Let $X$, $Y$ be reflexive Banach spaces, and let $\imath:X\hookrightarrow Y$ be a bounded inclusion with dense image. Then for any domain $\Omega\subset\Bbb R^n$ we may define $C\_{\mathrm c}^\infty(\Omega;X)$ to be the (Fréchet)-smooth functions $\Omega\to X$ with compact support, with the usual inductive limit topology, and likewise for $C\_{\mathrm c}^\infty(\Omega;Y)$. Then $\imath$ clearly induces a continuous inclusion $$\imath\_\*:C\_{\mathrm c}^\infty(\Omega;X)\hookrightarrow C\_{\mathrm c}^\infty(\Omega;Y)$$ however, it is not at all clear to me if $\imath\_\*$ will have dense image. I suspect it won't, though I have yet to find a counterexample. So, does it have dense image, and if so, why?



Let $X\subset Y$ be a dense inclusion of reflexive Banach spaces. Then is $C_c^\infty(\Omega;X)\subset C_c^\infty(\Omega;Y)$ dense?
$C^\infty\_c(\Omega, X) = C^\infty\_c(\Omega,\mathbb R){\bar\otimes} X$ completed inductive tensor product which agrees with the projective tensor product since $C^\infty\_c(\Omega,\mathbb R)$ is nuclear. Do the same for $Y$. Finite rank tensors are dense in $C^\infty\_c(\Omega,\mathbb R){\bar\otimes} Y$ and these can be approximated by finite rank tensors in $C^\infty\_c(\Omega,\mathbb R){\bar\otimes} X$.


