How to be sure that a series isn't a fourier series of a derivable function

[Telemachus]

Hi there. I have this interesting problem which I don't know how to solve. I'll post it here because I think more people will se it, but I'm not sure if this is the proper subforum.



The problem says: How can be sure that $$\sum_{n = 1}^\infty \frac{1}{n}\sin (nx)$$ isn't the Fourier series of a derivable function?



I thought that it doesn't accomplish the Diritchlet postulates, but it actually doesn't mean that it isn't a Fourier series.



Does anyone know how to solve this?



Bye there and thanks.

 

[micromass]

Remember that coefficients of Fourier series are in $\ell^2$...

 

[Telemachus]

Sorry, I don't know what you mean with $$l^2$$

 

[micromass]

$$\ell^2=\{(x_n)_n~\vert~\sum_{n=1}^{+\infty}{|x_n|^2}<+\infty\}$$

So basically, the coefficients of a Fourier series must be square summable. In fact, a sequence forms the coefficients of a Fourier series if and only if they are square summable!

 

[Telemachus]

Thanks. I think I'm not familiar with this definition. We've defined in class that a Fourier series it's a Fourier series if it accomplish Diritchlet conditions, and that the Fourier series are the series that minimizes the quadratic error.

 

[Dickfore]

Suppose:

$$f(x) \equiv \sum_{n = 1}^{\infty}{\frac{\sin n \, x}{n}}$$

Take the derivative term by term:

$$f'(x) = \sum_{n = 1}^{\infty}{\cos n \, x}$$

Do the coefficients of this Fourier series form a convergent sum?

 

[Telemachus]

Thanks :)