).$$

show that $F$ on $\mathbb{H}$ is well-defined and continuous. moreover, show that for every $\alpha \in (0,\pi)$

$$\lim_{r\to\infty} F(r\cos\alpha,r\sin\alpha)=0.$$

a hint is given to show that for $x \in (x_1,x_2)$ and $y \ge y_0$ the inequality

$$\frac{1}{(t-x)^2+y^2} \le \frac{1}{(t-x_1)^2+y_0^2} \chi_{(-\infty,-1)}(t) + \frac{1}{y_0^2} \chi_{[x_1-1,x_2+1]}(t) + \frac{1}{(t-x_2)^2+y_0^2} \chi_{(x_2+1,+\infty)}(t)$$


~ what i think to know

it is $(t-x)^2+y^2 \neq 0$. due to $\lim_{t \to \pm\infty}f(x,y,t)=1$ there is a $t_0 \in \mathbb{R^+}$ so that the function $f$ is bounded on $(-\infty,-t_0) \cup (+t_0,+\infty)$, and on the compact set $[-t_0,+t_0]$ it is bounded since $f$ is continuous. $\mu$ is a finite measure, hence the integrability.

~ what i want to know

how does one show the continuity of the function $F$? i suppose i have to use the given hint, but neither do i know how to show it nor how to use it. it looks somehow confusing :S

it is

$$\lim_{r \to \infty} F(r\cos\alpha,r\sin\alpha)=\lim_{r \to \infty} \int_{\mathbb{R}} \frac{1+t^2}{t^2-2tr\cos\alpha + r^2}\,d\mu(t).$$

is it possible to show that the integrand is bounded as above for every $r_n$, where $(r_n)_{n \in \mathbb{N}}$ with $r_n \to \infty$, and then use lebesgue's dominated convergence theorem?


  • 1
    $\begingroup$ that really needs to be $f(x,y,t).$ $\endgroup$
    – zhw.
    Jul 27 '15 at 20:09

Let $E_R =[-R,R]\times [1/R,\infty).$ Claim: There exists a constant $C_R>0$ such that

$$f(x,y,t)\le C_R, \text {for}\,(x,y) \in E_R, t\in \mathbb {R}.$$

Suppose the claim is true. Fix $(a,b)$ in the upper half plane. Then there is $R> 0$ such that $D((a,b),b/2)\subset E_R.$ Let $(x_n,y_n) \to (a,b).$ Then for large $n, (x_n,y_n) \in E_R.$ For such $n$ the functions

$$g_n(t) = f(x_n,y_n,t) \le C_R, t\in \mathbb {R}.$$

Note that $g_n(t) \to f(a,b,t)$ pointwise for each $t\in \mathbb {R}.$ Since the constant function $C_R$ is in $L^1(\mu),$ we can apply the DCT to see

$$F(x_n,y_n) = \int_{\mathbb {R}}g_n(t)\,d\mu(t) \to \int_{\mathbb {R}}f(a,b,t)\,d\mu(t) = F(a,b).$$

That shows $F$ is continuous at $(a,b)$ as desired.

Try to see if you can prove the claim.


The continuity of $F$ is proved by zhw. As you probably know, the key point is to show that for any compact set $K\subseteq \mathbb H$, one can find an integrable function $g_K$ such that $\vert f(x,y,t)\vert\leq g_K (t)$ when $(x,y)\in K$. Then you can apply the "standrad" theorem on continuity for integrals depending on a parameter; or reprove this theorem as zhw did.

To show that $F(r\cos\alpha, r\sin\alpha)\to 0$ as $r\to\infty$, one can indeed proceed as you wanted to do: by the dominated convergence theorem and since $\mu$ is a finite measure, it is enough to show that one can find some constant $C$ such that $\vert f(r\cos\alpha, r\sin\alpha, t)\vert\leq C$ for all $r\geq 1$ and all $t\in\mathbb R$.

By definition, $$\vert f(r\cos\alpha, r\sin\alpha, t)\vert=\frac{1+t^2}{(t-r\cos\alpha)^2+r^2\sin^2\alpha}\cdot $$

If $\vert t\vert\geq 2r\vert \cos\alpha\vert$, then $ \vert t-r\cos\alpha\vert\geq \frac{\vert t\vert}2$, so that (recall that $r\geq 1$) $$\vert f(r\cos\alpha, r\sin\alpha, t)\vert\leq \frac{1+t^2}{t^2/4+\sin^2\alpha}\, , $$ which is bounded by some constant $A_\alpha$.

If $\vert t\vert\leq 2r\vert \cos\alpha\vert$, then $$\vert f(r\cos\alpha, r\sin\alpha, t)\vert\leq\frac{1+r^2\cos 2\alpha}{r^2\sin^2\alpha} \, ,$$ which is bounded by some constant $B_\alpha$ because $r\geq 1$.

Altogether, we get $\vert f(r\cos\alpha, r\sin\alpha, t)\vert\leq C_\alpha:=\max(A_\alpha, B_\alpha)$ for all $r\geq 1$ and $t\in\mathbb R$, and the result follows.


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