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*

*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. □