My next acknowledgment goes to my respected guide Dr. Daniel Sukumar, because of whom, I had the opportunity to learn, understand my subject properly. I am very much indebted to him for his rigorous efforts in the realization of this project. I would like to thank my classmates and seniors who have always supported me in every matter.
In my project I am doing my work on the Korovkin's theorem and some standard version of Korovkin's theorem and their applications. If a series of linear positive functionals onCb(R) defines for which conditions it will be convergent and where converges. In this statement, it says that if a sequence of positive linear operators defined onC[a, b] satisfies certain conditions, then it is uniformly convergent onC[a, b].
The existence of such a relationship is explained by the fact that many important problems of approximation theory have been formulated. In fact, all known methods of approximating functions by means of algebric or trigonometric polynomials (which are partial sums of Taylor's series, interpolating polynomials, Bernstein and Landau polynomials, partial sums of the Fourier series, Fejer sums, etc.) linear comparators. .
Some definitions and formulas
Theorem
Lemma
We will now consider operators in function space that help us understand linear functionals. Theorem: Let (Tn(f)) be a sequence of positive linear functionals on the domain of existence of all functionals of this sequence. Proof: We will first construct two inequalities f(x) with the conditions that it is bounded on the real axis and continuous at x=c.
Since the function f(x) is bounded on the real axis, there exists a positive real number M such that. Since > 0 is arbitrary and the sequence (Tn) of positive linear functionals converges to 1 and 0 respectively at the two functions 1 and ψforn→. According to this equality, a function ψ(x) =H(f;x) is associated with each function f(t) given on the set of pointst1, t2, .., tn.
We say that an operator H(f;x) =H(f(t);x) is given on the setF of functions f(t) if the function ψ(x) is associated with every functionf(t) of the setF . If we accept that the conclusion of the theorem does hold, we get a function f(t) that will satisfy the conditions of this theorem, for which the sequence Ln(f;x) does not converge uniformly to the function f(x) in the interval [ will not converge. a, b]. This means that there exists a >0, a range of points (xk) for alk∈N, a≤x≤band a range of numbersnk, for alk∈Nand lim.
Since the sequence (xk) is bounded, so by the Bolzano-Weierstrass theorem we obtain a subsequence (xns) from the sequence (xk) which is convergent and converges to a point, say in the interval [a, b]. We will show that the sequence of functions Lnks(f;xks) satisfies the conditions of this theorem. Q = Q(δ) be the largest of the numbers p and q, since the function ψ(x) reaches its maximum value atx= 1, then the function ψ(x) satisfies the inequality.
Since Q1 > Q, if we take the limit as n→ ∞, we get the limit on the right-hand side of 1 and use the Sandwich theorem for the above inequality. Proof Before proving this theorem, we want to recall the theorem that says: “If a sequence of linear functionals (Tn) converges to 1 for the function 1(t) and converges to 0 for the function (t−α)2.
Application of this theorem
Approximation of functions by trigonometric polynomials
Therefore, this sequence of functions (fn) is uniformly convergent to the function f (x) on [0, a]. Theorem3: Let D be a subset of R and a sequence of functions (fn) be uniformly convergent on D to a function f. Let D ⊂ R and the sequence of bounded functions (fn) on D such that the sequence of functions (fn) is pointwise convergent to a function f on D. Then the limit function may not be bounded on D.
Now for all n ∈ N the sequence of functions is bounded functions on (0, 1). But the limit function f is not bounded on (0, 1). Thm4: Let D ⊂ R and let (fn) be sequence of bounded functions on D. If the sequence (fn) is uniformly convergent to a function f on D, then f is also bounded on D. Every fn is continuous on [0 , 1]. The sequence of functions (fn) is pointwise convergent to the function f on D defined by.
Theorem(Dini): Let D be a compact subset of R and the sequence of continuous functions (fn) converge pointwise to a continuous function f on D. If the sequence of functions (fn) is a monotone sequence of functions on D, is the convergence of the sequence of functions (fn) is uniform with the function f on D. Proof: If the sequence of functions (fn) is monotonically increasing, let us consider a sequence of functions (gn) on D such that gn = f − fn .If the sequence of functions (fn) is monotonically decreasing, then gn = fn− f. Therefore the sequence of functions (gn) converges uniformly on D and therefore the sequence of functions (fn) converges uniformly to the function f on D.
This says that if the convergence of the set of functions (fn) is uniform on the interval [a, b], we can interchange the limit and the integration. Let us consider a set of functions (fn) on [a, b], such that for all n ∈ N the function fn is differentiable on [a, b). Let the set of functions (fn) converge pointwise to a function f on [a, b]. Then dxdf (x) may not exist for all x ∈ [a, b]. Let us consider a set of functions (fn) on a closed and bounded interval [a, b]. Let us converge pointwise the set of functions (fn) to a function f on [a, b].
Theorem 7: Let (fn) be a sequence of functions on the closed and bounded interval [a, b] such that for all n ∈ N the function fn is differentiable on [a, b]. If the sequence of functions (dxdfn) is uniformly convergent on [a, b] and the sequence of functions (fn) converges at least at one point x0 ∈ [a, b], then the sequence of functions (fn) is uniformly convergent on [a, b] and let the function f be the limit, then the sequence of functions (dxdfn) converges to the function dxdf on [a, b]. Therefore, according to Cauchy's principle, the sequence of functions (fn) is uniformly convergent on [a, b]. Let f be the limit function, then lim. Therefore, according to the Cauchy criterion, the sequence of functions (gn) is uniformly convergent [a, b] for t 6= x. Since the sequence of functions (fn) converges to f, we conclude that.
For a sequence of functions (fn), where each function f is differentiable on a closed and bounded interval [a, b] and the sequence of functions (fn) is pointwise convergent on [a, b], the uniform convergence of the sequence of functions dxdfn is only sufficient, but not also a necessary condition for uniform convergence of a sequence of functions on [a, b]. Since the limit function g is not continuous on [0, 1]. Therefore, the sequence of continuous functions (dxdfn) is not uniformly convergent on [0, 1]. Thus, the sequence of functions (fn) is uniformly convergent on [0, 1], but the sequence of functions (dxdfn) is not uniformly convergent on [0, 1] and our claim is confirmed.
Now if the set of functions (fn) is bounded pointwise on D and D1 is a countable subset of D, it is always possible to find a subset of functions (fnk) such that the subset (fnk) converges for any point x ∈ D1 This can be done by a diagonal process.
