Consider the following multiple Fourier series $f(x_1,...,x_N)=\sum_m a^me^{2\pi i(m_1x_1+...+m_{N}x_{N})}$ where $m=(m_1,...,m_{N})$ is an integers vector. Moreover, suppose that $f$ is an infinitely differentiable function.
Then is it true that the series is uniformly convergent?
It seems to be true but I cannot find a good reference online about multiple Fourier series.
Thank you for your help
Sure. It works exactly the same way when $N= 1$ : the Fourier series of $f''(x) $ is $\sum_m a_m (2i \pi m)^2 e^{2i \pi m x}$ (we don't know yet if it converges) where $a_m (2i \pi m)^2 = \int_0^1 f''(x)e^{-2i \pi m x}dx \le \int_0^1 |f''(x)|dx= C$ where $C$ is finite since you assume $f$ is $C^\infty$ thus $C^2$. Therefore for $ m\ne 0$ : $|a_m| \le \frac{C}{4 \pi^2 m^2}$ and $\sum_m a_m e^{2i \pi m x}$ converges absolutely and uniformly.
For $N > 1$ you can look at the partial $2^{N}$-th derivatives $\prod_{j=1}^N \partial_{x_j}^2 f$ to obtain $|a_m | \le \frac{C}{\prod_{j=1}^N |m_j|^2}$ where $C = \int_{[0,1]^N} |\partial_{x_j}^2 f(x)|dx$ so that $\sum_{m \in \mathbb{Z}^N} a_m e^{2i \pi \langle m, x \rangle}$ converges absolutely.
Note smooth and $1$-periodic is not the same as smooth on $(0,1)$ and $1$-periodic, in the second case $a_m (2i \pi m)^2 = \int_0^1 f''(x)e^{-2i \pi m x}dx$ is not true.