Evaluation of convolutions by Fejér kernels have finite-dimentional range? Consider $\mathbb{R}^N = (\mathbb{R}^N,\|\cdot\|_\infty)$  with $N\in\mathbb{N}$. Let $\mathrm{Lip}_0(\mathbb{R}^N, \mathbb{R})$, or just $\mathrm{Lip}_0(\mathbb{R}^N)$, be the Banach space of real-valued Lipschitz functions on $\mathbb{R}^N$ that vanish at $0$, equipped with the norm of the Lipschitz constant, i.e. $\|f\|_{\mathrm{Lip}} = L(f)$. Now define $\phi_n: \mathbb{R}^N\to (\mathrm{Lip}_0(\mathbb{R}^N))^\ast$ such that, for every $x\in \mathbb{R}^N$, $$\phi_n(x) : f\mapsto (f\ast F_n^N)(x) - (f\ast F_n^N)(0),$$
where $F_n^N$ are the Fejér kernels (convoluted on the cube $\pi B_\mathbb{R^N}$) defined by the product of the standard Fejér kernels for each coordinate, that is,
$$F_n^N(x) = \prod_{k=1}^N F_n(x_k) = \frac{1}{n^N}\prod_{k=1}^N \frac{\sin^2(nx_k/2)}{\sin^2(x_k/2)}.$$
How to prove that $\phi_n(\mathbb{R}^N)$ is a finite-dimensional subspace of $(\mathrm{Lip}_0(\mathbb{R}^N))^\ast$? I have no clue why this is true. Note that $\phi_n$ is not linear.
 A: Ok, finally got it. Define, for every $k\in \{0,\dots, n\}^N$,
            \begin{align*}
   r_k: \mathbb{R}^N& \to (\mathrm{Lip}(\mathbb{R}^N))^\ast\\
   x & \mapsto
   \begin{aligned}[t]
   r_k(x) : \mathrm{Lip}_0(\mathbb{R}^N) &\to \mathbb{R}\\
   f&\mapsto \hat{f}(k)(\exp\left(i\langle k, x\rangle_2\right) - 1),
   \end{aligned}
   \end{align*}
where
            \begin{equation*}
   \hat{f}(k) = \frac{1}{(2\pi)^N}\int_{\pi B_\infty} f(t) \exp\left(-i \langle k, t\rangle_2\right)\mathrm{d} t.
   \end{equation*}
            Note that for any $x,y\in \mathbb{R}^N\setminus \{0\}$, there exists $\alpha\in \mathbb{R}$ such that $\alpha r_k(x) = r_k(y)$, i.e. $\exists \alpha :(\alpha r_k(x))(f) = (r_k(y))(f), \forall f\in\mathrm{Lip}_0(\mathbb{R}^N)$, which means $r_k$ has $1$-dimensional range. Consider the exponential form of the convolutions by Fejér kernels on cubes, that is
            \begin{equation}\label{eq:convdophin}
   f\ast F_n^N(x) = \sum_{\|k\|_\infty \leq n} \left( 1 + \frac{\|k\|_\infty}{n}\right)\hat{f}(k) \exp\left(i \langle k. x\rangle_2\right)
   \end{equation}
            It follows that $\phi_n(x)$ is written as a linear combinations of $r_k(x)$'s. Therefore $\dim \phi_n(\mathbb{R}^N) \leq |\{0,\dots,n\}^N| = (2n +1)^N$.
