Fejer's theorem says that for any continuous function $f \colon S^1 \to \mathbb C$ with Fourier coefficients $a=(a\_n)\_{n \in \mathbb Z}$ the sequence


$$\sigma\_n(a) := \frac1n \sum\_{k=1}^n \sum\_{l=-k}^k a\_l \exp(2\pi i \cdot l\phi)$$


convergences uniformly to $f$. Moreover, by the Riemann-Lebesgue Lemma, the sequence $a=(a\_n)\_{n \in \mathbb Z}$ is in $c\_0(\mathbb Z)$.



> 
> **Question:** Let $a=(a\_n)\_{n \in \mathbb Z}$ be in $c\_0(\mathbb Z)$.
>  Does $\sigma\_n(a)$ converge in measure to some measurable function on $S^1$? 
> 
> 
> 


More generally, is there any summing procedure (or in fact any assignment whatsoever), which leads to a linear map $\Phi \colon c\_0(\mathbb Z) \to M(S^1)$, which extends Fourier summation on finitely supported functions (and preferably also Fejer summation for Fourier series of continuous functions). Here, $M(S^1)$ denotes the space of measurable functions on $S^1$ (up to measure zero) with the usual measure topology given by the metric
$$d(f,g) := \inf\lbrace\varepsilon \mid \mu(\lbrace x \mid |f(x)-g(x)|\geq \varepsilon \rbrace \leq \varepsilon \rbrace.$$



Fejer's theorem and convergence of Fourier series in measure
There is no continuous linear operator from $c\_0$ to $M(S^1)$ that maps the unit vector basis to the characters. In fact, any continuous linear operator from $c\_0$ to $M(S^1)$ maps the unit vector basis to a sequence which converges to zero at a good rate. To see this, note that by Maurey-Nikishin, the operator factors through $L\_p$ for all $0<p<1$, and these spaces have cotype 2. 


Does this answer your question? If not, I don't see a good formulation for your question. If there is anything like a summation method, the Banach-Steinhaus theorem would imply that the operator is continuous. 


