= Solution
For <positive linear operators on continuous functions>, a general <Korovkin theorem>, in its <compact Korovkin test space> form, can be stated using a test space $H\subset C(K,\mathbb R)$, where $K$ is a <compact Hausdorff space>. Assume $H$ contains a strictly positive function $h_0$. Require that, for every $x\in K$, the only <positive linear functional> $L$ on $C(K)$ with $L(h)=h(x)$ for all $h\in H$ is evaluation at $x$. Then \b[convergence on the test space implies convergence on every continuous function]:
$$
\boxed{\|U_nh-h\|_\infty\longrightarrow0\ (h\in H)
\quad\Longrightarrow\quad\|U_nf-f\|_\infty\longrightarrow0\ (f\in C(K)).}
$$
By the <Riesz-Markov-Kakutani representation theorem>, the functional condition equivalently says that a positive measure with these test moments must be $\delta_x$. This permits arbitrary test families, not just the interval tests $1,x,x^2$.
For completeness, let $c=\min_Kh_0>0$. Positivity gives $U_n1\leq c^{-1}U_nh_0$, so the <operator norms> are uniformly bounded. If the conclusion failed for some $f$, choose $x_n$ where its error is bounded away from zero. The evaluation functionals $L_n(g)=U_ng(x_n)$ have uniformly bounded norm. Compactness of $K$ and the <Banach-Alaoglu theorem> give a subnet on which $x_n\to x$ and $L_n$ converges in the <weak-star topology> to a positive $L$. For every $h\in H$, uniform test convergence gives $L(h)=h(x)$. The hypothesis forces $L(f)=f(x)$, contradicting the chosen errors. Complex-valued functions follow by applying the real result to their real and imaginary parts.
A useful concrete sufficient condition is that $1\in H$ and for each $x$ there is $g_x\in H$ with $g_x\geq0$ and zero set exactly $\{x\}$. Its zero moment forces the representing measure to be supported at $x$, while the constant moment fixes its mass.
For the circle, take $H=\operatorname{span}\{1,\sin t,\cos t\}$. The nonnegative function $g_x(t)=1-\cos(t-x)$ belongs to $H$ and vanishes only at $x$ on the circle. Thus \b[the periodic Korovkin test set is]
$$
\boxed{1,\ \sin x,\ \cos x.}
$$
<Uniform convergence> on these three functions is sufficient for <uniform convergence> on $C(\mathbb T)$.
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