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Let $Q = [0,1]^2$. For sake of notation, let $$ f^{(i,j)}(x,\xi) = \frac{\partial^{i+j}}{\partial x^i \partial \xi^j}f(x,\xi). $$ Fix some non-negative integer $k$. Moreover let $f\in C^k(Q)$ if $$ \|f\|_{C^k} = \sum_{i+j \leq k}\sup_{(x,\xi)\in Q}|f^{(i,j)}(x,\xi)| < \infty. $$ Additionally, let $m = (m_1,m_2) \in \mathbf{Z}^2$. Define $$ E_m = E_m(x,\xi) = e^{2\pi i m_1 x}e^{2\pi i m_2 \xi}. $$ Question. Is $\{E_m\}_{m\in \mathbf{Z}^2}$ dense in $C^k(Q)$ for $k\geq 0$.

Ideas. The question can be rephrased as follows. For any $f\in C^k(Q)$ and $\epsilon > 0$ does there exist some double trigonometric polynomial $$ p_k(x,\xi) = \sum_{m\in F}c_m E_m,\quad F \subseteq \mathbf{Z},\; |F| <\infty. $$ such that $$ \|f-p_k\|_{C^k} < \epsilon. $$ I am confident that for $k=0$ this result is already been established elsewhere. I've tried a similar technique in this Q&A. However, the multi-index derivatives make it complicated to carry over. I've also thought about some potential "convolution with some kernel" argument, but I haven't given it much time. Any advice would be appreciated or a counterexample if this does not hold.

Secondary Question. I actually want to prove or dispove the following (I think the aforementioned claim helps). Let $$ \frac{\partial^{i+j}}{\partial x^i \partial \xi^j}f(x_l,\xi_l) = \langle f, \delta^{(i,j)}_{x_l,\xi_l}\rangle. $$ Morever, let $C^k(Q)$ be the set of $k$ continuously differentiable periodic functions. Hence, $\delta^{(i+j)}_{x_l,\xi_l}\in C^k(Q)'$ for $i+j \leq k$.

Claim: $\delta^{(i+j)}_{x_l,\xi_l}$ is uniquely determined by its Fourier coefficients.

I believe this is true when $k=0$, however, it seems less trivial for arbitrary $k\geq 1$. Any assistance would be greatly appreciated.

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    $\begingroup$ You ask whether the linear span of the exponentials is dense. Since they are periodic they only can approximate periodic $C^k$-functions. It is thus better to look at $C^k(\mathbb T)$ for the torus. You might be interested in Nachbin's theorem, discussed, e.g., here math.stackexchange.com/questions/2150294 $\endgroup$ Commented Apr 21 at 7:11

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In my opinion this is a question of MSE level.

For the first question the answer is yes. Because finite trigonometric polynomials is dense in $C^k(\Bbb T^n)$. This is essentially another Weierstrass approximation theorem.

Note that it is possible that for a $f(x,y)\in C^k(\Bbb T^2)$ neither $\sum_{|j|+|k|\le N}\hat f(j,k)e^{2\pi i(jx+ky)}$ or $\sum_{|j|^2+|k|^2\le N^2}\hat f(j,k)e^{2\pi i(jx+ky)}$ converges to $f$ in $C^k$. It is fail even in 1-dimension. You can read Stein and Shakarchi's Fourier Analysis: An Introduction for a proof on $k=0$. The larger $k$ can be modified similarly.

Just a remark the lim $N\to\infty$ do not converge in $C^k(\Bbb T^2)$. But if you only consider a suitable subsequence $N_l\to\infty$ it may still converges. And when we just prove Weierstrass approximation we don't even need the $p_k$ to be the partial Fourier series.

For your second question we prefer to use the notation $\partial^\alpha f(x_0)=(-1)^{|\alpha|}\langle f,\partial^\alpha\delta_{x_0}\rangle$. Here $\alpha=(\alpha_1,\dots,\alpha_n)$, $|\alpha|=\alpha_1+\dots+\alpha_n$ and $f\in C^k(\Bbb T^n)$ (yes your formula misses a $\pm$ sign).

$\partial^\alpha\delta_{x_0}$ is a distribution, whose Fourier expansion converges also in the space of distributions. I don't have an immediate reference on hand. You can read Folland's Real Analysis Chapter 9.2 for a brief discussion.

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    $\begingroup$ So you're saying that the trigonometric polynomials are dense, yet there formal Fourier series that do not converge to some specific function under those two orderings, correct? I've thought about the Weierstrass approximation, but the antiderivatives don't make sense. For example, if you were to integrate $e^{2\pi i n x}$ with respect to $y$ you will get a $y$ term and the function is no longer a trigonometric polynomial. Lastly, I thought the $(-1)$ term came about due to the boundary conditions equaling 0, but what if $x_0 \in \delta \mathbf{T}^n$? $\endgroup$ Commented Apr 21 at 20:15
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    $\begingroup$ @Doofenshmert I think yes. But what is the two ordering? $\endgroup$
    – Liding Yao
    Commented Apr 21 at 20:17
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    $\begingroup$ The two orderings you included. $|j| + |k| \leq N$ and $|j|^2 + |k|^2 \leq N^2$. I was just confirming that I was understanding your answer. $\endgroup$ Commented Apr 21 at 20:19
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    $\begingroup$ These are not ordering. These are how you consider the partial sums! The first one is the sum taking on interor of a diamond with diagonal length $2N$. The second one is the that on the interior of a disk of radius $N$. $\endgroup$
    – Liding Yao
    Commented Apr 21 at 20:21
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    $\begingroup$ I understand. Thanks for the clarification. $\endgroup$ Commented Apr 21 at 20:26