Fourier Coefficients of Continuous functions are square summable.

[Kreizhn]

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.

 

[mighty2000]

One way to approach this problem is by using the Cauchy-Schwarz inequality. We know that for any two sequences of complex numbers, a and b, we have
|\sum_n a_n \overline{b_n}| \leq (\sum_n |a_n|^2)^{1/2} (\sum_n |b_n|^2)^{1/2}.

Applying this to the case at hand, we have
|\sum_n n^2 \hat f(n) \overline{\hat f(n)}| \leq (\sum_n n^4 |\hat f(n)|^2)^{1/2} (\sum_n |\hat f(n)|^2)^{1/2}.

Now, since f is continuously differentiable, f' is also continuous and hence integrable. This means that the Fourier coefficients of f' are also square summable. Therefore, we have
\sum_n n^4 |\hat f(n)|^2 < \infty.

Combining this with the previous inequality, we get
\sum_n n^2 |\hat f(n)|^2 < \infty.

This shows that the Fourier coefficients of f are indeed square summable, as desired.