# Extension of universal approximation theorem

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 $$C_1(I_d)$$.

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