> [!theoremb] Theorem > > Let $p, q, r \in (0, \infty]$ such that $p < q < r$, then > $ > L^p \cap L^r \subset L^q \subset L^p + L^r > $ > where the inclusion maps $L^p \cap L^r \to L^q$ and $L^q \to L^p + L^r$ are [[Bounded Linear Map|bounded]]. # The Intermediate Spaces > [!definition] > > Define the following norm > $ > \norm{f}_{L^p \cap L^r} = \norm{f}_p + \norm{f}_q > $ > on intersection [$L^p$](Lp%20Space) space $L^p \cap L^r$, which makes it a [[Banach Space|Banach space]]. > > *Proof*. Let $\seq{f_n} \subset L^p \cap L^r$ be an absolutely convergent series, then $f_n$ converges absolutely [[Pointwise Convergence|pointwise]] [[Almost Everywhere|a.e.]], with respect to both norms. > [!definition] > > Let $f \in L^p + L^r$, then > $ > \norm{f}_{L^p + L^r} = \inf\bracs{\norm{g}_p + \norm{h}_r: g \in L^p, h \in L^r, f = g + h} > $ > is a norm. > > > *Proof*. Let $f_1, f_2 \in L^p + L^r$ and $\lambda \in \complex$, then > $ > \begin{align*} > \norm{\lambda f} &= \inf\bracs{\norm{g}_p + \norm{h}_r: \lambda f = g + h} \\ > &= \inf\bracs{\norm{\lambda g}_p + \norm{\lambda h}_r: \lambda f = \lambda g + \lambda h} \\ > &= \inf\bracs{\abs{\lambda}(\norm{g}_p + \norm{g}_r): f = \lambda g + \lambda h} \\ > &= \abs{\lambda} \norm{f} > \end{align*} > $ > and for any $g_1, h_1: f_1 = g_1 + h_1$ and $g_2, h_2: f_2 = g_2 + h_2$, > $ > \begin{align*} > \norm{f_1 + f_2} &\le \norm{g_1 + g_2}_p + \norm{h_1 + h_2}_r \\ > &\le \paren{\norm{g_1}_p + \norm{h_1}_r} + \paren{\norm{g_2}_p + \norm{h_2}_p} > \end{align*} > $ > Since this applies to all such functions, $\norm{f_1 + f_2} \le \norm{f_1} + \norm{f_2}$. > > Suppose that $f \ne 0$, then there exists $E \in \cm: \mu(E) > 0$ and $\varepsilon > 0$ such that $\abs{f} \ge \varepsilon$ on $E$. If $f = g + h$, then > $ > \underbrace{\bracs{x: \abs{f(x)} \ge \varepsilon}}_{F} \subset \underbrace{\bracs{x: \abs{g(x)} \ge \varepsilon/2}}_G \cup \underbrace{\bracs{x: \abs{h(x)} \ge \varepsilon/2}}_H > $ > meaning that $\mu(G) \ge \mu(E)/2$ or $\mu(H) \ge \mu(E)/2$. > > If $\mu(G) \ge \mu(E)/2$, then > $ > \norm{g}_p \ge \norm{\frac{\varepsilon \cdot \chi_G}{2}}_p = \frac{\varepsilon}{2} \cdot \mu(G)^{1/p} \ge \frac{\varepsilon}{2}\paren{\frac{\mu(E)}{2}}^{1/p} > $ > > If $\mu(H) \ge \mu(E)/2$, then > $ > \norm{h}_r \ge \begin{cases} > \norm{\frac{\varepsilon \cdot \chi_{H}}{2}}_r \ge \frac{\varepsilon}{2}\paren{\frac{\mu(E)}{2}}^{1/r} &r < \infty > \\ > \frac{\varepsilon}{2} &r = \infty > \end{cases} > $ > > Either way, since the inequalities hold for all decompositions, > $ > \norm{f} \ge \min\paren{ > \frac{\varepsilon}{2}\paren{\frac{\mu(E)}{2}}^{1/p}, \frac{\varepsilon}{2}\paren{\frac{\mu(E)}{2}}^{1/r}, \frac{\varepsilon}{2} > } > 0 > $ > we have found a lower bound for $\norm{f}$. > [!theorem] > > The norm on $L^p + L^r$ makes it a Banach space. > > *Proof*. Let $g_n, h_n$ be such that $f_n = g_n + h_n$ and > $ > \norm{g_n}_p + \norm{h_n}_r < \norm{f_n} + \frac{1}{2^n} > $ > Then $\seq{g_n} \subset L^p$ and $\seq{h_n} \subset L^r$ are absolutely convergent series, meaning that there exists $g \in L^p, h \in L^r$ such that $\sum_{k = 1}^{n}g_k \to g$ in $L^p$ and $\sum_{k = 1}^{n}h_k \to h$ in $L^r$. Importantly, since > $ > \begin{align*} > \norm{\phi} &\le \norm{\phi}_p + \norm{0}_r = \norm{\phi}_p\\ > \norm{\phi} &\le \norm{0}_p + \norm{\phi}_r = \norm{\phi}_r > \end{align*} > $ > we have $\sum_{k = 1}^{n}g_k \to g$ and $\sum_{k = 1}^{n}h_k \to h$ in $L^p + L^r$. Therefore $\sum_{k = 1}^{n}f_n \to g + h = f$. # First Interpolation > [!theorem] > > $L^q \supset L^p \cap L^r$ and $\norm{f}_q \le \norm{f}_p^\lambda \cdot \norm{f}_r^{1 - \lambda}$, where > $ > \frac{1}{q} = \frac{\lambda}{p} + \frac{1 - \lambda}{r} \quad \lambda = \frac{1/q - 1/r}{1/p - q/r} > $ > *Proof*. First suppose that $r = \infty$, then $\lambda = p/q$, $\abs{f}^q \le \norm{f}_\infty^{q - p} \cdot \abs{f}^p$ [[Almost Everywhere|a.e.]], and > $ > \begin{align*} > \abs{f}^q &\le \norm{f}_\infty^{q - p} \cdot \abs{f}^p \\ > \int \abs{f}^q &\le \norm{f}_\infty^{q - p} \cdot \int \abs{f}^p \\ > \braks{\int \abs{f}^q}^{1/q} &\le \braks{\norm{f}_\infty^{q - p} \cdot \int \abs{f}^p}^{1/q} \\ > \norm{f}_q &\le \norm{f}_\infty^{1 - p/q} \cdot \norm{f}_p^{p/q} \\ > \norm{f}_q &\le \norm{f}_p^\lambda \cdot \norm{f}_\infty^{1 - \lambda} > \end{align*} > $ > If $f \in L^p$ and $f \in L^r$, then $\norm{f}_q < \infty$ and $f \in L^q$. > > > Now suppose that $r < \infty$, then > $ > \begin{align*} > \frac{1}{q} &= \frac{\lambda}{p} + \frac{1 - \lambda}{r} \\ > 1 &= \frac{\lambda q}{p} + \frac{(1 - \lambda)q}{r} > \end{align*} > $ > where $p/\lambda q$ and $r/(1 - \lambda)q$ are Hölder conjugates. > > Using [[Hölder's Inequality]], > $ > \begin{align*} > \int \abs{f}^q &= \int \abs{f}^{\lambda q} \cdot \abs{f}^{(1 - \lambda)q} = \norm{\abs{f}^{\lambda q} \cdot \abs{f}^{(1 - \lambda)q}}_1 \\ > &\le \norm{\abs{f}^{\lambda q}}_{p/\lambda q} \cdot \norm{\abs{f}^{(1 - \lambda q)}}_{r/(1 - \lambda)q} \\ > &= \braks{\int\abs{f}^{p}}^{\lambda q/p} \cdot \braks{\int \abs{f}^{r}}^{(1 - \lambda) q/r} \\ > \norm{f}_q^q &\le \norm{f}_p^{\lambda q} \cdot \norm{f}_{r}^{(1 - \lambda)q}\\ > \norm{f}_q &\le \norm{f}_p^{\lambda} \cdot \norm{f}_{r}^{(1 - \lambda)} > \end{align*} > $ > [!theorem] > > The inclusion map $L^p \cap L^r \to L^q$ is [[Bounded Linear Map|bounded]]. > > *Proof*. From earlier results, > $ > \begin{align*} > \norm{f}_q &\le \norm{f}_p^{\lambda} \cdot \norm{f}_r^{1 - \lambda} \\ > &\le (\norm{f}_p + \norm{f}_r)^{\lambda + (1 - \lambda)} \\ > &= \norm{f} > \end{align*} > $ > making the inclusion map bounded. # Second Interpolation > [!theorem] > > The inclusion map $L^q \to L^p + L^r$ is continuous with > $ > \norm{f} \le 2\norm{f}_q \quad \forall f \in L^q > $ > *Proof*. First assume that $r < \infty$. Let $f \in L^q$ such that $\norm{f}_q = 1$ and $E = \bracs{x: \abs{f(x)} > 1}$, then > $ > \begin{align*} > \int \abs{f \cdot \chi_{E}}^p &\le \int \abs{f \cdot \chi_{E}}^q \\ > \int \abs{f \cdot \chi_{E^c}}^r&\le \int \abs{f \cdot \chi_{E^c}}^q > \end{align*} > $ > since $f = f \cdot \chi_E + f \cdot \chi_{E^c}$, > $ > \begin{align*} > \norm{f} &\le \norm{f \cdot \chi_E}_p + \norm{f \cdot \chi_{E^c}}_r \\ > &\le \braks{\int \chi_E \cdot \abs{f}^q}^{1/p} + \braks{\int \chi_E \cdot \abs{f}^q}^{1/r} \\ > &\le \braks{\int\abs{f}^q}^{1/p} + \braks{\int \abs{f}^q}^{1/r} \\ > &= \norm{f}_q^{q/p} + \norm{f}_q^{q/r} \\ > &= 2\norm{f}_q > \end{align*} > $ > > Now suppose that $r = \infty$, then the same decomposition yields > $ > \begin{align*} > \norm{f} &\le \norm{f \cdot \chi_E}_p + \norm{f \cdot \chi_{E^c}}_{\infty} \\ > &\le \norm{f}_q^{q/p} + \norm{f \cdot \chi_{E^c}}_{\infty} \\ > &\le 2 = 2\norm{f}_q > \end{align*} > $