> [!definition] > > Let $(X, \cm, \mu)$ be a [[Measure Space|measure space]]. If $f$ is a [[Measurable Function|measurable function]] and $p \in (0, \infty)$, define > $ > \norm{f}_p = \braks{\int \abs{f}^p d\mu}^{1/p} > $ > then > $ > L^p(X, \cm, \mu) = \bracs{f: X \to \complex: f\text{ measurable}, \norm{f}_p < \infty} > $ > where two functions define the same element of $L^p$ if they are equal [[Almost Everywhere|a.e.]] > [!definition] > > Let $I$ be a non-empty [[Set|set]], and $\mu$ be the counting measure, then > $ > l^p(I) = L^p(I, \pow{I}, \mu) > $ > is the $l^p$ space of $I$-valued [[Sequence|sequences]]. > [!theorem]- Vector Space > > $L^p$ is a [[Vector Space|vector space]]. > > *Proof*. Let $f, g \in L^p$, then > $ > \begin{align*} > \norm{f + g}_p^p &= \int\abs{f + g}^p \\ > &\le \int\abs{2\max(\abs{f}, \abs{g})}^p \\ > &\le \int 2^p(\abs{f}^p + \abs{g}^p) < \infty > \end{align*} > $ > and for any $\lambda \in \complex$, > $ > \begin{align*} > \norm{\lambda f}_p &= \braks{\int\abs{\lambda f}^p}^{1/p} \\ > &= \braks{\abs{\lambda}^p\int\abs{f}^p}^{1/p} \\ > &= \abs{\lambda}\braks{\int\abs{f}^p}^{1/p} \\ > &= \abs{\lambda} \cdot \norm{f}_p > \end{align*} > $ > [!theorem]- Normed Space > > $L^p$ is a [[Normed Vector Space|normed space]] when $p \ge 1$. > > *Proof*. Firstly, $\norm{f}_p = 0$ if and only if $f = 0$ a.e. Then, $\norm{\lambda f}_p = \abs{\lambda} \norm{f}_p$. Lastly by [[Minkowski's Inequality]], $\norm{\cdot}_p$ satisfies the triangle inequality. > [!theorem] > > For $1 \le p < \infty$, $L^p$ is a [[Banach Space|Banach space]]. > > *Proof*. Let $\seq{f_k} \subset L^p$ such that $\sum_{k \in \nat}\norm{f_k}_p = B < \infty$. Let $G_n = \sum_{k = 1}^{n}\abs{f_k}$ and $G = \sum_{n \in \nat}\abs{f_n}$, then > $ > \norm{G_n}_p \le \sum_{k = 1}^{n}\norm{f_k}_p \le B \quad \forall n \in \nat > $ > By the [[Monotone Convergence Theorem]], $\int G^p = \limv{n}\int G_n^p \le B^p$, and $G \in L^p$. Importantly, $G(x) < \infty$ [[Almost Everywhere|a.e.]] and $\sum_{k = 1}^{\infty}f_k(x)$ converges a.e. Let $F = \sum_{k = 1}^{\infty}f_k$, then the absolute sum $G$ can be used to dominate $F$ with > $ > \abs{F(x)} = \abs{\sum_{k = 1}^{\infty}f_k(x)} \le \sum_{k = 1}^{\infty}\abs{f_k(x)} = G(x) > $ > By the [[Dominated Convergence Theorem]], $F \in L^p$. Moreover, $2G$ dominates the difference $\abs{F - \sum_{k = 1}^{n}f_k}$ for all $n \in \nat$. With $\abs{F - \sum_{k = 1}^{n}f_k}^p \le (2G)^p \in L^1$, > $ > \begin{align*} > \norm{F - \sum_{k = 1}^{n}f_k}_p^p &= \int\abs{F - \sum_{k = 1}^{n}f_k}^p \to 0 > \end{align*} > $ > applying the dominated convergence theorem again yields convergence in $L^p$. > [!theorem] > > The collection > $ > \Phi = \bracs{\phi \in L^p: \phi \text{ simple}} > $ > of [[Simple Function|simple functions]] in $L^p$ is [[Dense|dense]] in $L^p$ for $p \ge 1$. > > *Proof*. Let $f \in L^p$, then there exists a [[Sequence|sequence]] of simple functions $\seq{\phi_n} \subset \Phi$ such that $\phi_n \to f$ [[Pointwise Convergence|pointwise]] and $\abs{\phi_n} \le \abs{f}$. Using this bound, dominate $\abs{\phi_n}^p$ with $\abs{f}^p$ and > $ > \abs{\phi_n - f}^p \le (\abs{\phi_n} + \abs{f})^p \le 2^p\abs{f}^p > $ > By the dominated convergence theorem $\norm{\phi_n - f}_p \to 0$. > > For each $\phi = \sum_{k = 1}^{n}a_k\chi_{E_k}$ where the $E_k$s are disjoint and $\abs{\phi} \le \abs{f}$, > $ > \int \abs{\phi}^p = \sum_{k = 1}^{n}\abs{a_k}^p\mu(E_k) < \infty > $ > meaning that $\mu(E_k) < \infty$ for all $k$. > [!theorem] > > Let $(X, \cm, \mu)$ be a semi-finite measure space, $1 \le p \le \infty$, and $\phi \in L^\infty(\mu)$. If $Tf: L^p \to L^p$ is a bounded linear map defined by $f \mapsto \phi f$, then $\norm{T} = \norm{\phi}_\infty$. > > *Proof*. Since $\abs{\phi f} \le \norm{\phi}_\infty \abs{f}$ a.e., $\norm{\phi f}_p \le \norm{\phi}_\infty \cdot \norm{f}_p$. > > Suppose that $p < \infty$. Let $\eps > 0$ and $E = \bracs{\abs{\phi} > \norm{\phi}_\infty - \eps}$, then $\mu(E) > 0$. Let $F \subset E$ such that $0 < \mu(F) < \infty$, then $\one_F \in L^p$, so > $ > \norm{\phi f}_p^p \ge (\norm{\phi}_\infty - \eps)^p \mu(F) = (\norm{\phi}_\infty - \eps)^p\norm{\one_F}_p^p > $ > since $\eps$ was arbitrary, $\norm{T} = \norm{\phi}_\infty$. If $p = \infty$, then $\abs{\phi \one_E} \ge \norm{\phi}_\infty - \eps$ on $E$, so $\norm{T} \ge \norm{\phi}_\infty - \eps$. As $\eps$ was arbitrary, $\norm{T} = \norm{\phi}_\infty$. # Translation > [!theorem] > > Let $z \in \real^n$, define $\tau_z$ be the translation map > $ > \tau_z(f)(x) = f(x - z) > $ > then $\tau_z$ is continuous in the $L^p$ norm, that is > $ > \lim_{y \to 0}\norm{\tau_{y + z}f - \tau_{z}f}_p = 0 > $ > > *Proof*. By decomposing $\tau_{y + z} = \tau_y\tau_z$, it's sufficient to assume that $z = 0$. First suppose that $g \in C_c$. If $\norm{y} < 1$, then > $ > \supp{\tau_y g} \subset \bracs{z \in \real^d: d(z, \supp{g}) \le 1} = K > $ > thus the functions are supported on a common compact set. Moreover, since compactly supported continuous functions are uniformly continuous, > $ > \int \abs{\tau_yg - g}^p \le \norm{\tau_y - g}_u^p m(K) \to 0 \quad y \to 0 > $ > Suppose that $f \in L^p$ and let $\eps > 0$, then there exists $g \in C_c$ such that $\norm{g - f}_p < \eps/2$. Therefore > $ > \norm{\tau_yf - f}_p \le \norm{\tau_yf - \tau_yg}_p + \norm{f - g}_p + \norm{\tau_yg - g}_p\le \eps + \norm{\tau_yg - g}_p \to \eps > $ > as $y \to 0$. Since $\eps$ was arbitrary, $\norm{\tau_y f - f}_p \to 0$.