Let $p, q$ be conjugate exponents with $1 < p < q < \infty$, and $\phi \in (L^p)^*$, then there exists a $\sigma$-finite set $F$ such that $ \supp{f} \subset F \quad \forall f \in L^p: \phi(f) \ne 0 $ ### Restriction of Functional Let $F \in \cm$ be a measurable set and define $\phi_F(f) = \phi(f \cdot \chi_F)$, then $ \abs{\phi_F(f)} = \abs{\phi(f \cdot \chi_F)} \le \norm{\phi} \cdot \norm{f \cdot \chi_F}_p \le \norm{\phi} \cdot \norm{f}_p $ Importantly, if $F \subset G$, then $\norm{\phi_F} \le \norm{\phi_G}$ by the above argument. ### Construction There exists a $\sigma$-finite set $F$ such that $\norm{\phi_F} = \norm{\phi}$. *Proof*. Let $\seq{f_n} \subset L^p$ be a sequence of functions such that $\norm{f_n}_p = 1$ for all $n \in \nat$, and $\phi(f_n) \upto \norm{\phi}$. Take $F_n = \bracs{x: \abs{f_n(x)} > 0}$, then each $F_n$ is $\sigma$-finite, and $F = \bigcup_{n \in \nat}F_n$ is also $\sigma$-finite. Since $ \phi(f_n) = \phi(f_n \cdot \chi_{F_n}) = \phi(f_n \cdot \chi_{F}) = \phi_F(f_n) $ for all $n \in \nat$, we have $\norm{\phi_F} \ge \norm{\phi} - \varepsilon$ for all $\varepsilon > 0$. Therefore $\norm{\phi_F} = \norm{\phi}$ ### Combination of Restricted Norms Let $\phi$ be a bounded linear functional and $F, G$ be disjoint measurable sets where $\norm{\phi_F}, \norm{\phi_G} > 0$. Then there exists $f \in L^p(F \cup G)$ such that $\norm{f} = 1$ and $\phi(f) > \norm{\phi_F}$. In other words, $ \norm{\phi_{F \cup G}} > \norm{\phi_F} $ *Proof*. Assume without loss of generality that $\norm{\phi_F} = 1$ (otherwise scale it). Let $\varepsilon \in (0, \norm{\phi_g})$, $f \in L^p(F)$ such that $\norm{f}_p = 1$ and $\phi(f) > 1 - \delta$. Since $\delta$ can be arbitrarily small and there would still be an $f$ satisfying the given criterion, the value of $\delta$ will be determined later. Let $g \in L^p(G)$ such that $\norm{g}_p = 1$ and $\phi(g) > \varepsilon$. Now, consider the linear combination $ \begin{align*} h(t) &= (1 - t^p)^{1/p}f +t g \\ \norm{h(t)}_p^p &= (1 - t^p)\norm{f}_p^p + t^p\norm{g}_p^p \\ &= t^p + (1 - t^p) = 1 \end{align*} $ which has a norm of $1$ for all $t > 0$. Apply $\phi_{F \cup G}$ on it, then $ \begin{align*} \phi(h(t)) &= (1 - t^p)^{1/p}\phi(f) + t\phi(g) \\ &> (1 - t^p)^{1/p}(1 - \delta) + t\varepsilon \end{align*} $ Differentiating with respect to $t$ yields $ \phi(h(t))' = \varepsilon - t^{p - 1}(1 - t^p)^{1/p - 1}(1 - \delta) $ where if $p > 1$, then $t^{p - 1}(1 - t^p)^{1/p - 1} \to 0$ as $t \to 0$. This means that there exists $t_0 > 0$ such that, $\phi(h(t))' > \varepsilon/2$ for all $t \in [0, t_0)$. Choose $\delta$ such that $ (1 - \delta) + \frac{t_0\varepsilon}{2} > 1 $ then $ \phi(h(t_0/2)) \ge \phi(h(0)) + \frac{t_0\varepsilon}{2} = (1 - \delta) + \frac{t_0\varepsilon}{2} > 1 $ Since $\norm{h(t)} = 1$ for all $t$, we have found $h(t_0/2)$ such that $\phi(h(t_0/2)) > 1$, and $\norm{\phi_{G \cup H}} > \norm{\phi_F}$. ### Combination of Restricted Norms (Stupid Version) Let $f \in L^p(F)$ and $g \in L^P(G)$ such that $\norm{f}_p = \norm{g}_p = 1$ and $\phi(f) > \norm{\phi_F} - \delta$ and $\phi(g) > \varepsilon > 0$. Then $ \norm{\phi}^q > (\norm{\phi_F} - \delta)^q + \varepsilon^q $ *Proof*. The map to $\complex^2$ with the counting measure $ \psi: \span(f, g) \to L^p(\complex^2) \quad \alpha f + \beta g \mapsto (\alpha, \beta) $ is an isometric isomorphism as $ \begin{align*} \norm{\alpha f + \beta g}_p^p &= \norm{\alpha f}_p^p + \norm{\beta g}_p^p \\ &= \alpha^p + \beta^p \\ \norm{\alpha f + \beta g}_p &= (\alpha^p + \beta^p)^{1/p} \end{align*} $ We can now represent $\phi$ as $\psi^{-1} \circ \phi' \circ \psi$ where $ \phi': L^p(\complex^2) \to \complex \quad (\alpha, \beta) \mapsto \alpha \phi(f) + \beta\phi(g) $ Then represent $\phi'$ as $h = (\alpha, \beta) \in L^q(\complex^2)$, which allows $ (\alpha^q + \beta^q)^{1/p} = \norm{h}_q = \norm{\phi'} $ Meaning that there exists $(\alpha, \beta) \in L^p(\complex^2): \norm{(\alpha, \beta)}_p = 1$ such that $ \phi'(\alpha, \beta)^q > (\norm{\phi_F} - \delta)^q + \varepsilon^q $ Translating this back to the original $L^p$, $ \phi(\alpha f + \beta g)^q > (\norm{\phi_F} - \delta)^q + \varepsilon^q $ with $\norm{\alpha f + \beta g}_p = 1$. ### Controlling the Norm Let $F$ be a $\sigma$-finite set such that $\norm{\phi_F} = \norm{\phi}$, then $\norm{\phi_{F^c}} = 0$. *Proof*. If $\norm{\phi_{F^c}} > 0$, then $\norm{\phi} > \norm{\phi_F} = \norm{\phi}$, which is a contradiction. Therefore $\norm{\phi_{F^c}} = 0$.