January 01, 1970
We prove two Korovkin-type approximation theorems for sequences of positive linear operators acting on continuous functions on \([0,\infty)\). Under the assumption of pointwise convergence on suitable test functions, we establish pointwise convergence for all functions with polynomial or exponential growth. As direct applications, we obtain convergence results for the classical Baskakov and Szász–Mirakjan operators. The proposed method offers an elementary framework that can be applied to a broad class of positive linear operators.
Korovkin-type theorem, positive linear operators, polynomial growth, exponential growth
41A36.
Let \(L_n : D \to C[0,\infty)\) be a sequence of positive linear operators, where \(D\) is a linear subspace of \(C[0,\infty)\) containing the space \(\Pi\) of all polynomial functions. For example, if \(V_n\) is the sequence of Baskakov operators, then \(D\) can be taken as the subspace of \(C[0,\infty)\) consisting of all functions with at most polynomial growth. In the case of Szász–Mirakjan operators \(S_n\), polynomial growth should be replaced by exponential growth.
For operators \(L_n\) as above, many results are known, guaranteeing a certain type of convergence of \(L_n f\) toward \(f\), for \(f\) in suitably restricted subspaces of \(D\). In this paper we give sufficient conditions entailing the pointwise convergence of \(L_n f\) toward \(f\), for all functions \(f \in D\).
Theorem 1. Let \(D\) be a linear subspace of \(C[0,\infty)\), \(\Pi \subset D\), \(L_n : D \to C[0,\infty)\) positive linear operators, \(L_n p(x) \to p(x)\), for all \(p \in \Pi\), \(x \geq 0\). Let \(f \in D\) such that there exists \(p \in \Pi\) with \(|f(x)| \leq p(x)\), \(x \geq 0\). Then \(L_n f(a) \to f(a)\), \(a \geq 0\).
For the proof we need the following lemma.
Lemma 1. Let \(f \in C[0,\infty)\), \(p \in \Pi\), \(p(x) \geq f(x)\), \(x \in [0,\infty)\). Let \(a \geq 0\), \(\varepsilon > 0\). Then, there exists \(P \in \Pi\): \[\label{equ1} P(a) = f(a) + \varepsilon, \quad P(x) \geq f(x), \quad x \geq 0.\qquad{(1)}\]
Proof Let \(a>0\). Then, there exists \(0 < \ell \leq \dfrac{a}{2}\), such that \[\label{equ2}f(a) + \varepsilon \geq f(x), \,\,x \in (a - 2\ell, a + 2\ell).\tag{1}\]
Let \(m \in \mathbb{N}\), \(2m > \deg p\). Then, there exists \(M > 0\) with \[\left( \frac{x - a}{\ell} \right)^{2m} + f(a) + \varepsilon \geq p(x), \quad x > M,\] which implies that \[\label{equ3} \left( \frac{x - a}{\ell} \right)^{2m} + f(a) + \varepsilon \geq f(x), \quad x > M.\tag{2}\]
Set \(U := [0, a - 2\ell] \cup [a + 2\ell, M]\). Take \(K \geq 1\) with \[\label{equ4} \min \left\{ K \left( \frac{x - a}{\ell} \right)^{2m} + f(a) + \varepsilon \mid x \in U \right\} \geq \max \left\{ p(x) \mid x \in U \right\}.\tag{3}\]
Define \[P(x) := K \left( \frac{x - a}{\ell} \right)^{2m} + f(a) + \varepsilon.\]
Using (1 )–(3 ) we see that \(P\) satisfies (?? ). The proof is similar for \(a = 0\).
Proof of Theorem 1 Let \(a \geq 0\), \(\varepsilon > 0\). Lemma 1 implies the existence of \(P, Q \in \Pi\), such that \[Q(x) \leq f(x) \leq P(x), \quad x \geq 0; \quad Q(a) = f(a) - \varepsilon,\;\;P(a) = f(a) + \varepsilon.\] Now \[L_n Q(a) \leq L_n f(a) \leq L_n P(a); \quad f(a) - \varepsilon = Q(a) = \lim_{n\to \infty} L_n Q(a)\] \[\leq \liminf_{n\to\infty} L_n f(a) \leq \limsup_{n\to\infty} L_n f(a) \leq \lim_{n\to\infty} L_n P(a) = P(a) = f(a) + \varepsilon.\] Therefore, for each \(\varepsilon > 0\) we have \[f(a) - \varepsilon \leq \liminf_{n\to\infty} L_n f(a) \leq \limsup_{n\to\infty} L_n f(a) \leq f(a) + \varepsilon.\] It follows that \(L_n f(a) \to f(a)\) as \(n \to \infty\).
Corollary 1. Let \(V_n\) be the classical Baskakov operators and \(S_n\) the Szász-Mirakjan operators. Let \(f \in C[0,\infty)\) such that there exists \(p \in \Pi\) with \(|f(x)| \leq p(x)\), \(x \geq 0\). Then, \(V_n f \to f\) and \(S_n f \to f\), pointwise on \([0,\infty)\).
Proof The Baskakov and the Szász–Mirakjan operators are exponential-type operators (see [1]). From [1] it follows that \(V_n p \to p\) and \(S_n p \to p\) pointwise on \([0,\infty)\), for all \(p \in \Pi\). An application of Theorem 1 concludes the proof.
Let \(\exp_{\lambda}(x)=e^{\lambda x}\), \(\lambda,x\in{\mathbb{R}}\). Consider also the function \(\alpha(x)=\cosh(x)-1\), \(x\in {\mathbb{R}}\).
Theorem 2. Let \(D\) be a linear subspace of \(C[0,\infty)\) such that \(\exp_\lambda \in D\), \(\lambda \in \mathbb{R}\). Let \(L_n : D \to C[0,\infty)\) be positive linear operators, \(L_n \exp_\lambda \to \exp_\lambda\) pointwise on \([0,\infty)\), \(\lambda \in \mathbb{R}\). Let \(f \in C[0,\infty)\), \(|f(x)| \le A \exp_{\omega}(x),\, x\ge 0\) for some \(A>0\) and \(\omega>0\). Then \(L_n f \to f\) pointwise on \([0,\infty)\).
For the proof we need the following lemma.
Lemma 2. Let \(A>0\), \(\omega>0\), \(f \in C[0,\infty)\), such that \[A \exp_{\omega}(x) \ge f(x), \quad x \ge 0.\] Fix \(a \ge 0\), \(\varepsilon > 0\). Then there exist \(\ell>0\), \(K \in \mathbb{N}\), \(m \in \mathbb{N}\) such that \[K\, \alpha^{m}\!\left(\frac{x-a}{\ell}\right) + f(a) + \varepsilon \ge f(x), \quad x \ge 0. \label{ew1}\qquad{(2)}\]
Proof Let \(a>0\). There exists \(0<\ell \le \dfrac{a}{2}\) such that \[\label{ew2} f(a)+\varepsilon \ge f(x), \quad x \in (a-2\ell, a+2\ell).\tag{4}\]
Let \(m \in \mathbb{N}\), \(m > \omega \ell\). Then \[\lim_{x \to \infty} \frac{\alpha^{m}\!\left(\frac{x-a}{\ell}\right) + f(a)+\varepsilon}{A \exp_{\omega}(x)} = \infty.\]
So, there exists \(M>0\) such that \[\label{ew3} \alpha^{m}\!\left(\frac{x-a}{\ell}\right) + f(a)+\varepsilon \ge f(x), \quad x > M.\tag{5}\]
Set \[U := [0, a-2\ell] \cup [a+2\ell, M].\] Take \(K \in \mathbb{N}\) with \[\label{ew4} \min \left\{ K\, \alpha^{m}\!\left(\frac{x-a}{\ell}\right) + f(a)+\varepsilon \; \middle|\; x \in U \right\} \ge \max \left\{ A \exp_{\omega}(x) \; \middle|\; x \in U \right\}.\tag{6}\]
Then (?? ) is a consequence of (4 )–(6 ). The proof is similar for \(a = 0\).
Proof of Theorem 2 Fix \(a\ge 0\) and \(\varepsilon>0\).
Since \[f(x)\le A\exp_\omega(x),\qquad x\ge 0,\] Lemma 3.2 yields a function \[G(x)=K\alpha^m\!\left(\frac{x-a}{\ell}\right)+f(a)+\varepsilon\] such that \(G(x)\ge f(x)\) for all \(x\ge 0\).
Since \(\alpha(x)=\cosh(x)-1\), the function \(G\) is a finite linear combination of exponentials, hence \(G\in D\). By positivity, \[L_n f(a)\le L_n G(a).\] Using the assumption \(\displaystyle\lim_{n\to\infty}L_n\exp_\lambda(a)= \exp_\lambda(a)\) for every \(\lambda\in\mathbb{R}\), we obtain \[\displaystyle\lim_{n\to\infty}L_nG(a)= G(a)=f(a)+\varepsilon,\] and therefore \[\limsup_{n\to\infty}L_nf(a)\le f(a)+\varepsilon.\] Applying Lemma 3.2 to \(-f\), we similarly obtain \[\liminf_{n\to\infty}L_nf(a)\ge f(a)-\varepsilon.\] Hence \[f(a)-\varepsilon \le \liminf_{n\to\infty}L_nf(a) \le \limsup_{n\to\infty}L_nf(a) \le f(a)+\varepsilon.\] Letting \(\varepsilon\to0\), we conclude that \[\displaystyle\lim_{n\to\infty}L_nf(a)= f(a).\] Since \(a\ge0\) was arbitrary, the convergence is pointwise on \([0,\infty)\).
Corollary 2. Let \(f \in C[0,\infty)\), \(|f(x)| \le A \exp_{\omega}( x)\), \(x \ge 0\), for some \(A>0\) and \(\omega>0\). Then \(S_n f \to f\) pointwise on \([0,\infty)\).
Proof For all \(\lambda\in {\mathbb{R}}\) and \(x\geq 0\), we have \[S_n \exp_{\lambda}(x) = e^{n x (e^{\lambda/n}-1)} \to \exp_{\lambda}(x)\quad (n \to \infty).\] An application of Theorem 2 concludes the proof.