Here is my problem : Let $K$ be a convex set of $\mathbb{R}^m$ ($m\in \mathbb{N}^*$), such that $0$ belongs to the interior of K, I want to approximate (in $L^2(\mathbb{R}^m,\mathbb{R}^m)$) a function $f\in L^2(\mathbb{R}^n,K)$ ($n\in \mathbb{N}^*$) by a sequence of functions $(f_n)_{n\in \mathbb{N}}$ such that :
$$\forall n\in \mathbb{N},\quad f_n \in H^1(\mathbb{R}^n,K).$$
I know that I can find $\mathcal{C}^{\infty}_{c}(\mathbb{R}^m,\mathbb{R}^m)$ (set of smooth functions with compact support) that approximate $f$ in $L^2(\mathbb{R}^m,\mathbb{R}^m)$ and $f_n\to f$ almost everywhere. Then for almost every $x\in \mathbb{R}^n$ and $\epsilon>0$, there exists $N_x\in \mathbb{N}$ such that
$$|f_n(x)-f(x)| \le \epsilon$$
Thanks to the fact that $0$ belongs to the interior of K, we can say that $f_n(x) \in K$, but this is not a uniform result so with this fact I can't say that $f_n \in H^1(\mathbb{R}^n,K)$ for sufficiently large $n$.
Any advice ?
Thank you
My idea/sketch for a proof would be the following: Take a mollifier $\psi$ and the convolution $\psi \star f$. Then, $$(\psi \star f)(x) = \int_{\mathbb R^n} \psi(y) \, f(x-y) \, dy.$$ Since $\int_{\mathbb R^n} \psi \, dy = 1$, the right-hand side is actually a (infinite) convex combinations of points of $K$, hence still belongs to $K$.
Finally, one can show $\psi \star f \in C^\infty$ and the convergence $\psi \star f \to f$ as $\psi$ approximates the dirac delta.