Fourier Coefficients of Continuous functions are square summable.

  • Thread starter Kreizhn
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Homework Statement



If [itex] C^1(\mathbb T) [/itex] denotes the space of continuously differentiable functions on the circle and [itex] f \in C^1(\mathbb T) [/itex] show that
[tex] \sum_{n\in\mathbb Z} n^2 |\hat f(n)|^2 < \infty[/tex]
where [itex] \hat f(n) [/itex] is the Fourier coefficient of f.

The Attempt at a Solution



Since f is continuous it is integrable and so [itex] \widehat{f'}(n) = in \hat f(n) [/itex]. Thus
[tex] \sum_n n^2 |\hat f(n)|^2 = \sum_n |\widehat{f'}(n)|^2 [/tex]
so this boils down to showing that the Fourier coefficients of a continuous function are square summable.

Now not all continuous functions need to have absolutely convergent Fourier series, so somehow this result implies that all continuous functions have square-summable series? This is not clear to me.
 

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