I have the following problem that I am struggling to prove :
If $K \subset \Omega$ and $K$ is a compact set, then there exists $\varphi \in C_{0}^{\infty}(\Omega)$ such that $0 \leq \varphi(x) \leq 1$ for all $x \in \Omega$ and $\varphi(y)=1$ for all $y \in K$
I am completely clueless in proving it. I have another lemma that states that If $\varphi \in C_{0}^{m}\left(\mathbb{R}^{n}\right), m \geq 0, f \in L_{l o c}^{1}\left(\mathbb{R}^{n}\right),$ then $\varphi * f \in C^{m}\left(\mathbb{R}^{n}\right)$ and $$ D^{\alpha}(\varphi * f)=\left(D^{\alpha} \varphi\right) * f $$ I don't know how to use this in the proof as well.
If $K=\emptyset$, then take $\varphi=0$ and we are done. Now Assume that $K\neq\emptyset$ and define : $$ d:= \begin{cases} 1&\text{if $\Omega=\mathbb{R^{n}}$} \\ \displaystyle\frac{\text{dist}(K,\Omega^{c})}{3}&\text{if otherwise} \end{cases} $$
Let $f$ be the characteristic function of the set $C=K+B(0, d)$ and define $\varphi:=J_{d} * f,$ where $J_{d}$ is a mollifier. We have that $\varphi \in C^{\infty}\left(\mathbb{R}^{n}\right)$ by the lemma you provided and by : $$ \varphi(x)=\int_{B(x, d) \cap C} J_{d}(x-y) d y, \quad x \in \mathbb{R}^{n} $$ since $B(x, d) \cap C=B(x, d)$ for $x \in K$ and $B(x, d) \cap C=\emptyset$ when : $$ x \notin K+B(0,2 d) \subset \Omega $$ $$ $$
[Note] : We say that $J_{d}$ is a mollifier, if for $d\in(0,+\infty)$ we have : $$J_{\varepsilon} \in C_{0}^{\infty}\left(\mathbb{R}^{n}\right), \quad J_{\varepsilon} \geq 0, \quad J_{\varepsilon}(x)=0 \text { for }|x| \geq \varepsilon, \quad \int_{\mathbb{R}^{n}} J_{\varepsilon}=1$$ I assume you are aware of this and of the method involvded in constructing mollifiers. Also note that your definition did not specify what the range of $\alpha$ is but your lemma would be valid if $$|\alpha|\leq m$$