# On convergence of multi-dimensional Fourier series

(Definition)

Let $m, n$ be positive integers with $m \le n.$

Let $f:\Bbb R^n \rightarrow \Bbb C$ be a fucntion in $n$ variables $x_1, x_2, \ldots , x_n$.

Suppose that $f$ is $1$-periodic in $x_1, \ldots, x_m$, i.e., $$f(x_1 + c_1,\ldots,x_m + c_m, x_{m+1},\ldots, x_n) = f(x_1,\ldots,x_n)$$ for $(c_1,\ldots,c_m) \in \{0,1\}^m.$

Let $$a(k_1,\ldots,k_m;x_{m+1},\ldots,x_n)=\int_{0}^1 \cdots \int_{0}^1 f(x_1,\ldots,x_n) \mathrm{exp}(-2 \pi i (k_1 x_1 + \cdots + k_m x_m)) dx_1 \cdots dx_m$$ be the Fourier coefficients of $f$.

(Problem)

I would like to know when the Fourier series $$F(f) = \sum_{k_1,\ldots,k_m \in \Bbb Z} a(k_1,\ldots,k_m;x_{m+1},\ldots,x_n) \mathrm{ exp }(2\pi i(k_1 x_1 + \cdots + k_m x_m))$$ of $f$ converges absolutely to $f$ (because the summation may be multi-dimensional, I impose absolute convergence). To be concrete, I have two problems: (1) When does $F(f)$ converge absolutely to $f$? (2) When does $F(f)$ converge absolutely to $f$ and $$\sum_{k_1,\ldots,k_m \in \Bbb Z} \lvert a(k_1,\ldots,k_m;x_{m+1},\ldots,x_n) \mathrm{ exp }(2\pi i(k_1 x_1 + \cdots + k_m x_m)) \rvert = \sum_{k_1,\ldots,k_m \in \Bbb Z} \lvert a(k_1,\ldots,k_m;x_{m+1},\ldots,x_n) \rvert$$ converge uniformly on any compact subsets in $\Bbb R^{n-m}$?

If $m=n=1$, the condition for convergence is well know. But I cannot find any textbook dealing with the general case. I guess that (1) and (2) hold if $f$ is smooth ($C^{\infty}$-function), but I cannot prove it. Please tell me any sufficent condition for the problem (1) or (2). If you know a good reference about this problem, please tell me as well.

• it works the same for $m > 1$, with the Dirichlet kernel corresponding the order of summation you are considering for your Fourier series, for example $\displaystyle D_N(x) = \sum_{n_1 = -N}^N \ldots \sum_{n_m = -N}^N e^{2 i \pi n_1 \ldots n_m x}$, and from the $\alpha$-Hölder continuity you should get the same result of uniform (absolute when $\alpha > 1/2$) convergence, as for $m=1$ – reuns Sep 15 '16 at 14:29
• Are you interesting mostly in pointwise convergence, or are you okay with $L^2$ convergence? $L^2$ convergence is easier because this is reduced to Parseval's equality, which is not so bad to verify for the products of the various basis functions, one in each variable. You can verify Parseval's equality on a dense set of functions, such as smooth, continuous, or $C^{\alpha}$ as suggested above. Then Parseval's equality extends automatically to the closure of such functions in the full space, which would typically be everything. – DisintegratingByParts Sep 25 '16 at 2:31
• @user1952009 Thank you for your explanation. I understood that the case $m \gt 1$ is not so different from the case $m = 1$. – user356126 Oct 2 '16 at 12:09
• @TrialAndError I am mainly interested in the pointwise convergence. It is because I want to know the values of some given points. – user356126 Oct 2 '16 at 12:12