Let $C(\mathbb{R}^n;\mathbb{R}^d)$ be the space of continuous functions with the uniform-convergence on compacts topology. What are function spaces $X$ building on $C(\mathbb{R}^n;\mathbb{R}^d)$?


* $X$ properly contains $C(\mathbb{R}^n;\mathbb{R}^d)$,
* The subspace topology on $C(\mathbb{R}^n;\mathbb{R}^d)$ (with respect to $X$) agrees with the topology of uniform convergence on compact subsets,
* The elements of $X$ are functions from $\mathbb{R}^n$ to $\mathbb{R}^d$
* $C(\mathbb{R}^n;\mathbb{R}^d)$ is dense in $X$.



Topological spaces containing paths
Any such topology will be fairly unpleasant. For instance, the topology of $X$ cannot be induced by any translation-invariant metric $d$.


**Lemma.** Let $Y\_1, Y\_2$ be two topological vector spaces whose topologies are induced by translation-invariant metrics $d\_1, d\_2$, and let $T : Y\_1 \to Y\_2$ be a continuous linear map. Then $T$ is uniformly continuous.


*Proof.* Since $T$ is continuous at 0, for any $\epsilon > 0$ there exists $\delta > 0$ such that if $d\_1(x, 0) < \delta$ then $d\_2(Tx, 0) < \epsilon$. Now if $d\_1(x,y) < \delta$, then $d\_1(x-y, 0) = d\_1(x,y) < \delta$ and we have
$d\_2(Tx, Ty) = d\_2(Tx-Ty, 0) = d\_2(T(x-y), 0) < \epsilon$.


Now recall that $C(\mathbb{R}^n; \mathbb{R}^d)$ is a Fréchet space, so its usual topology is induced by a complete translation-invariant metric $d\_0$. By assumption, the identity map $id$ from $(C(\mathbb{R}^n; \mathbb{R}^d), d\_0)$ to $(C(\mathbb{R}^n; \mathbb{R}^d), d)$ is a homeomorphism, and so by our lemma, $id$ and $id^{-1}$ are uniformly continuous. In particular, $C(\mathbb{R}^n; \mathbb{R}^d)$ is complete with respect to $d$, and therefore closed in $X$.




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**Edit.** Indeed, $X$ cannot even be a sequential Hausdorff topological vector space. In particular, assuming it is a TVS, its topology cannot be induced by any metric, translation-invariant or not.


In the following, let for brevity $Y = C(\mathbb{R}^n; \mathbb{R}^d)$; the same argument works for any Fréchet space. 


Suppose that $Y \subset X$ and that the subspace topology on $Y$ equals the usual topology induced by the complete translation-invariant metric $d\_0$ on $Y$. I claim $Y$ is closed in $X$.


Suppose $x$ is in the $X$-closure of $Y$, so that there is a sequence $y\_n \in Y$ converging to $x$ in the topology of $X$. Let $\epsilon > 0$ and let $B$ be the open $\epsilon$-ball of the metric $d\_0$ centered at $0$. By assumption $B$ is open in the subspace topology of $Y$ inherited from $X$, so there is an $X$-open set $U$ such that $B = U \cap Y$. In particular, $0 \in U$. Now since subtraction is jointly continuous in $X$, there is another $X$-open neighborhood $V$ of $0$ such that for all $a,b \in V$ we have $a-b \in U$.


Since $y\_n - x \to 0$ in $X$, there exists $N$ so large that for all $n \ge N$ we have $y\_n - x \in V$ (using again the fact that $X$ is a topological vector space). Now if $n,m \ge N$, we have $y\_n - x, y\_m - x \in V$, so that $y\_n - y\_m = (y\_n - x) - (y\_m - x) \in U$. Moreover, $y\_n - y\_m \in Y$ because $Y$ is a vector space. So $y\_n - y\_m \in U \cap Y = B$, meaning that $d\_0(y\_n, y\_m) = d\_0(y\_n - y\_m, 0) < \epsilon$, using the fact that $d$ is translation invariant. 


Hence $y\_n$ is Cauchy in the complete metric $d\_0$, so converges in $d\_0$-metric to some $y \in Y$. Thus we also have $y\_n \to y$ in the topology of $X$. Since the latter is Hausdorff, $x=y$ and thus $x \in Y$.


