Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 69 / 1

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 convergence on the test space implies convergence on every continuous function: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 $c={\min }_K{h}_0>0$. Positivity gives $U_{n}1\le c^{-1}{U}_n{h}_0$, 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 $L_n(g)=U_{n}g(x_n)$ 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 $g_x\ge 0$ 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 $H=\mathrm{span}\{1,\sin t,\cos t\}$. The nonnegative function $g_x(t)=1-\cos (t-x)$ belongs to $H$ and vanishes only at $x$ on the circle. Thus the periodic Korovkin test set isUniform convergence on these three functions is sufficient for uniform convergence on $C(\mathbb{T})$.

Averaging the given Dirichlet kernels gives the normalized Fejér kernelThe geometric-series identity, or summing the sine terms, proves the second equality. At multiples of $2\pi$ use its removable value $n/(2\pi )$. Therefore $K_n\ge 0$, andis a positive linear operator on continuous functions. The Fourier partial sums, viewed as orthogonal projections, fix $1$, so ${\sigma }_{n}1=1$ and $\int K_n=1$. They also fix $\sin x$ and $\cos x$ for degrees at least one, whereas $s_0$ annihilates them. ConsequentlyAll three periodic Korovkin theorem tests converge uniformly. Hence Fejér sums converge uniformly for every continuous periodic function:

Apply the periodic Korovkin theorem to the constant sequence $U_n=U$. Since it fixes the three test functions, the theorem gives $Uf=f$ for every $f$.

One can also see the rigidity directly. For fixed $x$, the positive linear functional $L_x(f)=(Uf)(x)$ has $L_x(1)=1$ and is represented by a probability measure. The fixed sine and cosine imply $L_x(1-\cos (t-x))=0$. The integrand is nonnegative with unique zero $x$ on the circle, so the measure is ${\delta }_x$. Thus $(Uf)(x)=f(x)$ at every point. The operator is uniquely the identity: $\boxed{U=I}$.