# Degree of smoothness of real functions and Fourier series

I was wondering, why is the degree of smoothness $S$ of functions an integer? Why can't there be functions with $S=2/3$ ? The way we determine how smooth a function is by how many continuous derivatives it has, and that's an integer. In Fourier series context, if a periodic funtion has a discontinuity the Fourier coefficients $a_n\sim 1/n$ as $n \to \infty$. For a continuous function $a_n\sim 1/n^2$ and in general if $f$ has $k$ continuous derivatives, $a_n\sim n^{-(k+2)}$. Is there any example where e.g. $a_n\sim 1/n^{3/2}$ or even slower, $1/(n\log{n})$ ? If there is, how does $f$ look like?

For functions in $C^\infty$ that have an infinite number of continous derivatives the coefficients fall off faster than any negative power of $n$. My simplest (non-trivial) example for those is

$$f(x)=\frac{1}{a+\cos{x}}$$

where $a>1$ is a constant, which is infinitely differentiable in $[-\pi,\pi]$ and hence $a_nn^p \to 0$ for every power $p$ no matter how large. I 've tried to evaluate the Fourier series for this, there doesn't seem to be a closed form for them and Mathematica takes an exponential time to calculate each coefficient, and the output is enormous! I would bet exponential convergence, and perhaps even faster. Has anybody an example of convergence according to fractional power law and/or exponential in closed form?

I look forward to that. Many thanks, I appreciate.

By looking at Fourier coefficients $c_n$, we can easily see the $L^2$ norm of the derivative of any order $k$: it is $\sqrt{\sum |n|^{2k} |c_k|^2}$. Nothing here restricts $k$ to integers. So, for any $k>0$ we can define $H^k$, the space of functions with square integrable $k$th order derivative, by the requirement that $\sum |n|^{2k} |c_k|^2<\infty$. An important example is $H^{1/2}$, the functions with square integrable half-derivative. This space is the trace class of $H^1$ functions in the plane, and has some other uses (random example).
You can construct explicit functions with exponential convergence of Fourier series by restricting any holomorphic function in a neighborhood of the circle $\{z:|z|=1\}$. Simplest example is $$\frac{1}{1-z/2} = \sum_{n=0}^\infty \frac1{2^n}z^n$$ which with $z=e^{it}$ becomes $$\frac{1}{1-e^{it}/2} = \sum_{n=0}^\infty \frac1{2^n}e^{int}$$
To have slower order of decay, you need a singularity on the circle. A convenient example is $f(z)=(1+z)^\alpha$ with fractional $\alpha$, which, when expanded into binomial series, has coefficients of order $1/n^{1+\alpha}$. Plugging in $z=e^{it}$ as above, you get $f(t) = (1+e^{it})^{\alpha}$, with Fourier coefficients of order $1/n^{1+\alpha}$.