For $1＜p＜\infty$ and the conjugate $p'$, show that the $p$- norm of $f$ is written as $$\| f \|_p=\sup\left\{\left|\int_{\mathbb R^n}fg\right| : g \in \mathcal S(\mathbb R^n), \|g\|_{p'}=1\right\}\cdots (\ast)$$ The hint is to use $\mathcal S(\mathbb R^n)\subset \mathcal L^{p'}(\mathbb R^n)$ dense, where $\mathcal S(\mathbb R^n)$ is Schwartz class.
I know that a similar representation of $\|f\|_{p}$ : $$\| f \|_p=\sup\left\{\left|\int_{\mathbb R^n}fg\right| : g \in \mathcal L^{p'}(\mathbb R^n), \|g\|_{p'}= 1\right\}.$$
Let $$A_1=\left\{\left|\int_{\mathbb R^n}fg\right| : g \in \mathcal S(\mathbb R^n), \|g\|_{p'}=1\right\},$$ $$ A_2=\left\{\left|\int_{\mathbb R^n}fg\right| : g \in \mathcal L^{p'}(\mathbb R^n), \|g\|_{p'}=1\right\}$$
Since $A_1 \subset A_2$, $\sup A_1\leqq\sup A_2 $ is OK. I have to see the opposite.
To see that $\sup A_1$ is an upper bound of $A_2$, let $g \in \mathcal L^{p'}$ with $\|g\|_{p'}=1.$
Since $\mathcal S(\mathbb R^n)\subset \mathcal L^{p'}(\mathbb R^n)$ dense, there is $\{g_k\}\subset\mathcal S$ s.t. $\|g_k-g\|_{p'}\to 0.$
From here, I don't know how to deduce $(\ast)$.
Thanks for the help. Other approach is also welcomed.