Korovkin's theorem

Remarks

The idea of the proof: fix $f$ and $\varepsilon > 0$. By uniform continuity there is $\delta > 0$ with $\abs{f(t) - f(x)} \le \varepsilon + 2 \norm{f}_\infty (t - x)^2 / \delta^2$ for all $t, x \in [a, b]$. Apply the positive operator $L_n$ in the variable $t$ and evaluate at $x$. This bounds $\abs{L_n f(x) - f(x)}$ by $\varepsilon$ plus terms in $L_n e_0 - e_0$, $L_n e_1 - e_1$ and $L_n e_2 - e_2$, which tend to $0$ uniformly.

Example: the Bernstein operators $B_n$ on $[0, 1]$ satisfy $B_n e_0 = e_0$, $B_n e_1 = e_1$ and $B_n e_2 = e_2 + x(1 - x)/n$. So Korovkin's theorem gives $B_n f \to f$ uniformly for every continuous $f$.

Positivity is needed. In the periodic setting the test functions are $1, \cos x, \sin x$. The partial sums of Fourier series reproduce these test functions for $n \ge 1$, but they are not positive operators. And there are continuous periodic functions whose Fourier series do not converge uniformly.

Going deeper

For a student

The theorem is the standard tool to prove convergence of the Bernstein, Szász–Mirakyan, Baskakov and Kantorovich operators and of Fejér means. The periodic version uses the test functions $1, \cos x, \sin x$, and it proves Fejér's theorem on Cesàro means of Fourier series. Quantitative forms bound $\norm{L_n f - f}_\infty$ by the modulus of continuity of $f$ and the errors on the test functions. The theorem also shows a rigidity: a single positive linear operator on $C[a, b]$ that fixes $1$, $x$ and $x^2$ must be the identity.