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Consider any continuous $2\pi$ periodic function (of bounded variation) $f : \mathbb{R} \to \mathbb{R}$ and its fourier series given as

$f(\theta) = \frac{a_o}{2} + \sum\limits_{n = 1}^{\infty} (a_n\cos{n\theta} + b_n\sin{n\theta})$

The above series is the real part of the series

$F(z) = \frac{a_o}{2} + \sum\limits_{n=1}^{\infty} (a_n-jb_n)z^n$

It is stated in this link that $F(z)$ is holomorphic on the unit disc without any restrictions on $f(\theta)$. I suspect that $F(z)$ is not holomorphic for any $f(\theta)$ and request you to clarify the validity of this statement.

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2 Answers 2

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In the given link it is only stated that $F(z)$ is a "power series along the unit circle" which says no more than that the coordinate $\theta$ on $\partial D$ has been replaced by the new variable $z$.

Now to your question: If the function $f(\theta)$ is continuous and of bounded variation on ${\mathbb R}/(2\pi{\mathbb Z})$ then its complex Fourier coefficients $c_k={1\over 2}(a_k-i b_k)$ $\ (k\geq0)$ are of order $O({1\over k})$. This implies that the radius of convergence of the power series $\sum_{k=0}^\infty c_k z^k$ is at least $1$. Therefore the function $F(z)$ is holomorphic in $D$.

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please read the first sentence on the link, I suppose by a holomorphic function, they are referring to $F(z)$. –  Rajesh D Jun 10 '11 at 9:01
    
please clarify on this doubt, Is $F(z)$ analytic on $|z| = 1$ also ? or just inside the disc ?? –  Rajesh D Jun 10 '11 at 10:15
    
@Rajesh D: A function $F$ is considered analytic on an ${\it open}$ set. If this set is $D$ then one may ask about the boundary values for $z\to \zeta\in \partial D$. This is a difficult matter. If, e.g., $\sum_k |c_k| < \infty$ then $F$ can be continued continuously to $\partial D$, and its real part there is equal to the given $f$. –  Christian Blatter Jun 10 '11 at 11:37
    
based on the analyticity of $F(z)$ i intend to evaluate a complex integral using residue theorem. Please see this post and let me know if the analyticity of $F(z)$ is sufficient for the purpose. post : math.stackexchange.com/questions/44267/… –  Rajesh D Jun 10 '11 at 11:44
    
Say $f(\theta) = |\theta|$ on $[-\pi,\pi]$ and extend periodically. Then this $f$ gives you the boundary values of $F$. Since $f$ is not differentiable at $\theta=0$, the function $F$ is certainly not analytic at $z=1$. –  GEdgar Jun 10 '11 at 14:10

It can be easily seen that $F(z)$ is always holomorphic because it is defined by a Taylor expansion and Taylor-expanded functions are always holomorphic within the disk where the Taylor expansion converges: the only thing we have to avoid are singularities.

And the radius of convergence is surely bigger than $1$ because $f(\theta)$ which is the value of $F(z)$ on the unit circle is, by assumption, convergent and well-defined. (If the real part of Taylor expansion is convergent, the imaginary part has to be as well.) Because the Taylor expansion converges on the unit circle, it must also converge for all smaller values of $|z|$ i.e. inside the unit disk.

The statement would actually work even for distributions, not just "genuine" functions $f(\theta)$. For example, if $f(\theta)=\delta(\theta)$, then the Fourier transform is proportional to $1+2\sum_{n\in Z^+} \cos n\theta$ and the corresponding Taylor expansion for $F(z)$ will be $\sum_{n\in Z+} z^n$ which is geometric series converging to $1/(1-z)$.

One picks the delta-function using the usual rules for the distributions, $1/(\theta+i\epsilon)$ is equal to the principal value minus $i\pi$ times $\delta(\theta)$ as long as the imaginary value is picked from the expansion. So distributions will produce singularities of $F(z)$ at the unit circle - but nothing strictly inside the unit disk.

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both the answers are equally helpful but choosing one for the sake of it –  Rajesh D Jun 10 '11 at 9:16
    
please clarify on this doubt, Is $F(z)$ analytic on $|z| = 1$ also ? or just inside the disc ?? –  Rajesh D Jun 10 '11 at 10:16

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