Theorem 7.4.1 (Korovkin).label For each $x \in [0, 1]$, let $g_{x}(t) = (t - x)^{2}$, then
- (1)
For each $f \in C([0, 1]; \complex)$ and $\eps > 0$, there exists $C > 0$ such that
\[|f(t) - f(x)| \le \eps + Cg_{x}(t)\]for all $x, t \in [0, 1]$.
- (2)
Let $f \in C([0, 1]; \complex)$ and $\eps > 0$, then there exists $C > 0$ such that for any positive map $\phi: C([0, 1]; \complex) \to C([0, 1]; \complex)$ and $x \in [0, 1]$,
\[-\eps\norm{\phi(1)}_{u} - C\phi(g_{x})(x) \le \phi(f)(x) - f(x) \le \eps\norm{\phi(1)}_{u} + C\phi(g_{x})(x)\]and
\[\norm{\phi(f) - f}_{u} \le \eps \norm{\phi(1)}_{u} + C\sup_{x \in [0, 1]}|\phi(g_{x})(x)|\] - (3)
For each $n \in \natp$, let $\phi_{n}: C([0, 1]; \complex) \to C([0, 1]; \complex)$ be a positive map. If $\phi_{n}(1) \to 1$, $\phi_{n}(t) \to t$, and $\phi_{n}(t^{2}) \to t^{2}$ uniformly as $n \to \infty$, then $\phi_{n}(f) \to f$ uniformly as $n \to \infty$ for all $f \in C([0, 1]; \complex)$.
Proof, [Exercise 2.20, Po02]. (1): By Proposition 6.4.5, there exists $\delta > 0$ such that $|f(t) - f(x)| < \eps$ for all $x, t \in [0, 1]$ with $|x - t| < \delta$. Thus for each $x, t \in [0, 1]$,
Let $C = \eps/\delta^{2}$, then for each $s \ge 0$,
so
(3): Let $f \in C([0, 1]; \complex)$ and $\eps > 0$. By (2), there exists $C > 0$ such that
for all $n \in \natp$, where
which tends to $0$ as $n \to \infty$. Thus
As the above holds for all $\eps > 0$, $\phi_{n}(f) \to f$ uniformly.$\square$
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