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Let $K$ be a compact Hausdorff space and $A$ be a subset of $C(K)$. $A$ is said to be a Korovkin set if for every sequence $(T_n)$ of positive linear operators on $C(K)$, the condition $\|T_n(f)- f\|_\infty\to 0$ for all $f\in A$ ensures that $\|T_n(g)- g\|_\infty\to 0$ for all $g\in C(K)$.

The famous Korovkin Theorem says $\{1,x,x^2\}$ is a Korovkin set for $C_{\mathbb{R}}[0,1]$.

My question is is it true that $\{1,z,z^2\}$ is Korovkin set for $C_{\mathbb{C}}(\mathbb{T})$?. Here $\mathbb{T}$ is the unit circle in the complex plane.

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    $\begingroup$ Working backwards from the statement of Korovkin's theorem encyclopediaofmath.org/index.php/Korovkin_theorems I assume that Tanmoy's use of "uniformly" should be "uniformly on $K$". My own preference would be to say $\Vert T_n(f)-f\Vert_\infty \to 0\forall\,f\in A \implies \Vert T_n(g)-g\Vert_\infty\to 0 \forall\,g\in C(K)$. $\endgroup$
    – Yemon Choi
    May 20, 2019 at 16:28
  • $\begingroup$ Yes, I mean, whether $\|T_n(f)-f\|_\infty\to 0$ for all $f\in A$ implies $\|T_n(g)-g\|_\infty\to 0$? $\endgroup$
    – Tanmoy Paul
    May 20, 2019 at 16:32
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    $\begingroup$ Note that, since all $T_n$ are positive, they map real-valued functions to real-valued functions. This implies that $\overline{T_nf} = T_n\overline{f}$ for each $n$ and each $f$. Hence, the answer is "yes", as follows from the link provided by @YemonChoi (the corresponding result is called "Korovkin's second theorem" there). By the way, this even shows that $\{1, z\}$ is a Korovkin set for $C_{\mathbb{C}}(\mathbb{T})$. $\endgroup$
    – Jochen Glueck
    May 20, 2019 at 16:49

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