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Let $I_d:=[0,1]^d$ with $d\ge 2$. Define $C(I_d):=\{F: I_d\to\mathbb R \mbox{ is continuous}\}$ and

$$N(I_d):=\{F\in C(I_d): F(x)=\sum_{k=1}^n f_k(v_k\cdot x), \mbox{ where } n\ge 1 \mbox{ and } f_1,\ldots, f_n:\mathbb R\to\mathbb R \mbox{ are continuous and bounded}\}$$.

The universal approximation theorem states that $N(I_d)$ is dense in $C(I_d)$ w.r.t. the uniform norm. With $C_1(I_d):=\{F\in C(I_d): F \mbox{ is } 1-\mbox{Lipschitz}\}$, could we find $L>0$ and some topology s.t. $N_L(I_d)$ is dense in $C_1(I_d)$? Here $N_L(I_d):=\{F\in N(I_d): F \mbox{ is } L-\mbox{Lipschitz}\}$. Any answers, references or comments are highly appreciated!

PS: According to Stone–Weierstrass Theorem (Lattice version), see e.g. https://en.wikipedia.org/wiki/Stone–Weierstrass_theorem one must have $\bigcup_{L>0}N_L(I_d)$ is dense in $C(I_d)$ and thus in $C_1(I_d)$. But it's not clear for me to find out $L>0$ s.t. $N_L(I_d)$ is dense in $C_1(I_d)$.

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  • $\begingroup$ The universal approximation theorem could be found in en.wikipedia.org/wiki/Universal_approximation_theorem $\endgroup$ – Neymar Nov 17 '19 at 20:47
  • $\begingroup$ On $C_1(I_d)$, you are still using the uniform norm? $\endgroup$ – PhoemueX Nov 19 '19 at 10:18
  • $\begingroup$ @PhoemueX Yes. Still with the uniform norm. $\endgroup$ – Neymar Nov 20 '19 at 11:10

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