We put, $L^{p}=L^{p}(\mathbb R), L^{q}=L^{q}(\mathbb R);$ $\frac{1}{p}+\frac{1}{q}=1;$ ($p$ and $q$ are conjugate exponents); and $<f,g> =\int_{\mathbb R} f(x)g(x) dx.$
Fix $g\in L^{q}, (1\leq q<\infty).$
Consider the set, $ A:= \{|<f, g>|: f\in L^{p}, \|f\|_{L^{p}} \leq 1 \};$ It is well-known that supremum of the set exist and, let us say it is, $\alpha= \sup A.$ (Infect, $\alpha= \|g\|_{L^{q}}$, (dual of $L^{p}$- spaces))
We put $C^{\infty}_{c}(\mathbb R)=\{f\in C^{\infty}(\mathbb R): \text{support of } \ f \ \text{is a compact set} \}.$
Next, we consider the set, $B:=\{|<f, g>|: f\in C_{c}^{\infty}(\mathbb R), \|f\|_{L^{p}} \leq 1 \}.$
Since $C^{\infty}_{c}(\mathbb R)\subset L^{p},$ we have, $B\subset A,$ and hence, $\sup B \leq \alpha.$
We also note that, $C_{c}^{\infty}(\mathbb R)$ is dense in $L^{p}(1\leq p <\infty).$
My Question is: Can we expect, $\sup B = \alpha,$ if yes how to prove it?
My attempt: Let $\epsilon >0 .$ I must prove that, there is a member in the set $B$(that is, some $|<f, g>|$ with $f\in C_{c}^{\infty},$ and $\|f\|_{L^{p}}\leq 1.$), so that, $$\alpha -\epsilon < |<f, g>|.$$
Now, since $\alpha = \sup A,$ there is $f\in L^{p}$ with $\|f\|_{L^{p}}\leq 1$ so that, $\alpha- \epsilon <|<f, g>|;$ if $f\in C_{c}^{\infty},$ then we are done.
If $f\in L^{p}\setminus C_{c}^{\infty}(\mathbb R),$ then I guess we need to use the fact that, $C_{c}^{\infty}$ is dense in $L^{p}(1\leq p <\infty),$ so there is sequence of functions $\{f_{n}\}\subset C_{c}^{\infty}$ so that, $\|f_{n}-f\|_{p} \to 0$ as $n\to \infty.$ From this, I don't know how to handle the situation; (or may be I have to argue slightly different way )
Thanks,
The denseness of $C_c^\infty(\def\R{\mathbb R}\R)$ in $L^p(\R)$ does the trick. Just use Hölder's inequality. For $f \in L^p(\R)$, choose $f_n \in C^\infty_c(\R)$ with $\def\norm#1{\left\|#1\right\|}\norm{f_n -f}_p \to 0$. Then $$ \left|\left<f_n - f, g\right>\right| \le \norm{f_n - f}_p \norm g_q \to 0$$
Addendum: We will show $\sup B = \alpha$. Given $\varepsilon > 0$, choose $f \in A$ with $\def\<#1>{\left<#1\right>}\def\abs#1{\left|#1\right|}\abs{\<f,g>} \ge \alpha - \frac{\def\eps{\varepsilon}\eps} 2$. By the above, there is a function $f_n \in C^\infty_c(\R)$ with $\norm{f-f_n}_p \le \frac \eps {2\norm g_q}$. We may assume that $\norm{f_n}_p \le \norm f_p \le 1$ (by construction of the approximating functions). Then $$ \abs{\<f_n,g> - \<f, g>} \le \norm{f_n-f}_p \norm g_q \le \frac\eps 2$$ Hence $$ \abs{\<f_n, g>} \ge \abs{\<f,g>} - \frac \eps 2 \ge \alpha - \eps $$