Let $\Omega$ be an open subset of $\mathbb R^d$ and $f \in L^1_{\text{loc}} (\Omega)$. I'm trying to do below exercise, i.e.,
Prove that $f \in L^1 (\Omega)$ IFF $$ A := \sup \left \{\int f \varphi : \varphi \in \mathcal C^\infty_c(\Omega) \text{ s.t. } \|\varphi\|_\infty \le 1 \right\} < \infty. $$ If $f \in L^1 (\Omega)$ then $A = \|f\|_1$.
Could you have a check on my below attempt?
First we assume $f \in L^1 (\Omega)$. We define a linear map $T:L^\infty (\Omega) \to \mathbb R, \varphi \mapsto \int f \varphi$. Then $T$ is continuous w.r.t. the norm topology of $L^\infty (\Omega)$. We have $$ \begin{align*} \infty > \|f\|_1 = \|T\|_{L^\infty (\Omega)^*} &:= \sup \left \{ T\varphi : \varphi \in L^\infty(\Omega) \text{ s.t. } \|\varphi\|_\infty \le 1 \right\} \\ &\ge \sup \left \{T\varphi : \varphi \in \mathcal C^\infty_c(\Omega) \text{ s.t. } \|\varphi\|_\infty \le 1 \right\} \\ &= A. \end{align*} $$
Now we assume $A < \infty$. Let $B_n := B(0, n)$ be the open ball centered at $0 \in \mathbb R^d$ with radius $n>0$. Then $$ \|f\|_1 = \int |f| = \lim_n \int_{B_n} f \operatorname{sgn} f. $$ Notice that $f \in L^1 (B_n)$ and $\operatorname{sgn} f \in L^\infty (B_n)$. Fix $\varepsilon >0$. I already proved that there is $\varphi_n \in \mathcal C_c^{\infty} (B_n)$ such that $\|\varphi_n\|_\infty \le \|\operatorname{sgn} f\|_\infty \le 1$ and $|\int_{B_n} f \operatorname{sgn} f - \int_{B_n} f \varphi_n | < \varepsilon$. We define $\overline{\varphi_n} : \Omega \to \mathbb R$ by $\overline{\varphi_n} (x) := \varphi_n(x)$ if $x \in B_n$ and $0$ otherwise. Then $\overline{\varphi_n} \in \mathcal C_c^{\infty} (\Omega)$ with $\operatorname{supp} \overline{\varphi_n} \subset B_n$. It follows that $\int_{B_n} f \varphi_n = \int f \overline{\varphi_n} \le A$. So $\int_{B_n} f \operatorname{sgn} f \le A + \varepsilon$ and thus $\|f\|_1 \le A + \varepsilon < \infty$. This completes the proof.
