Korovkin-type results on convergence of sequences of positive linear maps on function spaces

In this paper we deal with the convergence of sequences of positive linear maps to a (not assumed to be linear) isometry on spaces of continuous functions. We obtain generalizations of known Korovkin-type results and provide several illustrative examples.

1 of Korovkin's theorem. In particular, we obtain proper generalizations of [7,Theorems 3.1 and 4.1] and of several classical Korovkin-type results, and provide several illustrative examples.

Preliminaries
For any compact Hausdorff space X, let C(X) denote the space of continuous real or complexvalued functions on X, equipped with the uniform norm · . Note that we write C R (X) instead of C(X) when we want to consider only real-valued case. A unital subspace S of C(X) is called a function space on X if S separates the points of X in the sense that for each x, x ′ ∈ X with x = x ′ there exists a function f ∈ S such that f (x) = f (x ′ ).
Let S be a subspace of C(X), which we always assume to be linear. We denote by B S * the closed unit ball of the dual space of (S, · ). A nonempty subset E of X is called a boundary for S if each function in S attains its maximum modulus within E. The Choquet boundary Ch(S) of S is the non-empty set of all points x ∈ X for which δ x , the evaluation functional at x, is an extreme point of the closed unit ball B S * . Namely, we have ext(B S * ) = T Ch(S) = {αx : α ∈ T and x ∈ Ch(S)}, where T = {z ∈ C : |z| = 1}. It is known that Ch(S) is a boundary for S. In particular, one can obtain the following remark immediately: In the sequel, unless otherwise stated, it is assumed that X and Y are compact Hausdorff spaces, M is a self-conjugate subspace of C(X) in the sense thatf ∈ M whenever f ∈ M , and S is a Given f, g ∈ C(X), we shall write f ⊗ 1 + 1 ⊗ g to denote the function in C(X × X) such that Finally let us state the following lemma which is used in the proofs of our results.

Results
Theorem 3.1. Suppose that {T n } is a sequence of positive linear maps from M into C(Y ), and T ∞ is an isometry from M onto a subspace T ∞ (M ) of C(Y ).
Proof. We will base the proof of (a) through the following steps.
Since M is a self-conjugate function space we can find a real-valued function f ∈ M such that f (x) = 1 and f (x ′ ) = 0. Now we consider the following cases based on the value of f at z: In this case we choose a non-negative function g in M with g(x), g(x ′ ) > 3 and is the desired function. Otherwise, we can see that h = 2f + g − g(z) − 1 satisfy the requested properties.
Step 2. T ∞ is a linear isometry.
Note that T ∞ 0 = lim T n 0 = 0. Then according to the Mazur-Ulam theorem [10], T ∞ is a reallinear isometry. Hence now we only need to consider the complex case. Let us point out that . Hence taking into account that Ch(T ∞ (M )) is a boundary for T ∞ (M ), we deduce that T ∞ is a linear isometry.
. 3 Let f ∈ M and ǫ > 0. Then we can define a function in C(X × X) as F : Let y ′ ∈ Ch(T ∞ (S)) and In other words, Hence for each y ∈ X we get Since {T n } is a sequence of linear positive maps, it follows that for each y ∈ X. Now, from the representation of T ∞ on M ( Step 2), we deduce that for any z ∈ Y and z ′ ∈ Ch(T ∞ (M )). Thus, again since T ∞ 1 = 1, T ∞ is a positive linear map and also Ch(T ∞ (M )) is a boundary for T ∞ (M ), it is observed that the above relation holds for all z, z ′ ∈ Y . Therefore, especially we get Rewriting the above inequality adopted to our notation in Section 2 we have equivalently, Consequently, from the fact that each T n is a positive linear map and the representation of T ∞ , it follows that which is to say, Thus, from the latter inequality, the representation of T ∞ and for any sufficiently large integer n, we get Hence , which completes the proof of part (a).
(b) We first claim that T n ≤ √ 2 T n 1 , where T n is the operator norm of T n (for each n ∈ N).
To see this, assume that g ∈ M is real-valued and has supremum norm at most 1. Then −1 ≤ g ≤ 1 and thus, −T n 1 ≤ T n g ≤ T n 1, which implies that T n g ≤ T n 1 . In the real case, this shows that T n is continuous and the claim holds. In the complex case, from this argument and the fact that M is self-conjugate, it easily follows that T n ≤ √ 2 T n 1 .
Let f ∈ M . Taking into account the above claim and the boundedness of {T n 1 : n ∈ N}, we deduce that the set {T n f : n ∈ N} is bounded. Now one can follow the last part of the proof of [7, Theorem 3.3] to conclude that T n f −→ T ∞ f on Y and we include it for completeness. Assume that ∼ is the equivalence relation on Y defined by The quotient space of Y by ∼ is denoted by Y / ∼, andŷ will stand for the image of y ∈ Y under the canonical map· from Y onto Y / ∼. Moreover, we defineĝ(ŷ) = g(y) for all g in N and y in Therefore, T n f −→ T ∞ f , as desired.
Let us recall here the famous Arzela-Ascoli theorem, which will be used in the proof of the next result.
Theorem (Arzela-Ascoli). Given a subset A of C(X), the following statements are equivalent: (1) A is a compact subset of (C(X), · ).
(2) A is closed, bounded, and equicontinuous in the sense that for each x ∈ X and ǫ > 0, there exists a neighborhood V of x such that |f (y) − f (x)| < ǫ for all f ∈ A and y ∈ V . such that for all f ∈ S, T ∞ f (y) = f (ϕ(y)) (f ∈ S, y ∈ Ch(T ∞ (S))).
Suppose that K is a compact subset of Ch(T ∞ (S)). Let f ∈ M , y ′ ∈ K and ǫ > 0. Put F = f ⊗ 1 − 1 ⊗ f and x ′ = ϕ(y ′ ). As before, we choose an open neighborhood V x ′ of x ′ and a function < ǫ, and we also have on Y . Now, we prove the following claim.

Claim:
The set {T n f : n ∈ N} is equicontinuous at y ′ .
Since {T n f y ′ } and {T n 1} converge uniformly to T ∞ f y ′ and 1, respectively, there is an integer n 0 such that for each n ≥ n 0 , T n f y ′ − T ∞ f y ′ < ǫ and T n 1 − 1 < ǫ. On the other hand, ReT ∞ f y ′ (y ′ ) < ǫ and so, from the continuity of ReT ∞ f y ′ and T ∞ f , we can choose a neighborhood Hence, letting η = sup i∈N T i 1 , for each y ∈ W y ′ and n ≥ n 0 we get Now, from the continuity of T 1 f, ..., T n0 f , it follows that the set {ReT n f : n ∈ N} is equicontinuous at y ′ . Similarly, the set {ImT n f : n ∈ N} is equicontinuous at y ′ , and, as a consequence, {T n f : n ∈ N} is equicontinuous at y ′ , as claimed.
Moreover, as observed in the proof of Theorem 3.1(b), {T n f : n ∈ N} is bounded. Therefore, from the Arzela-Ascoli theorem and Theorem 3.1(a), it follows that each subsequence {T n f } has a uniformly convergent sequence to T ∞ f on K. This argument shows that {T n f } converges uniformly to T ∞ f on the compact set K.
(b) When either Ch(T ∞ (S)) or Ch(N ) is compact, then, from the above discussion, we deduce that {T n f } converges uniformly to T ∞ f on Ch(N ). Next, since Ch(N ) is a boundary for N , it is immediately seen that {T n f } converges uniformly to T ∞ f (on Y ).

Remark 3.3.
We would like to remark that the sequential version of Korovkin's theorem does not yield its net version (see [14]). However, it can be easily checked that our techniques hold true when we replace the sequence {T n } by a net of positive linear maps.

Examples
In this section we provide several examples which show how our results can be applied.
For each a ∈ I, the function h(x) = 1 − (x − a) 2 belongs to the function space S = Span{1, x, x 2 }.
Since h(a) = 1 and |h(y)| < 1 for any y = a, we infer Ch(S) = I, by Remark 2.1. Now from Theorem 3.1, we conclude that T n f −→ f for all f ∈ C (k) (I). Meantime, by Theorem 3.2, the same result holds true for "uniformly convergence" instead of "pointwise convergence", which can be also obtained from Korovkin's first theorem. 8 Example 4.2.
Let Ω be a non-empty open subset of R p and K be a compact subset of Ω. The term multi-index denotes an ordered p-tuple α = (α 1 , ..., α p ) of nonnegative integers α i . For each multi-index α, consider the differential operator all multi-index α. By D K we denote the space {f | K : f ∈ C ∞ (Ω)}. Since D K may be considered as a function space on K, from our results we deduce the following.
If {T n : D K −→ C(K) : n ∈ N} is a sequence of positive linear maps such that T n 1 −→ 1, then T n f −→ f for all f ∈ D K . A similar result holds true for "uniformly convergence" instead of "pointwise convergence".
The following example includes the complex Korovkin theorem. It should be noted that since T n is positive, it is easily seen that T nz = T n z, which yields , which belongs to S = Span{1, z,z, |z| 2 }, is the appropriate function for Remark 2.1.
The two above results holds true for "uniformly convergence" instead of "pointwise convergence".
Remark 4.4. From our theorems, one can obtain the Korovkin-type results of [11] and [15] (with respect to both "uniformly convergence" and "pointwise convergence"), which are generalizations of Korovkin's second theorem on convergence of a sequence of positive linear maps for the space of real-valued continuous 2π-periodic functions on R.