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Can you show an example of a function that does not satisfy pointwise convergence theorem hypotheses for Fourier series but that is still expressible as Fourier series?

[Added after comment] In particular, I want to know if there is a function whose Fourier series converges but which does not satisfy the hypotheses of pointwise convergence theorem:

Let $c \in \mathbb{R}$ and suppose that $f: \mathbb{R} \to \mathbb{R}$ has the following properties:

$f$ has period $2 \pi$;

$f$ is piecewise continuous on $[-\pi,\pi]$;

$D^{+}f$ and $D^{-}f$ exist.

If f is continuous at $c$, then its Fourier series is equal to $f(c)$, whereas if $f$ has a jump discontinuity at $c$, then its Fourier series is

$$\frac{1}{2} \left[f(c^{+}) + f(c^{-}) \right].$$

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    $\begingroup$ What do you mean "still expressible as a Fourier series"? If a function is continuous on the circle, it admits a Fourier series expansion. Such Fourier series expansion need not converge everywhere. Do you want to ask about a function whose Fourier series converges but which does not satisfy the hypotheses of a particular pointwise convergence theorem? If so please specify which pointwise convergence theorem you are thinking of. $\endgroup$ – Willie Wong Jul 12 '13 at 15:05
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The Fourier series $\sum_n {e^{inx}\over (1+|n|)^{{1\over 2}+\epsilon}}$, with any positive epsilon, converges in $L^2$, but certainly does not converge pointwise at $0$ when $0<\epsilon<{1\over 2}$.

The series does converge at other points, but certainly not absolutely, and not uniformly pointwise.

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  • $\begingroup$ What is the function from which this series comes? $\endgroup$ – pter26 Feb 8 '18 at 13:25
  • $\begingroup$ @user411485, the function probably has no simpler alternative description than its Fourier expansion. $\endgroup$ – paul garrett Feb 8 '18 at 13:30
  • $\begingroup$ So I'm sure it's a Fourier series because of its "shape"? $\endgroup$ – pter26 Feb 8 '18 at 13:39
  • $\begingroup$ @user411485, yes, the Fourier coefficients are square-summable, so, by Plancherel, it is/gives a function in $L^2$. $\endgroup$ – paul garrett Feb 8 '18 at 13:51

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