Suppose $f:\mathbb{R}^{n}\rightarrow \mathbb{C}$ is smooth and 1-periodic.


$$c_{k} = \int_{0}^{1} \dots \int_{0}^{1} f(\vec{y})\operatorname{exp}^{-2\pi i \langle \vec{y},\vec{k}\rangle}dy_{1}\dots dy_{n}$$ where $\langle \cdot,\cdot \rangle$ is taken to be the usual dot product.

How do you prove that

$$f(\vec{x}) = \sum_{k\in\mathbb{Z}^{n}} c_{k}\operatorname{exp}{2\pi i \langle \vec{x}, \vec{k}\rangle} \quad (1)$$

given the standard theorems for Fourier series in one dimension?

I have spent a great deal of time attempting this problem using induction, trying to use the smoothness of $f$ to get that it is rapidly decreasing in each argument, studying the projection-slice theorem among countless other approaches.

At this stage, I have failed so much I don't know what is right or wrong anymore. I think this result needs to be broken down into two parts, where first I show that the RHS of (1) converges to a function $g$, then show using some sort of inequality that $\lVert f-g\rVert_{2} = 0$ and so it must converge to $f(\vec{x})$. Even then, I have gotten more or less nowhere and this may be a very bad idea.

up vote 1 down vote accepted

Here a proof for $n=2$.

Denote $e^\alpha = \exp(2i\pi \alpha)$ and $a_n(x \mapsto f(x)) = \int_{x=0}^1 f(x)e^{-inx} dx$. Let $(x,y) \in \mathbb{R}^2$. Fixing $x$, and using 1-dimensional FT, we have $$f(x,y) = \sum_\ell \int_{y'} f(x,y')~e^{-\ell y'} dy' \cdot e^{\ell y}.$$ Then fixing $y'$, we have : $$f(x,y) = \sum_\ell \int_{y'} \left[ \sum_k a_k\big(x' \mapsto f(x',y')\big) e^{k x}~e^{-\ell y'} dy' \right] \cdot e^{\ell y}.$$ Now we have to prove that $$\int_{y'} \left[ \sum_k a_k \big(x' \mapsto f(x',y')\big) e^{k x}~e^{-\ell y'} dy' \right] = \sum_k \left[ \int_{y'} a_k \big(x' \mapsto f(x',y')\big) ~e^{-\ell y'} dy' ~ e^{k x} \right].$$ This is true because $a_k \big(x' \mapsto f(x',y')\big) \leq \frac{C}{k^2}$ (recall $a_n(\phi) = \frac{i}{n} a_n(\phi')$). But $\int_{y'} a_k \big(x' \mapsto f(x',y')\big) ~e^{-\ell y'} dy' = c_{k,\ell}(f)$, hence you get your formula (1). Moreover since $c_{k,\ell}(f) = \frac{1}{k^2.\ell^2}(\partial_{x^2,y^2} f)$, the convergence is normal.

  • Thank you! It looks so obvious in hindsight. I just want to double check that the second $e^{-\ell y^{\prime}}$ is off by a negative sign? – JessicaK Feb 9 '15 at 13:08
  • I've fixed the typos. It should be correct now. – user10676 Feb 9 '15 at 13:16

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.