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.


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.

  • $\begingroup$ 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? $\endgroup$ – JessicaK Feb 9 '15 at 13:08
  • $\begingroup$ I've fixed the typos. It should be correct now. $\endgroup$ – user10676 Feb 9 '15 at 13:16

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