Topological spaces containing paths 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 $X$.

 A: 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 $X$.

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 $x \in Y$.
