5.1 Proof of Theorem 3
We prove first that the statement of the theorem is true in the case \(n=0\). In the case \(n=0\), we write \(\varphi_{r,f}\) instead of \(\varphi _{r,0,f}\) and \(G_{r,f,M}\) instead of \(G_{r,0,f,M}\). To prove the theorem, we need the following lemma.
We will proceed using induction on m in order to prove the first inequality.
Let
\(m=1\).
\(P_{1}(t)=-A_{0}+g_{1}(t)\). Since conditions (
6) hold and
\(P_{1}(0)=-A_{0}\), we have
\(P_{1}(\infty)=0\). Since the function
\(h_{1}(t)=g_{1}(t)+C\) has one zero,
\(C\in(-A_{0},0]\). This means that
\(|h_{1}(t)|<|P_{1}(t)|\) for all
\(t\in[0,\alpha_{1}]\).
Let the statement of the lemma hold in the case \(m=k\leq r-1\). We will show that it holds in the case \(m=k+1\) too.
To be definite, assume that the number
\(k+1\) is even. Then
\(P_{k+1}(0)>0\), and in view of Lemma
1,
Assume the converse, let a point
\(t_{0}\in[0,\alpha_{k+1})\) such that
\(h_{k+1}(t_{0})\geq P_{k+1}(t_{0})\) exist. Since the function
\(h_{k+1}(t)\) has
\(k+1\) zeroes, the function
\(h^{(k+1)}_{k+1}(t)= g(t)\) does not have zeroes, we have that the function
\(h_{k+1}'(t)\) has
k zeroes, and hence, in virtue of induction assumption and
\(P_{k+1}'(t)=P_{k}(t)\), we get that
\(|h_{k+1}'(t)|<|P_{k}(t)|\)
\(\forall t\in[0,\beta]\), where
β is the first zero of the function
\(h_{k+1}'(t)\). Hence, due to (
13) for all
\(t\in[0,\beta]\), the following inequality holds:
$$ 0\geq h_{k+1}'(t)>P_{k}(t). $$
(14)
In view of Rolle’s theorem,
\(\beta>\alpha_{k}\). This means that inequality (
14) holds for all
\(t\in[0,\alpha_{k}]\). Since
\(h_{k+1}(\alpha_{k+1})=0< P_{k+1}(\alpha_{k+1})\), we have
\(P_{k+1}(t_{0})-P_{k+1}(\alpha_{k+1})< h_{k+1}(t_{0})-h_{k+1}(\alpha _{k+1})\); on the other hand, inequality (
14) holds, and hence
$$\begin{aligned} P_{k+1}(t_{0})-P_{k+1}(\alpha_{k+1}) =&- \int_{t_{0}}^{\alpha _{k+1}}P_{k}(t)\,dt>-\int _{t_{0}}^{\alpha_{k+1}}{h_{k+1}' (t) \,dt} \\ =&h_{k+1}(t_{0})-h_{k+1}(\alpha_{k+1}). \end{aligned}$$
Contradiction. The first inequality is proved.
From Lemma
1 it follows that
\(\operatorname{sgn} h_{m}(0) = \operatorname{sgn}P_{m}(0)\). From the proved part of the lemma it now follows that the function
\(P_{m}(t)-P_{m}(0)+h_{m}(0)\) has exactly one zero. The second inequality in the statement of the lemma can be proved using arguments similar to the ones used in the proof of the first inequality. The lemma is proved.
Let us return to the proof of the theorem in the case when \(n=0\).
Let conditions (
6) and (
7) hold. Set
\(K_{r}:=\sup_{t\in [0,\infty)}\frac{|P_{r}(t)|}{f(t)}\). Assume the converse, let
\({\varphi _{r,f}(\infty)=\infty}\). This means that there exists
\(M>0\) such that
\(\varphi_{r,f}(M)>K_{r}\). Due to Lemma
2 the inequality
\(|G_{r,f,M}(t)|<|P_{r}(t)|\) holds on the interval
\([0,\alpha_{r}^{M}]\). Moreover,
\(|P_{r}(t)|\leq K_{r} f(t)<\varphi_{r,f}(M) f(t)\). Thus, on the interval
\([0,\alpha_{r}^{M}]\), the inequality
\(|G_{r,f,M}(t)|<\varphi _{r,f}(M) f(t)\) holds. However, in this case
\(G_{r,f,M}(t)\) has not more than
r oscillating points. Contradiction. Sufficiency is proved.
Let now
\(\varphi_{r,f}(\infty)< \infty\). This means that for all
\(a\geq0\),
\(t\in[0,a]\) we have
\(\vert G_{r,f,a}(t)\vert \leq\varphi_{r,f}(\infty) f(t) \leq\varphi _{r,f}(\infty) f(0)\). Passing to the limit as
\(a\to\infty\), we get existence of bounded on
\([0,\infty)\) primitive
\(Q_{r}\) of order
r of the function
\(g(t)\). Really, for each
\(a>0\), we have
\(G_{r,f,a}(t) = g_{r}(t) +\sum_{k=0}^{r-1}c_{k}(a)t^{k}\), where
\(c_{k}\) are real functions,
\(k=0,\ldots, r-1\), and
\(g_{r}\) is some (fixed) primitive of order
r of the function
g. For
\(a,t\geq0\), set
\(R(a;t):=\sum_{k=0}^{r-1}c_{k}(a)t^{k}\). From Markov’s inequality for polynomials we get that for
\(k=0,\ldots, r-1\) there exists a constant
\(N_{k}\) independent of
a such that
$$\begin{aligned} \bigl\vert c_{k}(a)\bigr\vert =& \bigl\vert R^{(k)}(a;0)\bigr\vert \leq\max_{t\in[0,1]}\bigl\vert R^{(k)}(a;t)\bigr\vert \leq N_{k} \max_{t\in[0,1]} \bigl\vert R(a;t)\bigr\vert \\ \leq& N_{k} \Bigl[ \max_{t\in[0,1]}\bigl\vert g_{r}(t)\bigr\vert + \varphi_{r,f}(\infty) f(0) \Bigr]. \end{aligned}$$
Hence the functions
\(c_{k}\) are bounded,
\(k=0,\ldots, r-1\). This means that there exists a sequence
\(\{a_{n}\}_{n=1}^{\infty}\),
\(\lim_{n\to\infty }a_{n}=\infty\) such that all the sequences
\(c_{k}(a_{n})\) have finite limits
\(c_{k}:=\lim_{n\to\infty}c_{k}(a_{n})\),
\(k=0,\ldots,r-1\). For the function
\(Q_{r}(t):=g_{r}(t) + \sum_{k=0}^{r-1}c_{k} t^{k}\), fixed
\(t\geq0\) and all natural
n such that
\(a_{n}>t\), we have
$$\begin{aligned} \bigl\vert Q_{r}(t)\bigr\vert =&\Biggl\vert g_{r}(t) + \sum_{k=0}^{r-1}c_{k}(a_{n}) t^{k}+\sum_{k=0}^{r-1}c_{k} t^{k}-\sum_{k=0}^{r-1}c_{k}(a_{n}) t^{k}\Biggr\vert \\ \leq& \Biggl\vert g_{r}(t) + \sum_{k=0}^{r-1}c_{k}(a_{n}) t^{k}\Biggr\vert +\Biggl\vert \sum_{k=0}^{r-1} \bigl(c_{k}-c_{k}(a_{n})\bigr) t^{k}\Biggr\vert \\ =&\bigl\vert G_{r,f,a_{n}}(t)\bigr\vert + \Biggl\vert \sum _{k=0}^{r-1}\bigl(c_{k}-c_{k}(a_{n}) \bigr) t^{k}\Biggr\vert . \end{aligned}$$
Since
n can be arbitrarily large and all the functions
\(G_{r,n,f,a}\),
\(a>0\) are bounded on
\([0,a]\), by an absolute constant
\(\varphi _{r,f}(\infty) f(0)\), we get that
\(Q_{r}\) is a bounded on
\([0,\infty)\) primitive of order
r of the function
g.
Since functions
\(f(t)\) and
\(g(t)\) are bounded, we get that all functions
\(Q_{r}^{(k)}(t)\),
\(k=1,\ldots,r-1\), are also bounded on
\([0,\infty)\). Note that the only bounded on
\([0,\infty)\) primitives of the function
\(g(t)\) of order
\(k\in{\mathbb{N}}\) are functions
\(P_{k}(t) + C_{k}\), where
\(C_{k}\in{\mathbb{R}}\), and only in the case when corresponding conditions (
6) hold. This means that conditions (
6) hold. Necessity of conditions (
6) are proved.
Note that from arguments above it follows that the following lemma holds.
We will prove that condition (
7) also holds. If
\(f(\infty)>0\), then condition (
7) holds always when conditions (
6) hold. Below we will assume that
From Lemma
2 we get
$$\begin{aligned} \biggl\vert \int_{0}^{\alpha_{r-1}^{M}}P_{r-1}(t) \,dt\biggr\vert \geq& \bigl\vert G_{r,f,{M}}(0)-G_{r,f,M}\bigl( \alpha_{r-1}^{M}\bigr)\bigr\vert \\ \geq&\biggl\vert \int_{0}^{\gamma_{M}} \bigl[P_{r-1}(t)-P_{r-1}(0)+G_{r, f,M}'(0) \bigr]\,dt\biggr\vert , \end{aligned}$$
where
\(\gamma_{M}\) is zero of the function
\(P_{r-1}(t)-P_{r-1}(0)+G_{r,f,M}'(0)\).
In view of Lemma
3 (with
\(k=r-2\),
\(M\to\infty \)) we get
$$\biggl\vert \int_{0}^{\alpha_{r-1}^{M}}P_{r-1}(t) \,dt\biggr\vert \to \biggl\vert \int_{0}^{\infty}P_{r-1}(t) \,dt\biggr\vert = A_{r-1};\qquad -P_{r-1}(0)+G_{r,f,M}'(0) \to0, $$
hence
$$\gamma_{M}\to \infty \quad \mbox{and}\quad \biggl\vert \int _{0}^{\gamma_{M}} \bigl[P_{r-1}(t)-P_{r-1}(0)+G_{r,f,M}'(0) \bigr]\,dt \biggr\vert \to A_{r-1}. $$
From Lemma
3 and equality (
15) we get
\(G_{r,f,M}(\alpha _{r-1}^{M})\to0\) and hence
$$ \bigl\vert G_{r,f,{M}}(0)\bigr\vert \to A_{r-1},\quad M\to\infty. $$
(16)
We will show that
\(\varphi_{r,f}(\infty)\geq\sup_{t\in[0,\infty)}\frac {P_{r}(t)}{f(t)}\). Assume the converse. Suppose that a point
\(t_{0}\) such that
\(\vert P_{r}(t_{0})\vert >\varphi_{r,f}(\infty)f(t_{0})\) exists. We can choose
\(\varepsilon>0\) in such a way that
$$ \bigl\vert P_{r}(t_{0})-\varepsilon \operatorname{sgn} \bigl[P_{r}(0) \bigr] \bigr\vert > \varphi_{r,f}(\infty)f(t_{0}). $$
(17)
In virtue of (
16) and Lemma
2, we can choose
\(M>0\) big enough, so that
$$\bigl\vert P_{r}(t)-{\varepsilon} \operatorname{sgn} \bigl[P_{r}(0) \bigr] \bigr\vert < \bigl\vert G_{r,f,M}(t) \bigr\vert < \bigl\vert P_{r}(t)\bigr\vert \quad \forall t\in[0, \gamma], $$
where
γ is zero of the function
\(P_{r}(t)-{\varepsilon} \operatorname{sgn} [P_{r}(0) ]\). Since inequality (
17) holds, we have
\(\gamma>t_{0}\), and hence
\(\vert G_{r,f,M}(t_{0})\vert >\varphi _{r,f}(\infty)f(t_{0})\). Contradiction. Thus condition (
7) is proved. The theorem is proved in the case when
\(n=0\).
Let n be an arbitrary natural number now.
We will prove that for all \(r,n\in{\mathbb{N}}\), \(\varphi_{r,n,f}(\infty )<\infty\) if and only if \(\varphi_{r,f}(\infty)<\infty\).
It is clear that \(\varphi_{r,f}(M)\geq\varphi_{r,n,f}(M)\) for all \(M>0\), and hence \(\varphi_{r,f}(\infty)<\infty\) implies \(\varphi _{r,n,f}(\infty)<\infty\).
Assume that \(\varphi_{r,n,f}(\infty)<\infty\). Denote by \(t_{n,k}^{M}\) the kth knot of the g-spline \(G_{r,n,f,M}(t)\), \(k=1,2,\ldots,n\). Set \(t_{n,0}^{M}:=0\), \(t_{n,n+1}^{M}:=M\). Let \(1\leq k\leq n+1\) be the smallest number of the knots of the g-spline \(G_{r,n,f,M}(t)\) for which the set \(\{t_{n,k}^{M}: M>0 \}\) is unbounded. We can choose an increasing sequence \(\{M_{l} \}_{l=1}^{\infty}\), \(M_{l}\to\infty\) as \(l\to\infty\) such that \(t_{n,s}^{M_{l}}\to t_{n,s}<\infty\), \(s\leq k-1\) and \(t_{n,k}^{M_{l}}\to\infty\) as \(l\to \infty\).
Denote by
\(G_{r,f,M}^{K}(t)\) the primitive of order
r of the function
g that least deviates from zero in norm
\(\|\cdot\|_{C[K,K+M],f}\). Set
$$\varphi_{r,f}^{K}(M):=\bigl\Vert G_{r,f,M}^{K} \bigr\Vert _{C[K,K+M],f}. $$
Then, for all
l,
\(\|G_{r,n,f,M_{l}}\| _{C[t_{n,k-1}^{M_{l}},t_{n,k}^{M_{l}}],f}\geq\varphi _{r,f}^{t_{n,k-1}}(t_{n,k}^{M_{l}}-t_{n,k-1}^{M_{l}})\). Passing to the limit as
\(l\to\infty\), we get
\(\varphi^{t_{n,k-1}}_{r,f}(\infty)\leq \varphi_{r,n,f}(\infty)<\infty\). Note that from the case when
\(n=0\) proved above, it follows that for all
\(K>0\),
\(\varphi_{r,f}^{K}(\infty )<\infty\) if and only if
\(\varphi_{r,f}(\infty)<\infty\). The theorem is proved.
Really, in the case when
\(f\equiv1\), condition (
7) holds always when conditions (
6) hold. If
$$g(t)= \bigl[-1+(t+\sqrt{3})^{2} \bigr]e^{-\frac{(t+\sqrt{3})^{2}}{2}} $$
and
$$f(t)=e^{-\frac{(t+\sqrt{3})^{2}}{2}}, $$
then
$$\begin{aligned}& P_{1}(t)=- \bigl[(t+\sqrt{3}) \bigr]e^{-\frac{(t+\sqrt{3})^{2}}{2}}, \\& P_{2}(t)=e^{-\frac{(t+\sqrt{3})^{2}}{2}}, \end{aligned}$$
and condition (
7) holds when
\(r=2\) and does not hold when
\(r=1\).
5.2 Proof of Theorem 4
To prove Theorem
4, it is sufficient to prove that for all
\(\delta>0\) there exists a
g-spline
\(G_{r,\delta}=G_{r,\delta}(\cdot ,f,g)\) of order
r defined on
\([0,\infty)\) with infinite number of knots
\(y_{k}\) (
\(k=1,2,\ldots\)),
\(0=y_{0}< y_{1}<\cdots<y_{k}<\cdots\),
\(y_{k}\to \infty\) (
\(k\to\infty\)), with the following properties:
1.
\(\|G_{r,\delta}\|_{C[0,\infty),f}= \delta\) and either \(G_{r,\delta }^{(r)}\equiv g\) or \(G_{r,\delta}^{(r)}\equiv-g\) on the intervals \((y_{k},y_{k+1})\) (\(k=0,1,2,\ldots\)).
2.
For all \(c>0\), the sequences \(\{ G_{r,n,f,\delta _{r,n}}^{(k)} \}_{n=0}^{\infty}\) (\(k=0,1,\ldots,r-1\)) (whose elements are defined on \([0,c]\) for big enough n) converge to \(G_{r,\delta}^{(k)}\) uniformly on \([0,c]\).
Really, from Theorem
2 it will follow that
\(\omega(D^{k},\delta )=\vert G^{(k)}_{r,\delta}(0)\vert \), and from condition 2 it will follow that
$$\lim_{n\to\infty}\bigl\vert G^{(k)}_{r,n,f,\delta_{r,n}}(0)\bigr\vert = \bigl\vert G^{(k)}_{r,\delta}(0)\bigr\vert . $$
\(\{\delta_{r,n} \}_{n=0}^{\infty}\) is a non-decreasing sequence. Moreover, this sequence is unbounded because otherwise we would get a perfect g-spline G with arbitrarily close oscillating points; this is impossible because the functions G and \(G^{(r)}\) (and hence \(G'\)) are bounded.
Denote by \(t_{n,k}\) (\(k=1,\ldots,n\), \(n=1,2,\ldots\)) the knots of the g-spline \(G_{r,n,f,\delta_{r,n}}\). We can choose a sequence \(n_{s}\) (\(n_{s}\to\infty\) as \(s\to\infty\)) such that every sequence \(\{t_{n_{s},k} \}_{n_{s}\geq k}^{\infty}\) (\(k=1,2,\ldots\)) has a (finite or infinite) limit.
Let \(0\leq y_{1}< y_{2}<\cdots\) be all distinct finite limits of these sequences, ordered in an ascending way. The number of the nodes \(y_{k}\) is infinite since from the statement of the theorem, we have \(\varphi_{r,n,f}(\infty)=\infty\) for all \(n\in{\mathbb{N}}\).
For all \(i\in{\mathbb{N}}\) and for all small enough \(\varepsilon>0\), there exists \(N=N(i,\varepsilon)\) such that for every \(n>N(i,\varepsilon )\), \(G_{r,n,f,\delta_{r,n}}^{(r)}\equiv g\) or \(G_{r,n,f,\delta _{r,n}}^{(r)}\equiv-g\) on \(I_{i}(\varepsilon):=(y_{i-1}+\varepsilon ,y_{i}-\varepsilon)\). In other words, for each \(i\in {\mathbb{N}}\) starting with some \(n = N(i,\varepsilon)\), the restriction of the g-spline \(G_{r,n,f,\delta_{r,n}}\) to the interval \(I_{i}(\varepsilon)\) is a primitive of order r of g or −g. Since \(\varepsilon>0\) is arbitrary, on each interval \((y_{i-1},y_{i})\) we get existence of point-wise limit \(\lim_{n\to \infty}G_{r,n,f,\delta_{r,n}}=:G_{r,\delta}\); moreover, on the intervals \((y_{i},y_{i+1})\), \(G_{r,\delta}^{(r)}\equiv g\) or \(G_{r,\delta }^{(r)}\equiv-g\) (\(i=1,2,\ldots\)). It is clear that \(\|G_{r,\delta}\| _{C[0,\infty),f}= \delta\). Using arguments similar to the ones used to prove that \(\lim_{n\to\infty}{\delta_{r,n}}=+\infty\), we can prove that \(y_{k}\to\infty\) (\(k\to\infty\)).
Let us fix some \(c>0\). Starting with some n, all g-splines \(G_{r,n,f,\delta_{r,n}}\) are defined on \([0,c]\). From \(\|G_{r,n,f,\delta _{r,n}}\|_{C[0,\delta_{r,n}],f}=\delta\) and the fact that f is non-increasing (and hence is bounded) it follows that the sequence \(\{G_{r,n,f,\delta_{r,n}} \} _{n=0}^{\infty}\) is uniformly bounded on \([0,c]\); from \(\vert G_{r,n,f,\delta_{r,n}}^{(r)}(t)\vert \leq g(t)\) almost everywhere on \([0,\infty)\) and the fact that g is non-increasing (and hence is bounded) it follows that sequences \(\{G_{r,n,f,\delta_{r,n}}^{(k)} \}_{n=0}^{\infty}\), \(k=0,\ldots ,r-1\), are uniformly bounded on \([0,c]\) and equicontinuous. The later implies uniform convergence on \([0,c]\) of the sequence \(G_{r,n,f,\delta_{r,n}}\) to \(G_{r,f}\). The theorem is proved.
5.3 Proof of Theorem 5
Let
\(n\geq0\). We can choose an increasing sequence
\(\{M_{k} \} _{k=1}^{\infty}\),
\(M_{k}\to\infty\) as
\(k\to\infty\) in such a way that all sequences
\(t_{n,s}^{M_{k}}\),
\(1\leq s\leq n\) (as above,
\(t_{n,s}^{M_{k}}\) is the
sth knot of the
g-spline
\(G_{r,n,f,M_{k}}\)) have limits (finite or infinite). Let
\(t_{n,1}<\cdots<t_{n,m}\) be all distinct finite limits of these sequences in ascending order. Analogously to the proof of Theorem
4, we get uniform on each segment
\([0,c]\),
\(c>0\) convergence of the sequence
\(G_{r,n,f,M_{k}}\) to the
g-spline
\(P_{r,n,f,\{M_{k}\}}\) with
m knots (defined on the whole half-line) together with all derivatives up to the order
\(r-1\) inclusively. For brevity we will write
\(P_{r,n,f}\) instead of
\(P_{r,n,f,\{M_{k}\}}\).
Let the function
\(x(t)\) be such that
$$ \begin{aligned} &\|x\|_{C[0,\infty),f}\leq\varphi_{r,n,f}( \infty), \\ &\bigl\Vert x^{(r)}\bigr\Vert _{L_{\infty}(0,\infty),g}\leq1. \end{aligned} $$
(18)
We will show that for all
\(s=1,2,\ldots,r-1\),
$$ \bigl\Vert x^{(s)}\bigr\Vert _{C[0,\infty)}\leq \bigl\Vert P^{(s)}_{r,n,f}\bigr\Vert _{C[0,\infty)}. $$
(19)
Assume the converse, let for some
s,
\(\Vert x^{(s)}\Vert _{C[0,\infty)}> \Vert P^{(s)}_{r,n,f}\Vert _{C[0,\infty)}\). Then there exists
\(\varepsilon>0\) such that
\(\Vert x^{(s)}\Vert _{C[0,\infty)}> (1+\varepsilon)\Vert P^{(s)}_{r,n,f}\Vert _{C[0,\infty)}\). We can suppose that
\(\vert x^{(s)}(0) \vert >(1+\varepsilon)\vert P_{r,n,f}^{(s)}(0)\vert \) (if this is not true, then there exists a point
\(t_{0}>0\) such that
\(\vert x^{(s)}(t_{0}) \vert >\Vert (1+\varepsilon)P_{r,n,f}^{(s)}\Vert _{C[0,\infty)}\) and instead of the function
\(x(t)\) we can consider
\(y(t):= x(t+t_{0})\). Moreover, since the functions
f and
g are non-increasing, conditions (
18) and (
19) are not broken and uniform norms of the function
x and its derivatives do not increase). Moreover, we can assume that
$$ x^{(s)}(0)>(1+\varepsilon)P_{r,n,f}^{(s)}(0)>0 $$
(20)
(if this is not true, we can multiply
x and (or)
\(P_{r,n,f}\) by −1). Set
\(\Delta_{k}(t):=x(t)-(1+\varepsilon)P_{r,n,f,M_{k}}(t)\),
\(t\in [0,M_{k}]\). We can choose
k so big that
$$ x^{(s)}(0)>(1+\varepsilon)P^{(s)}_{r,n,f,M_{k}}(0) $$
(21)
and
$$ (1+\varepsilon)\varphi_{r,n,f}(M_{k})> \varphi_{r,n,f}(\infty). $$
(22)
From Lemma
1 we get
$$ (-1)^{s} P_{r,n,f,M_{k}}(0)>0. $$
(23)
In view of (
22) and (
23)
\((-1)^{s}\Delta _{k}(0) < 0\), and hence due to Lemma
1 we get
\(\Delta_{k}^{(s)}(0) < 0\). However, this contradicts (
21).
In virtue of property (
19) proved above, the limit
\(\lim_{M_{k}\to\infty} \vert G^{(k)}_{r,n,f,M_{k}}(0) \vert \) does not depend on the choice of the sequence
\(\{M_{k}\}_{k=1}^{\infty}\). This finishes the proof of the theorem.
5.4 Proof of Theorem 6
We will need the following lemmas.
We will prove the lemma using induction. In the case
\(s=r\), equality (
24) holds. Let it be true for
\(s=k\geq2\). We will prove that it is true for
\(s=k-1\) too. In view of the induction assumption
\(Q_{r,n}^{(k)}(t)=\epsilon P_{r}^{(k)}(t)\),
\(t\geq t_{n}\). Moreover,
\(Q_{r,n}^{(k-1)}(\infty)=P_{r}^{(k-1)}(\infty)=0\). Then, for
\(t\geq t_{n}\),
$$-Q_{r,n}^{(k-1)}(t)=\int_{t}^{\infty}Q_{r,n}^{(k)}(s)\,ds=\epsilon\int_{t}^{\infty}P_{r}^{(k)}(s)\,ds= -\epsilon P_{r}^{(k-1)}(t). $$
The lemma is proved.
Let some \(n\in{\mathbb{N}}\) be fixed. Suppose that \(M>0\) is such that for all \(t>M\), \(\frac{f(t)}{P_{r}(t)} > \frac{2}{\varphi_{r,n,f}(\infty )}\). Let the g-spline \(P_{r,n,f}\) have k oscillating points \(0\leq a_{1}< a_{2}<\cdots<a_{k}\). Denote by \(0\leq b_{1}< b_{2}<\cdots<b_{n+r+1}\) all oscillation points of the g-spline \(G_{r,n,f,K}\), where K is chosen so big that \(\operatorname{sgn} G_{r,n,f,K}(b_{s})=\operatorname{sgn} P_{r,n,f}(b_{s})\), \(s=1,2,\ldots,k\), \(b_{k+1}>\max\{a_{k}, M\}\) and \(\varphi_{r,n,f}(K) > \frac{1}{2} \varphi_{r,n,f}(\infty)\). Choose \(\varepsilon>0\) so small that \((1-\varepsilon )\varphi_{r,n,f}(K) > \frac{1}{2} \varphi_{r,n,f}(\infty)\).
Set
\(\Delta(t):= P_{r,n,f}(t)- (1-\varepsilon )G_{r,n,f,K}(t)\). Then
\(\operatorname{sgn} \Delta(a_{s})=\operatorname{sgn} P_{r,n,f}(a_{s})\),
\(s=1,2,\ldots,k\), since
$$\bigl\vert P_{r,n,f}(a_{s})\bigr\vert = \varphi_{r,n,f}(\infty)f(a_{s}) > (1-\varepsilon) \varphi_{r,n,f}(K)f(a_{s}) \geq(1-\varepsilon) \bigl\vert G_{r,n,f,K}(a_{s})\bigr\vert . $$
Hence the function
\(\Delta(t)\) has
\(k-1\) sign changes on the interval
\([0,a_{k}]\). Moreover, for
\(s = k+1,\ldots, n+r+1\),
$$\bigl\vert P_{r,n,f}(b_{s})\bigr\vert < \frac{\varphi_{r,n,f}(\infty)}{2}f(b_{s}) < (1-\varepsilon)\varphi_{r,n,f}(K)f(b_{s}) = (1-\varepsilon)\bigl\vert G_{r,n,f,K}(b_{s})\bigr\vert . $$
This means that
\(\operatorname{sgn} \Delta(b_{s})=\operatorname{sgn} G_{r,n,f,K}(b_{s})\),
\(s=k+1,\ldots,n+r+1\). Hence the function
\(\Delta(t)\) has
\(n+r-k\) sign changes on the interval
\([b_{k+1},a]\), and hence at least
\(n+r-1\) sign changes on the whole interval
\([0,K]\). This means that the function
\(\Delta^{(r)}(t)\) has at least
\(n-1\) sign changes, and hence the
g-spline
\(P_{r,n,f}\) has at least
\(n-1\) knots.
Let \(t_{n,s}^{K}\), \(s = 1,\ldots, n\), be the knots of the g-spline \(G_{r,n,f,K}\). Note that for all \(s = 1,2,\ldots, n\), the g-spline \(G_{r,n,f,K}\) has at least s oscillating points on the interval \([0,t_{n,s}^{k}]\). Really, assume the converse, suppose for some \(1\leq s\leq n\) that the g-spline \(G_{r,n,f,K}\) has less than s oscillation points on the interval \([0,t_{n,s}^{K}]\). Then the g-spline \(G_{r,n,f,K}\) has more than \(n+r+1-s\) oscillation points on the interval \((t_{n,s}^{K}, K]\), and hence more than \(n+r-s\) sign changes. This means that the function \(G^{(r)}_{r,n,f,K}\) has more than \(n-s\) sign changes on the interval \((t_{n,s}^{K}, K]\). However, this is impossible.
This means that the limiting g-spline \(P_{r,n,f}\) has at least \(n-1\) oscillation points. The lemma is proved.
We will prove the statement of the lemma using induction. In the case
\(s=r\), inequality (
25) holds with equality sign. Let inequality (
25) hold with
\(s=k\geq2\). We will prove that it is true for
\(s=k-1\).
Assume the converse. Let
$$T:= \bigl\{ t\in[0,\infty):\bigl\vert P^{(k-1)}_{r,n,f}(t)\bigr\vert > \bigl\vert P^{(k-1)}_{r}(t)\bigr\vert \bigr\} \neq \emptyset. $$
Denote by
\(0< t_{1}<\cdots<t_{l}\) all knots of the
g-spline
\(P_{r,n,f}\). Then, due to Lemma
4,
\(T\subset[0,t_{l})\) and
$$ \bigl\vert P^{(k-1)}_{r,n,f}(t_{l}) \bigr\vert =\bigl\vert P^{(k-1)}_{r}(t_{l})\bigr\vert . $$
(26)
Let
\(a\in T\). Then
$$\begin{aligned} \bigl\vert P^{(k-1)}_{r}(t_{l})-P^{(k-1)}_{r}(a) \bigr\vert =&\biggl\vert \int_{a}^{t_{l}}P_{r}^{(k)}(t) \,dt\biggr\vert = \int_{a}^{t_{l}}\bigl\vert P_{r}^{(k)}(t)\bigr\vert \,dt \\ \geq&\int_{a}^{t_{l}} \bigl\vert P^{(k)}_{r,n,f}(t)\bigr\vert \,dt \geq\biggl\vert \int_{a}^{t_{l}}P^{(k)}_{r,n,f}(t) \,dt\biggr\vert \\ =& \bigl\vert P^{(k-1)}_{r,n,f}(t_{l})-P^{(k-1)}_{r,n,f}(a) \bigr\vert ; \end{aligned}$$
this is impossible in virtue of (
26), and the facts that
\(\operatorname{sgn} P_{r}^{(k-1)}(t_{l})=\operatorname{sgn} P_{r}^{(k-1)}(a)\), function
\(\vert P_{r}^{(k-1)}\vert \) is non-increasing and
\(a\in T\). Contradiction. Hence
\(T=\emptyset\) and the lemma is proved.
Let us return to the proof of the theorem.
Let
$$ \varliminf_{t\to\infty}\frac{f(t)}{\vert P_{r}(t)\vert }=:c< \infty. $$
(27)
Then, in view of Theorem
3,
\(c>0\). We will show that
\(\varphi _{r,n,f}(\infty)\geq\frac{1}{2c}\) ∀
n. Assume the converse, let a number
\(n_{0}\) such that
$$ \varphi_{r,n_{0},f}(\infty)< \frac{1}{2c} $$
(28)
exist. Denote by
\(0< t_{1}< t_{2}<\cdots<t_{k}\) all knots of
\(P_{r,n_{0},f}\). Then, due to Lemma
4,
\(P^{\prime}_{r,n_{0},f}(t)=\pm P_{r}^{\prime }(t)\),
\(t\geq t_{k}\). In virtue of (
28) we have
$$\bigl\vert P_{r,n_{0},f}(t)\bigr\vert < \frac{f(t)}{2c} $$
for all
\(t\geq0\).
\(f(\infty)=0\) since (
27) holds. Then
\(P_{r,n_{0},f}(\infty)=0\), and hence
$$P_{r,n_{0},f}(t)=\pm P_{r}(t), $$
\(t\geq t_{k}\). But then
\(|P_{r}(t)|<\frac{f(t)}{2c}\),
i.e.,
$$\frac{f(t)}{|P_{r}(t)|}>2c,\quad t\geq t_{n}, $$
which contradicts (
27). Sufficiency is proved.
We will prove the necessity now.
\(\lim_{n\to\infty} \varphi _{r,n,f}(\infty)=\delta>0\). Assume the converse, let
\(\varliminf_{t\to \infty}\frac{f(t)}{\vert P_{r}(t)\vert }=\infty\). Then there exists a number
\(M>0\) such that for all
\(t>M\) the inequality
$$\frac{f(t)}{|P_{r}(t)|}>\frac{1}{\delta} $$
holds; this is equivalent to
$$\bigl\vert P_{r}(t)\bigr\vert < \delta f(t). $$
In view of Lemma
6, for all
n, the following inequality holds:
$$ \int_{0}^{M}\bigl\vert P^{\prime}_{r,n,f}(t)\bigr\vert \,dt\leq\int_{0}^{M} \bigl\vert P_{r-1}(t)\bigr\vert \,dt. $$
(29)
Choose
n so big that
$$ n\cdot f(M)>\int_{0}^{M}\bigl\vert P_{r-1}(t)\bigr\vert \,dt. $$
(30)
Choose
m such that the
g-spline
\(P_{r,m,f}(t)\) has at least
\(n + 1 \) oscillation points (this is possible due to Lemma
5). Denote by
\(0\leq a_{1}< a_{2}<\cdots<a_{n+1}\) the first oscillation points of the
g-spline
\(P_{r,m,f}(t)\). Then, in virtue of (
30), and by the facts that
f is non-increasing function,
\(\bigvee_{0}^{M} P_{r,m,f}=\int_{0}^{M}\vert P^{\prime}_{r,m,f}(t)\vert \,dt \) and (
29), we get that
\(a_{n}>M\). Thus
\(a_{n+1}>a_{n}>M\) are oscillation points of the
g-spline
\(P_{r,m,f}(t)\), and we get
$$\begin{aligned} \bigl\vert P_{r,m,f}(a_{n+1})-P_{r,m,f}(a_{n}) \bigr\vert =&\varphi_{r,m,f}(\infty )\cdot\bigl(f(a_{n+1})+f(a_{n}) \bigr) \\ \geq&\delta\cdot\bigl(f(a_{n+1})+f(a_{n})\bigr)>\delta f(a_{n}) \\ >&\bigl\vert P_{r}(a_{n})\bigr\vert =\int _{a_{n}}^{\infty}\bigl\vert P_{r-1}(t)\bigr\vert \,dt. \end{aligned}$$
On the other hand, in view of Lemma
6,
$$\begin{aligned} \bigl\vert P_{r,m,f}(a_{n+1})-P_{r,m,f}(a_{n}) \bigr\vert \leq&\int_{a_{n}}^{a_{n+1}}\bigl\vert P^{\prime}_{r,m,f} (t)\bigr\vert \,dt \\ \leq&\int_{a_{n}}^{a_{n+1}}\bigl\vert P_{r-1}(t)\bigr\vert \,dt \\ < &\int_{a_{n}}^{\infty} \bigl\vert P_{r-1}(t)\bigr\vert \,dt. \end{aligned}$$
Contradiction. The theorem is proved.