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Is there an $f\in L^1(\mathbb{T})$ whose Fourier series converges a.e. on $\mathbb{T}$ but not a.e. to $f$?

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If the Fourier series of $f \in L^1(\mathbb T)$ converges to $g$ pointwise almost everywhere then it will converge in Cesàro mean as well to the same function. Furthermore the Cesàro mean of the Fourier series of $f$ will converge in $L^1(\mathbb T)$ to $f$.

Claim: $f$ and $g$ must coincide.

Edit: There is an error in the below argument, I will fix it tomorrow. I'm too tired now.

Let $f_n$ converge to $f$ in $L^1$ and let $f_n$ converge almost everywhere to $g$ where $g \neq f$ almost everywhere.

Let $\mu$ be the measure, note that

$$ \begin{align} \mu \{|f_n - f| \geqslant \varepsilon \} &\leqslant \frac1{\varepsilon} \int_{\{|f_n - f| \geqslant \varepsilon\}} \varepsilon \, \textrm{d}\mu\\ &\leqslant \frac1{\varepsilon} \int |f_n - f| \, \textrm{d}\mu.\tag{1} \end{align} $$ And the RHS goes to $0$ for all $\varepsilon > 0$. Now fix $\varepsilon > 0$ and consider $$M := \{x : |f(x) - g(x)| \geqslant \varepsilon \}.$$ We will show that this is a null set. Suppose as for contradiction that $M$ has positive measure. Furthermore, define $$M_m := \{x : |f_n(x) - g(x)| \leqslant \varepsilon \text{ for all } n \geq m \}.$$ These are clearly measurable increasing sets in $m$. As we have that $f_n \to g$ almost everywhere. So almost every $x$ in $M$ is contained in one of the $M_m$. Hence the union $$\bigcup_{n = 1}^\infty M_n$$ is equal to $M$ except for maybe a null set. This means that $$\mu \left( \bigcup_{n = 1}^\infty M_n \right ) > 0.$$ So there is at least one $M_m$ with positive measure. So we have $|f_n(x) - f(x)| > \epsilon$ for all $x$ in this $M_m$ and higher $M_m$. Thus the LHS of $(1)$ cannot go to $0$.

More cool things:

Kolmogorov (1923) - Une série de Fourier-Lebesgue divergente presqne partout gives an example of a $L^1$ function that has an almost everywhere divergent Fourier series.

For $1 < p < \infty$ this will not be the case due to the Carleson-Hunt theorem.

Pointwise but not uniform convergence of a Fourier series might also be of interest to you.

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Your argument is correct, but as the Fourier series may not converge to $f$ in $L^1$, this doesn't seem to solve my problem. –  Acky Dec 9 '11 at 3:59
    
@Acky Well it if converges, then so will its Cesaro mean to the same function. The Cesaro mean of the Fourier series will converge everywhere where $f$ is continuous to $f$. Now if $f$ is Riemann integrable then it is a.e. equal to a continuous function. –  Jonas Teuwen Dec 9 '11 at 10:58
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