Recall that in all assertions below condition (1.5) is assumed to be fulfilled.

**Lemma 4.1** *W* *is a Hilbert space*.

*Proof* The relation $u(x)={\int}_{0}^{x}z(s)\phantom{\rule{0.2em}{0ex}}ds$ establishes a bijection between *W* and the Hilbert space ${L}_{2}({R}_{+})$. □

**Lemma 4.2**
*The value*
${A}_{u}={\int}_{{R}_{+}}\frac{|u(s){u}^{\prime}(s)|}{s}\phantom{\rule{0.2em}{0ex}}ds$
*satisfies the inequality*
${A}_{u}\le 2[u,u].$

(4.1)

*Proof* This follows from the inequalities

$\begin{array}{rcl}{A}_{u}^{2}& \le & {\int}_{{R}_{+}}\frac{{u}^{2}}{{s}^{2}}\phantom{\rule{0.2em}{0ex}}ds{\int}_{{R}_{+}}{\left({u}^{\prime}\right)}^{2}\phantom{\rule{0.2em}{0ex}}ds=2{\int}_{{R}_{+}}\frac{ds}{{s}^{2}}{\int}_{0}^{s}u(t){u}^{\prime}(t)\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{0.2em}{0ex}}[u,u]\\ =& 2{\int}_{{R}_{+}}\frac{u(t){u}^{\prime}(t)}{t}\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{0.2em}{0ex}}[u,u]\le 2{A}_{u}\cdot [u,u].\end{array}$

□

**Remark 4.1** The estimate (4.1) is accurate, since if

$u={s}^{1/2-\epsilon}$,

$s\ge 1$, and

$u={s}^{1/2+\epsilon}$,

$s\ge 1$, then

$\frac{2}{1+2\epsilon}\le \frac{{A}_{u}}{[u,u]}.$

**Remark 4.2** If

$u(s)\ge 0$,

${u}^{\prime}(s)\ge 0$,

${u}^{\u2033}(s)\le 0$, then

In fact, denote ${B}_{u}={\int}_{{R}_{+}}\frac{{u}^{2}}{{s}^{2}}\phantom{\rule{0.2em}{0ex}}ds$. Then ${B}_{u}=2{A}_{u}$. Since $u(s)={\int}_{0}^{s}{u}^{\prime}(t)\phantom{\rule{0.2em}{0ex}}dt\ge s{u}^{\prime}(s)$, ${B}_{u}\ge {\int}_{{R}_{+}}{u}^{\prime}{(s)}^{2}\phantom{\rule{0.2em}{0ex}}ds=[u,u]$.

**Lemma 4.3** *Suppose* (1.5)

*holds*.

*Then* $T(W)\subset {L}_{2}({R}_{+},\rho )$,

*the operator* $T:W\to {L}_{2}({R}_{+},\rho )$ *is bounded*,

*and the norm of* *T* *satisfies the estimate* ${\parallel T\parallel}^{2}\le 4\underset{s\in {R}_{+}}{sup}(s{\int}_{s}^{\mathrm{\infty}}\rho (x)\phantom{\rule{0.2em}{0ex}}dx).$

(4.2)

*Proof* Since

$(Tu,Tu)={\int}_{{R}_{+}}{u}^{2}\rho \phantom{\rule{0.2em}{0ex}}dx=2{\int}_{{R}_{+}}\rho (x)\phantom{\rule{0.2em}{0ex}}dx{\int}_{0}^{x}u(s){u}^{\prime}(s)\phantom{\rule{0.2em}{0ex}}ds$, we can estimate

$(Tu,Tu)\le 2{\int}_{{R}_{+}}\rho (x)\phantom{\rule{0.2em}{0ex}}dx{\int}_{0}^{x}|u(s){u}^{\prime}(s)|\phantom{\rule{0.2em}{0ex}}ds=2{\int}_{{R}_{+}}\frac{|u(s){u}^{\prime}(s)|}{s}s{\int}_{s}^{\mathrm{\infty}}\rho (x)\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}ds.$

From this and (4.1)

$(Tu,Tu)\le 4\underset{s\in {R}_{+}}{sup}(s{\int}_{s}^{\mathrm{\infty}}\rho (x)\phantom{\rule{0.2em}{0ex}}dx)[u,u].$

(4.3)

□

**Remark 4.3** If

${lim}_{s\to \mathrm{\infty}}\mathrm{\Phi}(s)=\mathrm{\infty}$ (see (1.6)), then

$T(W)\not\subset {L}_{2}({R}_{+},\rho )$. In fact, if

$u(s){u}^{\prime}(s)\ge 0$, then

$(Tu,Tu)=2{\int}_{{R}_{+}}\frac{u(s){u}^{\prime}(s)}{s}\mathrm{\Phi}(s)\phantom{\rule{0.2em}{0ex}}ds$. It is possible to find a nonincreasing function

$\psi (s)$ such that

${\int}_{{R}_{+}}\frac{\psi (s)}{s}\phantom{\rule{0.2em}{0ex}}ds<\mathrm{\infty},\phantom{\rule{1em}{0ex}}\text{but}{\int}_{{R}_{+}}\frac{\psi (s)}{s}\mathrm{\Phi}(s)\phantom{\rule{0.2em}{0ex}}ds=\mathrm{\infty}.$

Now find *u* such that $u(s){u}^{\prime}(s)=\psi (s)$: $u{(s)}^{2}=2{\int}_{0}^{s}\psi (t)\phantom{\rule{0.2em}{0ex}}dt$. Since $u\ge 0$, ${u}^{\prime}(s)\ge 0$, and ${u}^{\prime}$ is nonincreasing, by Remark 4.2, $[u,u]\le 2{A}_{u}={\int}_{{R}_{+}}\frac{\psi (s)}{s}\phantom{\rule{0.2em}{0ex}}ds<\mathrm{\infty}$. But $(Tu,Tu)=\mathrm{\infty}$.

**Lemma 4.4** *The image* $T(W)$ *is dense in* ${L}_{2}({R}_{+},\rho )$.

*Proof* The proof is left to the reader. □

The following theorem [[5], p.318] can be used to show compactness.

**Theorem 4.1** (Gelfand)

*A set*
*E*
*from a separable Banach space*
*X*
*is relatively compact if and only if for any sequence of linear continuous functionals that converge to zero at each point*
${f}_{n}(x)\to 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall}x\in X,$

(4.4)

*the convergence* (4.4) *would be uniform on the* *E*.

**Lemma 4.5** *Suppose* (1.3)

*holds*.

*If* ${f}_{n}\in {L}_{2}({R}_{+},\rho )$ *is bounded*,

*then* $\underset{N\to \mathrm{\infty}}{lim}{\int}_{N}^{\mathrm{\infty}}{f}_{n}(x)u(x)\rho (x)\phantom{\rule{0.2em}{0ex}}dx=0$

*uniformly on the set* $\{(u,n):\parallel u\parallel \le 1,n\in \{1,2,\dots \}\}$.

*Proof* Since

${({\int}_{N}^{\mathrm{\infty}}{f}_{n}(x)u(x)\rho (x)\phantom{\rule{0.2em}{0ex}}dx)}^{2}\le {\int}_{N}^{\mathrm{\infty}}{f}_{n}{(x)}^{2}\rho (x)\phantom{\rule{0.2em}{0ex}}dx{\int}_{N}^{\mathrm{\infty}}u{(x)}^{2}\rho (x)\phantom{\rule{0.2em}{0ex}}dx$

it is sufficient to show that

${\int}_{N}^{\mathrm{\infty}}u{(x)}^{2}\rho (x)\phantom{\rule{0.2em}{0ex}}dx\to 0$

uniformly on

$\parallel u\parallel \le 1$. We have

${\int}_{N}^{\mathrm{\infty}}{u}^{2}\rho \phantom{\rule{0.2em}{0ex}}dx={\int}_{N}^{\mathrm{\infty}}\rho \phantom{\rule{0.2em}{0ex}}dx(u{(N)}^{2}+2{\int}_{N}^{x}u(s){u}^{\prime}(s)\phantom{\rule{0.2em}{0ex}}ds).$

The first term tends to zero because of the inequality

$u{(N)}^{2}={({\int}_{0}^{N}{u}^{\prime}(s)\phantom{\rule{0.2em}{0ex}}ds)}^{2}\le {\int}_{0}^{N}{u}^{\prime}{(s)}^{2}\phantom{\rule{0.2em}{0ex}}ds\cdot {\int}_{0}^{N}1\phantom{\rule{0.2em}{0ex}}ds\le N$

and (1.3). The second term is equal to

$2{\int}_{N}^{\mathrm{\infty}}\rho \phantom{\rule{0.2em}{0ex}}dx{\int}_{N}^{x}u(s){u}^{\prime}(s)\phantom{\rule{0.2em}{0ex}}ds=2{\int}_{N}^{\mathrm{\infty}}\frac{u(s){u}^{\prime}(s)}{s}\left(s{\int}_{s}^{\mathrm{\infty}}\rho \phantom{\rule{0.2em}{0ex}}dx\right)\phantom{\rule{0.2em}{0ex}}ds.$

This tends to zero because of (4.1) and (1.3). □

**Lemma 4.6** *Suppose* (1.4)

*holds*.

*If* ${f}_{n}\in {L}_{2}({R}_{+},\rho )$ *is bounded*,

*then* $\underset{a\to 0}{lim}{\int}_{0}^{a}{f}_{n}(x)u(x)\rho (x)\phantom{\rule{0.2em}{0ex}}dx=0$

*uniformly on the set* $\{(u,n):\parallel u\parallel \le 1,n\in \{1,2,\dots \}\}$.

*Proof* Since

$\begin{array}{rcl}{({\int}_{0}^{a}{f}_{n}(x)u(x)\rho (x)\phantom{\rule{0.2em}{0ex}}dx)}^{2}& \le & {\int}_{0}^{a}{f}_{n}{(x)}^{2}\rho (x)\phantom{\rule{0.2em}{0ex}}dx{\int}_{0}^{a}u{(x)}^{2}\rho (x)\phantom{\rule{0.2em}{0ex}}dx\\ \le & C{\int}_{0}^{a}u{(x)}^{2}\rho (x)\phantom{\rule{0.2em}{0ex}}dx\end{array}$

it is sufficient to show that

${\int}_{0}^{a}u{(x)}^{2}\rho (x)\phantom{\rule{0.2em}{0ex}}dx\to 0$

when

$a\to 0$ uniformly on

$\parallel u\parallel \le 1$. We have

${\int}_{0}^{a}{u}^{2}\rho \phantom{\rule{0.2em}{0ex}}dx={\int}_{0}^{a}(2{\int}_{0}^{x}u(s){u}^{\prime}(s)\phantom{\rule{0.2em}{0ex}}ds)\rho \phantom{\rule{0.2em}{0ex}}dx=2{\int}_{0}^{a}\frac{u(s){u}^{\prime}(s)}{s}(s{\int}_{s}^{a}\rho (x)\phantom{\rule{0.2em}{0ex}}dx)\phantom{\rule{0.2em}{0ex}}ds.$

Now we refer to (1.4) and (4.1). □

**Lemma 4.7** *If* (1.3) *and* (1.4) *hold*, *then* *T* *is compact*.

*Proof* Let $\mathrm{\Omega}=\{Tu:\parallel u\parallel \le 1\}$. We use the criterium of compactness of Gelfand (see Theorem 4.1). Let ${f}_{n}\in {L}_{2}({R}_{+},\rho )$ be a sequence of functionals such that ${f}_{n}(z)\to 0$ for any $z\in {L}_{2}({R}_{+},\rho )$. We have to show that ${f}_{n}(Tu)\to 0$ uniformly for $\parallel u\parallel \le 1$.

Let

$\epsilon >0$. Using Lemma 4.5 choose

*N* such that

$|{\int}_{N}^{\mathrm{\infty}}{f}_{n}(x)u(x)\rho (x)\phantom{\rule{0.2em}{0ex}}dx|<\epsilon /4$

for all *n* and for all $\parallel u\parallel \le 1$. The same we can do with ${\int}_{0}^{a}{f}_{n}(x)u(x)\rho (x)\phantom{\rule{0.2em}{0ex}}dx$ for sufficiently small $a>0$. For this aim we can use Lemma 4.6.

Now we only have to show that

${\int}_{a}^{N}{f}_{n}(x)u(x)\rho (x)\phantom{\rule{0.2em}{0ex}}dx\to 0$ uniformly on the set

$\parallel u\parallel \le 1$. Since

$\begin{array}{rcl}{({\int}_{a}^{N}{f}_{n}(x)u(x)\rho (x)\phantom{\rule{0.2em}{0ex}}dx)}^{2}& =& {({\int}_{a}^{N}{f}_{n}(x)({\int}_{a}^{x}{u}^{\prime}(s)\phantom{\rule{0.2em}{0ex}}ds)\rho (x)\phantom{\rule{0.2em}{0ex}}dx)}^{2}\\ =& {({\int}_{a}^{N}{u}^{\prime}(s)\phantom{\rule{0.2em}{0ex}}ds{\int}_{s}^{N}{f}_{n}(x)\rho (x)\phantom{\rule{0.2em}{0ex}}dx)}^{2}\\ \le & {\int}_{a}^{N}{u}^{\prime}{(s)}^{2}\phantom{\rule{0.2em}{0ex}}ds{\int}_{a}^{N}{\phi}_{n}{(s)}^{2}\phantom{\rule{0.2em}{0ex}}ds\\ \le & {\int}_{a}^{N}{\phi}_{n}{(s)}^{2}\phantom{\rule{0.2em}{0ex}}ds,\end{array}$

where

${\phi}_{n}(s)={\int}_{s}^{N}{f}_{n}(x)\rho (x)\phantom{\rule{0.2em}{0ex}}dx$, it suffices to show that

${\int}_{a}^{N}{\phi}_{n}{(s)}^{2}\phantom{\rule{0.2em}{0ex}}ds\to 0$. We show that

${\phi}_{n}(s)\to 0$ uniformly for

$s\in [a,N]$. Let

${z}_{s}(x)=\{\begin{array}{cc}1\hfill & \text{if}x\in [s,N],\hfill \\ 0\hfill & \text{if}x\notin [s,N].\hfill \end{array}$

Then ${\phi}_{n}(s)={f}_{n}({z}_{s})$. The set $\{{z}_{s}:s\in [a,N]\}$ is compact. Thus by Theorem 4.1 of Gelfand ${f}_{n}({z}_{s})$ converges to zero uniformly for these *s*. So, ${\phi}_{n}(s)\to 0$ uniformly for $s\in [a,N]$, and ${f}_{n}(Tu)\to 0$ uniformly for $\parallel u\parallel \le 1$. □

**Remark 4.4** It seems that condition (1.4) is necessary for compactness of *T*.

**Lemma 4.8** *Equation* (2.1) *is equivalent to problem* (2.2).

*Proof* Denote

${h}^{\prime}=-f\rho $. From (2.1) it follows that

${\int}_{0}^{\mathrm{\infty}}{u}^{\prime}{v}^{\prime}\phantom{\rule{0.2em}{0ex}}dx=-{\int}_{0}^{\mathrm{\infty}}{h}^{\prime}v\phantom{\rule{0.2em}{0ex}}dx=-vh{|}_{0}^{\mathrm{\infty}}+{\int}_{0}^{\mathrm{\infty}}h{v}^{\prime}\phantom{\rule{0.2em}{0ex}}dx$

(4.5)

(if we choose

*v* such that the corresponding limits exist). If

$v=0$ for

$x\notin [a,b]\subset (0,\mathrm{\infty})$, then

${\int}_{a}^{b}{u}^{\prime}{v}^{\prime}\phantom{\rule{0.2em}{0ex}}dx={\int}_{a}^{b}h{v}^{\prime}\phantom{\rule{0.2em}{0ex}}dx$. From this

${u}^{\prime}=h+\mathrm{const}$ on

$[a,b]$. Since the segment

$[a,b]$ is arbitrary, the relation

${u}^{\prime}=h+\mathrm{const}$ is fulfilled on the whole semiaxis

$(0,\mathrm{\infty})$. So,

${u}^{\u2033}={h}^{\prime}=-\rho f$. By the first equality in (4.5)

${\int}_{0}^{\mathrm{\infty}}{u}^{\prime}{v}^{\prime}\phantom{\rule{0.2em}{0ex}}dx=-{\int}_{0}^{\mathrm{\infty}}{u}^{\u2033}v\phantom{\rule{0.2em}{0ex}}dx=-{u}^{\prime}v{|}_{0}^{\mathrm{\infty}}+{\int}_{0}^{\mathrm{\infty}}{u}^{\prime}{v}^{\prime}\phantom{\rule{0.2em}{0ex}}dx.$

Now, ${u}^{\prime}v{|}_{x=\mathrm{\infty}}={u}^{\prime}v{|}_{x=0}=0$. Choosing *v* such that $v(\mathrm{\infty})\ge C>0$, we obtain ${u}^{\prime}(\mathrm{\infty})=0$. □