Note On Jakimovski-Leviatan Operators Preserving e^–x

If g(1) = 1 in (1), we obtain pk(x)=xkk! $\begin{array}{} \displaystyle p_{k}(x)=\frac{x^k}{k!} \end{array}$ and we get classical Szász-Mirakjan operator which is given by

Sn(f;x)=e−<!-- - -->nx∑<!-- ? -->k=0∞<!-- 8 -->nxkk!fkn. $$\begin{array}{} \displaystyle S_{n}(f;x)=e^{-nx}\sum_{k=0}^{\infty}\frac{\left(nx\right)^k}{k!}f\left(\frac{k}{n}\right). \end{array}$$

B. Wood in [6] proved that the operators P_n are positive if and only if ang(1)≥<!-- = -->0 $\begin{array}{} \displaystyle \frac{a_{n}}{g(1)}\geq 0 \end{array} $ for n ∈ ℕ. In [5], Ciupa studied the rate of convergence of these operators. The convergence of these operators in a weighted space of functions on a positive semi-axis and estimate the approximation by using a new type of weighted modulus of continuity introduced by A.D. Gadjiev and A. Aral in [8] were studied in [3].

The moment for the operator (3) may be given as

EneAt;x=enan(x)e A / n−<!-- - -->1. $$\begin{array}{} \displaystyle E_{n}\left(e^{At};x\right)=e^{na_{n}(x)\left[e^{A/n}-1\right]}. \end{array} $$(7)

From (5), we have

EneAt;x=e−<!-- - -->nan(x)g(1)∑<!-- ? -->k=0∞<!-- 8 -->aknan(x)kk!eAkn=1g(1)∑<!-- ? -->k=0∞<!-- 8 -->ake−<!-- - -->nan(x)nan(x)eA/nkk!=enan(x)eA/n−<!-- - -->1 $$\begin{array}{} \begin{split} \displaystyle E_{n}\left(e^{At};x\right)=&\frac{e^{-na_{n}(x)}}{g(1)}\sum_{k=0}^{\infty}a_{k}\frac{\left(na_{n}(x)\right)^k}{k!}e^{A\frac{k}{n}}\\ \displaystyle=&\frac{1}{g(1)}\sum_{k=0}^{\infty}a_{k}\frac{e^{-na_{n}(x)}\left(na_{n}(x)e^{A/n}\right)^k}{k!}\\ \displaystyle=&e^{na_{n}(x)\left[e^{A/n}-1\right]} \end{split} \end{array}$$(8)

□

If the operator E_n is defined by (3), then with e_r(t) = t^r, r = 0, 1, 2... the moments as follows;

Ene0;x=1Ene1;x=an(x)Ene2;x=an2(x)+an(x)/n $$\begin{array}{} \begin{split} \displaystyle E_{n}\left(e_{0};x\right)=&1\\ \displaystyle E_{n}\left(e_{1};x\right)=&a_{n}(x)\\ \displaystyle E_{n}\left(e_{2};x\right)=&a_{n}^2(x)+a_{n}(x)/n\\ \end{split} \end{array} $$(9)

By Lemma (2) the central moments for E_n operator

Enϕ<!-- ? -->xm(t);x=En(t−<!-- - -->x)m;x,m=0,1,2 $$\begin{array}{} \displaystyle E_{n}\left(\phi_{x}^m(t);x\right)=E_{n}\left((t-x)^m;x\right), m=0,1,2 \end{array} $$(10)

are given by

Enϕ<!-- ? -->x0(t);x=1Enϕ<!-- ? -->x1(t);x=an(x)−<!-- - -->xEnϕ<!-- ? -->x2(t);x=(an(x)−<!-- - -->x)2+an(x)/n. $$\begin{array}{} \begin{split} \displaystyle E_{n}\left(\phi_{x}^0(t);x\right)=&1\\ \displaystyle E_{n}\left(\phi_{x}^1(t);x\right)=&a_{n}(x)-x\\ \displaystyle E_{n}\left(\phi_{x}^2(t);x\right)=&(a_{n}(x)-x)^2+a_{n}(x)/n. \end{split} \end{array}$$(11)

([11]) If a sequence of linear positive operators L_n:C^*[0, ∞) → C^*[0, ∞) satisfy the equalities

Lne0−<!-- - -->1[0,∞<!-- 8 -->)=α<!-- a -->nLn(e−<!-- - -->t)−<!-- - -->e−<!-- - -->x[0,∞<!-- 8 -->)=β<!-- ? -->nLn(e−<!-- - -->2t)−<!-- - -->e−<!-- - -->2x[0,∞<!-- 8 -->)=γ<!-- ? -->n $$\begin{array}{c} \begin{split} \displaystyle \left\|L_{n}e_{0}-1\right\|_{[0,\infty)}=&\alpha_{n}\\ \displaystyle\left\|L_{n}(e^{-t})-e^{-x}\right\|_{[0,\infty)}=&\beta_{n}\\ \displaystyle\left\|L_{n}(e^{-2t})-e^{-2x}\right\|_{[0,\infty)}=&\gamma_{n}\\ \end{split} \end{array} $$(14)

then, for f ∈ C^*[0, ∞), we have

Lnf−<!-- - -->f[0,∞<!-- 8 -->)≤<!-- = -->α<!-- a -->nf[0,∞<!-- 8 -->)+(2+α<!-- a -->n)ω<!-- ? -->∗<!-- * -->(f,α<!-- a -->n+2β<!-- ? -->n+γ<!-- ? -->n), $$\begin{array}{} \displaystyle \left\|L_{n}f-f\right\|_{[0,\infty)}\leq \alpha_{n}\left\|f\right\|_{[0,\infty)}+(2+\alpha_{n})\omega^*(f,\sqrt{\alpha_{n}+2\beta_{n}+\gamma_{n}}), \end{array}$$(15)

where the modulus of continuity is defined as:

ω<!-- ? -->∗<!-- * -->(f,δ<!-- d -->)=supe−<!-- - -->x−<!-- - -->e−<!-- - -->t≤<!-- = -->δ<!-- d -->x,t><!-- > -->0f(t)−<!-- - -->f(x). $$\begin{array}{} \displaystyle \omega^*(f,\delta)=\sup_{\left|e^{-x}-e^{-t}\right|\leq \delta \atop x,t \gt0}\left|f(t)-f(x)\right|. \end{array}$$(16)

For f ∈ C^*[0, ∞), we get

Enf−<!-- - -->f[0,∞<!-- 8 -->)≤<!-- = -->2ω<!-- ? -->∗<!-- * -->f;2β<!-- ? -->n+γ<!-- ? -->n, $$\begin{array}{} \displaystyle \left\|E_{n}f-f\right\|_{[0,\infty)}\leq 2\omega^*\left(f;\sqrt{2\beta_{n}+\gamma_{n}}\right), \end{array} $$(17)

where

β<!-- ? -->n=En(e−<!-- - -->t)−<!-- - -->e−<!-- - -->x[0,∞<!-- 8 -->),γ<!-- ? -->n=En(e−<!-- - -->2t)−<!-- - -->e−<!-- - -->2x[0,∞<!-- 8 -->). $$\begin{array}{} \begin{split} \displaystyle \beta_{n}=& \left\|E_{n}(e^{-t})-e^{-x}\right\|_{[0,\infty)},\\ \displaystyle\gamma_{n}=& \left\|E_{n}(e^{-2t})-e^{-2x}\right\|_{[0,\infty)}. \end{split} \end{array}$$

Here, β_n and γ_n tend to zero as n goes to infinity, therefore E_n converges uniformly to f.

By using the equalities (3) and (5), we get

Ene−<!-- - -->λ<!-- ? -->t;x=e−<!-- - -->nan(x)g(1)∑<!-- ? -->k=0∞<!-- 8 -->aknan(x)kk!e−<!-- - -->λ<!-- ? -->kn=e−<!-- - -->nan(x)enan(x)e−<!-- - -->λ<!-- ? -->/n=enan(x)e−<!-- - -->λ<!-- ? -->/n−<!-- - -->1=exeλ<!-- ? -->/neλ<!-- ? -->/n−<!-- - -->1e−<!-- - -->1/n−<!-- - -->1. $$\begin{array}{} \begin{split} \displaystyle E_{n}\left(e^{-\lambda t};x\right)=&\frac{e^{-na_{n}(x)}}{g(1)}\sum_{k=0}^{\infty}a_{k}\frac{\left(na_{n}(x)\right)^k}{k!}e^{-\lambda \frac{k}{n}}\\ \displaystyle=&e^{-na_{n}(x)}e^{na_{n}(x)e^{-\lambda /n}}\\ \displaystyle=&e^{na_{n}(x)\left(e^{-\lambda /n}-1\right)}\\ \displaystyle=&e^{\frac{x}{e^{\lambda/n}}\left(\frac{e^{\lambda /n}-1}{e^{-1/n}-1}\right)}. \end{split} \end{array} $$

Firstly, let take λ = 1. Using the inequality

u−<!-- - -->vln⁡<!-- ? -->u−<!-- - -->ln⁡<!-- ? -->v<<!-- ? -->u+v2 $$\begin{array}{} \displaystyle \frac{u-v}{\ln u-\ln v}\lt\frac{u+v}{2} \end{array}$$

for 0 < v < u, we get

e−<!-- - -->xun−<!-- - -->e−<!-- - -->x<<!-- ? -->1−<!-- - -->un2xe−<!-- - -->xun+xe−<!-- - -->x. $$\begin{array}{} \displaystyle e^{-xu_{n}}-e^{-x}\lt\frac{1-u_{n}}{2}\left(xe^{-xu_{n}}+xe^{-x}\right). \end{array}$$

On the other hand, since

maxx><!-- > -->0xe−<!-- - -->bx=1eb $$\begin{array}{} \displaystyle \max_{x\gt0}xe^{-bx}=\frac{1}{eb} \end{array}$$

for every b > 0, we can write as

e−<!-- - -->xun−<!-- - -->e−<!-- - -->x<<!-- ? -->1−<!-- - -->un21eun+1e=1−<!-- - -->un22eun $$\begin{array}{} \begin{split} \displaystyle e^{-xu_{n}}-e^{-x}\lt&\frac{1-u_{n}}{2}\left(\frac{1}{eu_{n}}+\frac{1}{e}\right)\\ =&\frac{1-u_{n}^2}{2eu_{n}} \end{split} \end{array} $$

where un=−<!-- - -->1e1/ne1/n−<!-- - -->1e−<!-- - -->1/n−<!-- - -->1. $\begin{array}{} \displaystyle u_{n}=-\frac{1}{e^{1/n}}\left(\frac{e^{1/n}-1}{e^{-1/n}-1}\right). \end{array} $ Thus

En(e−<!-- - -->t;x)−<!-- - -->e −<!-- - --> x [0,∞<!-- 8 -->)=β<!-- ? -->n<<!-- ? -->1−<!-- - -->un22eun→<!-- ? -->0, $$\begin{array}{} \displaystyle \left\|E_{n}(e^{-t};x)-e^{-x}\right\|_{[0,\infty)}=\beta_{n}\lt \frac{1-u_{n}^2}{2eu_{n}}\rightarrow 0, \end{array}$$

as n → ∞.

For λ = 2, we have

e−<!-- - -->xvn−<!-- - -->e−<!-- - -->2x<<!-- ? -->2−<!-- - -->vn2xe−<!-- - -->xvn+xe−<!-- - -->2x<<!-- ? -->2−<!-- - -->vn21evn+12e=4−<!-- - -->vn24evn $$\begin{array}{} \begin{split} \displaystyle e^{-xv_{n}}-e^{-2x}\lt&\frac{2-v_{n}}{2}\left(xe^{-xv_{n}}+xe^{-2x}\right)\\ \displaystyle\lt&\frac{2-v_{n}}{2}\left(\frac{1}{ev_{n}}+\frac{1}{2e}\right)\\ \displaystyle=&\frac{4-v_{n}^2}{4ev_{n}} \end{split} \end{array} $$

where vn=−<!-- - -->1e2/ne2/n−<!-- - -->1e−<!-- - -->2/n−<!-- - -->1. $\begin{array}{} \displaystyle v_{n}=-\frac{1}{e^{2/n}}\left(\frac{e^{2/n}-1}{e^{-2/n}-1}\right). \end{array}$ Therefore

En(e−<!-- - -->2t;x)−<!-- - -->e−<!-- - -->2x[0,∞<!-- 8 -->)=γ<!-- ? -->n<<!-- ? -->4−<!-- - -->vn24evn→<!-- ? -->0, $$\begin{array}{} \displaystyle \left\|E_{n}(e^{-2t};x)-e^{-2x}\right\|_{[0,\infty)}=\gamma_{n}\lt \frac{4-v_{n}^2}{4ev_{n}}\rightarrow 0, \end{array}$$

as n → ∞. Hence the proof is complete.□

Let f', f˝ ∈ C^*[0, ∞). Then, we have

n[En(f;x)−<!-- - -->f(x)]−<!-- - -->x2f′(x)−<!-- - -->x2f″(x)≤<!-- = -->rn(x)f′(x)+qn(x)f″(x)+22gn(x)+x+sn(x)ω<!-- ? -->∗<!-- * -->(f″;1/n) $$\begin{array}{} \displaystyle \left|n[E_{n}(f;x)-f(x)]-\frac{x}{2}f'(x)-\frac{x}{2}f''(x)\right| \leq \left|r_{n}(x)\right|\left|f'(x)\right|+\left|q_{n}(x)\right|\left|f''(x)\right|+2\left(2g_{n}(x)+x+s_{n}(x)\right)\omega^*(f'';1/ \sqrt{n}) \end{array}$$

where

rn(x)=nEn(ϕ<!-- ? -->x1(t);x)−<!-- - -->x2qn(x)=12nEn(ϕ<!-- ? -->x2(t);x)−<!-- - -->xsn(x)=n2En(e−<!-- - -->x−<!-- - -->e−<!-- - -->t)4;xn2En(t−<!-- - -->x)4;x. $$\begin{array}{} \begin{split} \displaystyle r_{n}(x)=&nE_{n}(\phi_{x}^1(t);x)-\frac{x}{2} \\ q_{n}(x)=&\frac{1}{2}\left(nE_{n}(\phi_{x}^2(t);x)-x\right) \\ \displaystyle s_{n}(x)=&\sqrt{n^2E_{n}\left((e^{-x}-e^{-t})^4;x\right)}\sqrt{n^2E_{n}\left((t-x)^4;x\right)}. \end{split} \end{array}$$

By the Taylor’s expansion of f, we have

f(t)=f(x)+f′(x)(t−<!-- - -->x)+f″(x)2(t−<!-- - -->x)2+h(t,x)(t−<!-- - -->x)2, $$\begin{array}{} \displaystyle f(t)=f(x)+f'(x)(t-x)+\frac{f''(x)}{2}(t-x)^2+h(t,x)(t-x)^2, \end{array} $$(18)

where

h(t,x)=f″(η<!-- ? -->)−<!-- - -->f″(x)2 $$\begin{array}{} \displaystyle h(t,x)=\frac{f''(\eta)-f''(x)}{2} \end{array} $$

is a continous function and η is a number between x and t. Applying the E_n operator to both sides of equality (18), we get

En(f;x)−<!-- - -->f(x)−<!-- - -->f′(x)En(ϕ<!-- ? -->x1(t);x)−<!-- - -->f″(x)2En(ϕ<!-- ? -->x2(t);x)≤<!-- = -->En(h(t,x)ϕ<!-- ? -->x2(t);x). $$\begin{array}{} \displaystyle \left|E_{n}(f;x)-f(x)-f'(x)E_{n}(\phi_{x}^1(t);x)-\frac{f''(x)}{2}E_{n}(\phi_{x}^2(t);x)\right| \leq \left|E_{n}(h(t,x)\phi_{x}^2(t);x)\right|. \end{array} $$

n[En(f;x)−<!-- - -->f(x)]−<!-- - -->x2f′(x)−<!-- - -->x2f″(x)≤<!-- = -->nEn(ϕ<!-- ? -->x1(t);x)−<!-- - -->x2f′(x)+12nEn(ϕ<!-- ? -->x2(t);x)−<!-- - -->xf″(x)+nEn(h(t,x)ϕ<!-- ? -->x2(t);x). $$\begin{array}{} \begin{split} \displaystyle \left|n[E_{n}(f;x)-f(x)]-\frac{x}{2}f'(x)-\frac{x}{2}f''(x)\right|\leq& \left|nE_{n}(\phi_{x}^1(t);x)-\frac{x}{2}\right|\left|f'(x)\right|\\ \displaystyle+&\frac{1}{2}\left|nE_{n}(\phi_{x}^2(t);x)-x\right|\left|f''(x)\right|+\left|nE_{n}(h(t,x)\phi_{x}^2(t);x)\right|. \end{split} \end{array}$$

Take rn(x)=nEn(ϕ<!-- ? -->x1(t);x)−<!-- - -->x2 $\begin{array}{} \displaystyle r_{n}(x)=nE_{n}(\phi_{x}^1(t);x)-\frac{x}{2} \end{array} $ and qn(x)=12nEn(ϕ<!-- ? -->x2(t);x)−<!-- - -->x. $\begin{array}{} \displaystyle q_{n}(x)=\frac{1}{2}\left(nE_{n}(\phi_{x}^2(t);x)-x\right). \end{array}$ Thus, we get

n[En(f;x)−<!-- - -->f(x)]−<!-- - -->x2f′(x)−<!-- - -->x2f″(x)≤<!-- = -->rn(x)f′(x)+qn(x)f″(x)+nEn(h(t,x)ϕ<!-- ? -->x2(t);x) $$\begin{array}{} \displaystyle \left|n[E_{n}(f;x)-f(x)]-\frac{x}{2}f'(x)-\frac{x}{2}f''(x)\right| \leq \left|r_{n}(x)\right|\left|f'(x)\right|+\left|q_{n}(x)\right|\left|f''(x)\right|+\left|nE_{n}(h(t,x)\phi_{x}^2(t);x)\right| \end{array}$$(19)

In order to complete the proof of the theorem, we must find the last term of the inequality (19). Here, we also know that the equalities given in (12) and (13) r_n(x)→ 0, q_n(x)→ 0 as n → ∞ at any point x ∈ [0, ∞).

Using the property

f(t)−<!-- - -->f(x)≤<!-- = -->1+(e−<!-- - -->t−<!-- - -->e−<!-- - -->x)2δ<!-- d -->2ω<!-- ? -->∗<!-- * -->(f;δ<!-- d -->),δ<!-- d -->><!-- > -->0 $$\begin{array}{} \displaystyle \left|f(t)-f(x)\right|\leq\left(1+\frac{(e^{-t}-e^{-x})^2}{\delta^2}\right)\omega^*(f;\delta), \delta\gt0 \end{array} $$

we have

h(t,x)≤<!-- = -->1+(e−<!-- - -->t−<!-- - -->e−<!-- - -->x)2δ<!-- d -->2ω<!-- ? -->∗<!-- * -->(f″;δ<!-- d -->). $$\begin{array}{} \displaystyle \left|h(t,x)\right|\leq\left(1+\frac{(e^{-t}-e^{-x})^2}{\delta^2}\right)\omega^*(f'';\delta). \end{array}$$

If |e^–x–e^–t| ≤ δ, then |h(t,x)| ≤ 2ω^*(f˝;δ). In case |e^–x–e^–t|> δ, then we get |h(t, x)| ≤ 2(e−<!-- - -->t−<!-- - -->e−<!-- - -->x)2δ<!-- d -->2ω<!-- ? -->∗<!-- * -->(f″;δ<!-- d -->). $\begin{array}{} \displaystyle 2\frac{(e^{-t}-e^{-x})^2}{\delta^2}\omega^*(f'';\delta). \end{array}$ Hence

h(t,x)≤<!-- = -->21+(e−<!-- - -->t−<!-- - -->e−<!-- - -->x)2δ<!-- d -->2ω<!-- ? -->∗<!-- * -->(f″;δ<!-- d -->). $$\begin{array}{} \displaystyle \left|h(t,x)\right|\leq2\left(1+\frac{(e^{-t}-e^{-x})^2}{\delta^2}\omega^*(f'';\delta)\right). \end{array}$$

By using this inequality and applying Cauchy-Schwarz inequality, we get

n.Enh(t,x)ϕ<!-- ? -->x2(t);x≤<!-- = -->2nω<!-- ? -->∗<!-- * -->(f″;δ<!-- d -->)Enϕ<!-- ? -->x2(t);x+2nδ<!-- d -->2ω<!-- ? -->∗<!-- * -->(f″;δ<!-- d -->)En(e−<!-- - -->x−<!-- - -->e−<!-- - -->t)4;xEn(t−<!-- - -->x)4;x. $$\begin{array}{} \begin{split} \displaystyle n.E_{n}\left(\left|h(t,x)\right|\phi_{x}^2(t);x\right)\leq& 2n\omega^*(f'';\delta)E_{n}\left(\phi_{x}^2(t);x\right)\\ \displaystyle+&\frac{2n}{\delta^2}\omega^*(f'';\delta)\sqrt{E_{n}\left((e^{-x}-e^{-t})^4;x\right)}\sqrt{E_{n}\left((t-x)^4;x\right)}. \end{split} \end{array}$$

Choosing δ<!-- d -->=1/n $\begin{array}{} \displaystyle \delta=1/\sqrt{n} \end{array}$ and letting

sn(x)=n2En(e−<!-- - -->x−<!-- - -->e−<!-- - -->t)4;xn2En(t−<!-- - -->x)4;x, $$\begin{array}{} \displaystyle s_{n}(x)=\sqrt{n^2E_{n}\left((e^{-x}-e^{-t})^4;x\right)}\sqrt{n^2E_{n}\left((t-x)^4;x\right)}, \end{array} $$

we obtain our result which was claim in the theorem.□

Let f, f˝ ∈ C^*[0, ∞). Then we get

limn→<!-- ? -->∞<!-- 8 -->n[En(f;x)−<!-- - -->f(x)]=x2[f′(x)+f″(x)] $$\begin{array}{} \displaystyle \lim_{n\rightarrow \infty}n[E_{n}(f;x)-f(x)]=\frac{x}{2}[f'(x)+f''(x)] \end{array}$$

for any x ∈ [0, ∞).