The neural-operator input and output spaces have the approximation property

The listed Sobolev spaces are Banach under the stated Countable Choice hypothesis. Indeed, let $(u_n)$ be Cauchy in $W^{m,p}(D)$. By [F3], for each of the finitely many $\mathrm{\mid }\alpha \mathrm{\mid }\le m$, $D^{\alpha }{u}_n$ converges in $L^p$ to a class $v_{\alpha }$; put $u=v_0$. For any $\varphi \in C_c^{\mathrm{\infty }}(D)$, its derivatives are bounded with compact support, hence belong to the conjugate $L^{p^{\mathrm{\prime }}}$ space, including $p^{\mathrm{\prime }}=\mathrm{\infty }$ when $p=1$. By [F4], both integrals in ${\int }_D{u}_n{D}^{\alpha }\varphi =(-1{)}^{\mathrm{\mid }\alpha \mathrm{\mid }}{\int }_D{D}^{\alpha }{u}_n\varphi$ converge to the corresponding integrals of $u$ and $v_{\alpha }$. Thus $v_{\alpha }$ is the weak derivative $D^{\alpha }u$ by [F3], so $u\in W^{m,p}(D)$, and the finite norm sum gives $u_n\to u$ in that norm. This includes $m=0$, which gives $L^p(D)$.

The listed $C^m(\bar{D})$ spaces are Banach as well. If $(u_n)$ is Cauchy in their norm, every derivative sequence has a uniform limit $g_{\alpha }$ on $\bar{D}$: pointwise real completeness gives the limit, and the uniform Cauchy estimate gives uniform convergence. Each $g_{\alpha }$ is continuous, since for a positive error one first approximates it uniformly by a fixed continuous $D^{\alpha }{u}_n$ and then applies continuity of that fixed function. Put $u=g_0$. For $\mathrm{\mid }\alpha \mathrm{\mid }<m$, an interior point $x$, and a coordinate direction $e_i$, take $h$ small enough that the segment $x+[0,h]e_i$ lies in $D$. By [F5], $D^{\alpha }{u}_n(x+h{e}_i)-D^{\alpha }{u}_n(x)={\int }_0^h{D}^{\alpha +e_i}{u}_n(x+s{e}_i)ds$. Uniform convergence passes this identity to $g_{\alpha }(x+h{e}_i)-g_{\alpha }(x)={\int }_0^h{g}_{\alpha +e_i}(x+s{e}_i)ds$, since the integral error is bounded by $\mathrm{\mid }h\mathrm{\mid }$ times the uniform error. By [F5], ${\mathrm{\partial }}_i{g}_{\alpha }=g_{\alpha +e_i}$. Induction in $\mathrm{\mid }\alpha \mathrm{\mid }$ yields $D^{\alpha }u=g_{\alpha }$ in $D$ through order $m$, with the continuous extensions already supplied on $\bar{D}$. Thus $u\in C^m(\bar{D})$ and $u_n\to u$ in norm. For $m=0$ only the uniform-limit continuity argument is needed. The displayed maximum is a norm because its order-zero term detects zero functions and the remaining norm axioms hold termwise.

Fix one listed space $X$, a norm-compact $K\subseteq X$, and $\epsilon >0$. Its Banach-space status was established in the two preceding steps. Apply [F1], using $W^{0,p}=L^p$ in the Lebesgue case, to obtain a bounded finite-rank $F:X\to X$ with ${\sup }_{u\in K}\mathrm{\parallel }Fu-u{\mathrm{\parallel }}_X<\epsilon$. These are exactly the quantified requirements of [F2], so every listed space has AP, including every $p=1$ case.

The argument in step 2.1 holds for every Euclidean dimension, every nonnegative integer regularity and every finite exponent in the stated ranges. Applying it with the output dimension and output regularity parameters proves the output-space clause. The proof establishes AP for $C^m(\bar{D})$ without any basis assertion.