Muhammad Haris Rao
Let $(X, \mathcal{A}, \mu)$ be a $\sigma$-finite measure space. For a measurable function $f : X \longrightarrow \mathbb{R}$, we denote \begin{align*} L^p(X) &= \left\{ f : X \longrightarrow \mathbb{R} \mid \int_X |f|^p \, d \mu < \infty \right\} \end{align*} As is usual in measure theory, we are really looking at equivalence classes of functions on $X$ with two functions being equivalent if they are equal $\mu$-a.e. For $p > 1$, and $f : X \longrightarrow \mathbb{R}$ measurable, define \begin{align*} \Vert f \Vert_p &= \left( \int_X |f|^p \, d \mu \right)^{1/p} \in \mathbb{R}_{\ge 0} \cup \{ \infty \} \end{align*} This restricts to a function on $L^p(X)$ \begin{align*} \Vert \cdot \Vert_p : L^p(X) \longrightarrow \mathbb{R}_{\ge 0}, f \longmapsto \Vert f \Vert_p \end{align*}
Lemma (Young's Inequality): If $p, q > 1$ with $1/p + 1/q = 1$ and $a, b \ge 0$, then \begin{align*} ab \le \frac{a^p}{p} + \frac{b^q}{q} \end{align*}
Proof. The desired result clearly holds if any one of $a, b$ is zero. So assume $a, b > 0$. Since the logarithm function is concave, we have for any $t \in [0, 1]$ that \begin{align*} \log\left((1 - t) a^p + t b^q \right) \ge p (1 - t) \log{a} + q t \log{b} \end{align*} Taking $t = 1/q$, we have \begin{align*} \log\left( \frac{a^p}{p} + \frac{b^q}{q} \right) \ge \log{a} + \log{b} = \log{ab} \end{align*} Exponentiating gives the desired result.
Theorem (Hölder's Inequality): Suppose $p, q > 1$ with $1/p + 1/q = 1$. Then for any $f \in L^p(X)$ and $g \in L^q(X)$ it holds that $\Vert f g \Vert_1 \le \Vert f \Vert_p \Vert g \Vert_q$. That is, \begin{align*} \int_X |f g| \, d\mu \le \left( \int_X |f|^p \, d \mu \right)^{1/p} \left( \int_X |g|^q \, d \mu \right)^{1/q} \end{align*}
Proof. Assume for now that $\Vert f \Vert_p = \Vert g \Vert_q = 1$. Then, \begin{align*} \Vert f g \Vert &= \int_X |f| |g| \, d \mu \le \int_X \frac{|f|^p}{p} + \frac{|g|^q}{q} \, d \mu = \frac{1}{p} \Vert f \Vert_p^p + \frac{1}{q} \Vert g \Vert_q^q = \frac{1}{p} + \frac{1}{q} = 1 = \Vert f \Vert_p \Vert g \Vert_q \end{align*} For general $f \in L^p(X)$, $g \in L^q(X)$, we can rescale and apply the above special case to get the desired inequality.
This gives the following important result:
Theorem (Minkowski's Inequality): Suppose $p > 1$. If $f, g \in L^p(X)$, then $\Vert f + g \Vert_p \le \Vert f \Vert_p + \Vert g \Vert_p$. That is \begin{align*} \left( \int_X |f + g|^p \, d \mu \right)^{1/p} \le \left( \int_X |f|^p \, d \mu \right)^{1/p} + \left( \int_X |g|^p \, d \mu \right)^{1/p} \end{align*}
Proof. If $\Vert f + g \Vert_p^p = 0$ then the result is trivial, so assume $\Vert f + g \Vert_p^p \ne 0$. Let $q > 1$ be so that $1/p + 1/q = 1$. This implies that $p - 1 = p/q$ so \begin{align*} \int_X |f + g|^p \, d \mu &= \int_X |f + g| | f + g |^{q/p} \, d \mu \\ &\le \int_X |f| | f + g |^{p/q} \, d \mu + \int_X |g| | f + g |^{p/q} \, d \mu \\ &\le \left( \int_X |f|^p \, d \mu \right)^{1/p} \left( \int_X \left| | f + g |^{p/q} \right|^q \, d \mu \right)^{1/q} + \left( \int_X |g|^p \, d \mu \right)^{1/p} \left( \int_X \left| | f + g |^{p/q} \, d \mu\right|^q \right)^{1/q} \\ &= \left( \left( \int_X |f|^p \, d \mu \right)^{1/p} + \left( \int_X |g|^p \, d \mu \right)^{1/p} \right) \left( \int_X |f + g|^p \, d \mu \right)^{1 - \frac{1}{p}} \end{align*} This simply says that \begin{align*} \Vert f + g \Vert_p^p \le \left( \Vert f \Vert_p + \Vert g \Vert_p \right) \frac{\Vert f + g \Vert_p^p}{\Vert f + g \Vert_p} \end{align*} A simple rearrangment and cancellation gives \begin{align*} \Vert f + g \Vert_p \le \Vert f \Vert_p + \Vert g \Vert_p \end{align*} which is what we set out to prove.
Proposition: If $p > 1$, then the pair $\left( L^p(X), \Vert \cdot \Vert_p \right)$ is a normed $\mathbb{R}$-vector space.
Proof. The collection of real valued measurable functions on $X$ is clearly an $\mathbb{R}$-vector space under pointwise addition and scalar multiplication by $\mathbb{R}$. To show that $L^p(X)$ inherits this vector space structure, it needs to be shown that it contains the zero function and is closed under pointwise addition and scalar multiplication by $\mathbb{R}$. The first two properties are obvious, and the third follows from the Minkowski inequality, so $L^p(X)$ is a vector space under these operations.
To show that $\Vert \cdot \Vert$ is a norm on the vector space $L^p(X)$, there are three things which need to be shown. First, if $f \in L^p(X)$ then $\left\Vert f \right\Vert_p = 0$ if and only if $f = 0$. Second, if $\lambda \in \mathbb{R}$, $f \in L^p(X)$, then $\left\Vert \lambda f \right\Vert = |\lambda| \left\Vert f \right\Vert$. Third, the triangle inequality which is that for any $f, g \in L^p(X)$, $\left\Vert f + g \right\Vert_p \le \left\Vert f \right\Vert_p + \left\Vert g \right\Vert_p$.
If $f = 0$ then it is obvious that $\left\Vert f \right\Vert = 0$. So now suppose $f \ne 0$. Then there exists $A \in \mathcal{A}$ with $\mu(A) > 0$ such that if $x \in A$ then $|f(x)| > 0$. For $n \in \mathbb{Z}_{> 0}$, define \begin{align*} A_n &= \left\{ x \in A \mid |f(x)| > \frac{1}{n} \right\} \end{align*} Evidently, $A_1 \subseteq A_2 \subseteq A_3 \subseteq \cdots $ and $A_n \to A$ as $n \to \infty$. Then $\mu(A_n) \to \mu(A)$ as $n \to \infty$. Since $\mu(A) > 0$, There exists $N \in \mathbb{Z}_{> 0}$ such that if $n \in \mathbb{Z}_{\ge N}$, then $\mu(A_n) > 0$. Then we have \begin{align*} \int_X |f|^p \, d \mu &\ge \int_A |f|^p \, d \mu \ge \int_{A_N} |f|^p \, d \mu \ge \int{A_N} \frac{1}{N^p} \, d \mu = \frac{\mu(A_N)}{N^p} > 0 \end{align*} So if $f \ne 0$ then $\left\Vert f \right\Vert_p \ne 0$.
The second property is easy. Indeed, if $f \in L^p(X)$ and $\lambda \in \mathbb{R}$ then \begin{align*} \left\Vert \lambda f \right\Vert_p &= \left( \int_X | \lambda f |^p \right)^{1/p} = \left( |\lambda|^p \int_X | f |^p \right)^{1/p} = |\lambda| \left( \int_X | f |^p \right)^{1/p} = |\lambda| \left\Vert f \right\Vert_p \end{align*}
The triangle inequality is just a restatement of Minkowski's inequality, so we are done. The pair $\left( L^p(X), \Vert \cdot \Vert_p \right)$ is a normed $\mathbb{R}$-vector space.
Given a normed vector spaces $(V, \Vert \cdot \Vert_V)$, $(W, \Vert \cdot \Vert_W)$, and linear map $f : V \longrightarrow W$, the operator norm of $f$ is defined as \begin{align*} \Vert f \Vert &= \sup\left\{ \Vert f(v) \Vert_W \mid v \in V \text{ such that } \Vert v \Vert_V = 1 \right\} \end{align*} The dual space of the normed vector space $V$ is defined as \begin{align*} V^* &= \left\{ f : V \longrightarrow \mathbb{R} \mid \text{$f$ linear and $\Vert f \Vert < \infty$} \right\} \end{align*} We have the following remarkable result.
Theorem: Let $1 < p, q < \infty$ with $1/p + 1/q = 1$. Then we have an isomorphism of normed vector spaces \begin{align*} L^p(X) \cong L^q(X)^* \end{align*}
Proof. Define the mapping \begin{align*} \Phi : L^p(X) &\longrightarrow L^q(X)^* \\ f &\longmapsto \Phi_f \end{align*} with $\Phi_f$ defined by \begin{align*} \Phi_f : L^q(X) &\longrightarrow \mathbb{R} \\ g &\longmapsto \int_X fg \, d \mu \end{align*} If $f \in L^p(X)$, then $\Phi_f$ is well-defined as a function from $L^q(X)$ to $\mathbb{R}$ because if $g \in L^q(X)$ \begin{align*} \int_X |f g| \, d \mu &\le \left( \int_X |f|^p \, d \mu \right)^{1/p} \left( \int_X |g|^p \, d \mu \right)^{1/q} < \infty \end{align*} by Hölder's inequality. This means $fg \in L^1(X)$ so the integral defining $\Phi_f(g)$ is well-defined. The fact that $\Phi_f$ is linear is obvious. We check now that $\Vert \Phi_f \Vert < \infty$. If $g \in L^q(X)$ is such that $\Vert g \Vert_q = 1$ then by Hölder's inequality again, \begin{align*} \left| \Phi_f(g) \right| &= \left| \int_X f g \, d \mu \right| \le \int_X | f g | \, d \mu \le \left( \int_X |f|^p \, d \mu \right)^{1/p} \left( \int_X |g|^q \, d \mu \right)^{1/q} = \left( \int_X |f|^p \, d \mu \right)^{1/p} \end{align*} Hence we have \begin{align*} \Vert \Phi_f \Vert &= \sup\left\{ \left| \Phi_f(g) \right| \mid g \in L^q(X) \text{ such that } \Vert g \Vert_q = 1 \right\} \le \left( \int_X |f|^p \, d \mu \right)^{1/p} = \Vert f \Vert_p \end{align*} Since $f \in L^p(X)$, this means that $\Vert \Phi_f \Vert < \infty$ so that $\Phi_f$ really is a bounded linear functional of $L^q(X)$. So the map $\Phi : L^p(X) \longmapsto L^q(X)^*$ is well-defined. It is also easily seen to be linear. We will now prove that it is an isometry. We have already shown that if $f \in L^p(X)$ then $\Vert \Phi_f \Vert \le \Vert f \Vert_p$. We will now show that this is actually an equality. The equality is obvious in the case where $f = 0$ so assume $f \ne 0$ so that in particular, $\Vert f \Vert_p \ne 0$. Define the function \begin{align*} \text{sgn}(f) : X &\longrightarrow \mathbb{R} \\ x &\longmapsto \begin{cases} 1 &\text{ , if $f(x) \ge 0$} \\ -1 &\text{ , if $f(x) < 0$} \end{cases} \end{align*} so that $f \cdot \text{sgn}(f) = |f|$, and define $g = |f|^{p/q}/ \Vert f \Vert_p^{p/q}$. Then we have \begin{align*} \Vert g \Vert_q &= \left( \int_X |g|^q \, d \mu \right)^{1/q} = \left( \int_X \left| \frac{|f|^{p/q}}{\Vert f \Vert_p^{p/q}}\right|^q \, d \mu \right)^{1/q} = \frac{1}{\Vert f \Vert_p^{p/q}} \left( \int_X |f|^p \, d \mu\right)^{1/q} = \frac{\Vert f \Vert_p^{p/q}}{\Vert f \Vert_p^{p/q}} = 1 \end{align*} Furthermore, \begin{align*} \left| \int_X f \cdot \text{sgn}(f) g \, d \mu \right| = \frac{1}{\Vert f \Vert_p^{p/q}} \int_X |f|^{p/q + 1} \, d \mu = \frac{1}{\Vert f \Vert_p^{p/q}} \int_X |f|^p \, d \mu = \frac{\Vert f \Vert_p^p}{\Vert f \Vert_p^{p/q}} = \Vert f \Vert_p^{p - \frac{p}{q}} = \Vert f \Vert_p \end{align*} This means that $\Vert \Phi_f \Vert \ge \Vert f \Vert_p$ as well. Hence, we have $\Vert \Phi_f \Vert = \Vert f \Vert_p$, so $\Phi : L^p(X) \longrightarrow L^q(X)^*$ preserves the norm. Then $\Phi$ is also injective as all isometries are. The result that $\Phi$ is surjective is technical, and is called the Riesz representation thereom, the proof of which is omitted here. We conclude that $\Phi$ is a bijective linear isometry, and hence is the desired isomorphism of normed vector spaces.
Recall that a norm $\Vert \cdot \Vert_V : V \longrightarrow \mathbb{R}_{\ge 0}$ on a vector space $V$ induces a metric \begin{align*} d_V : V \times V &\longrightarrow \mathbb{R}_{\ge 0} \\ (v, w) &\longmapsto d_V(v, w) = \Vert v - w \Vert_V \end{align*} So for $p > 1$, $L^p(X)$ is a metric space. Recall also that a metric space is said to be complete if every Cauchy sequence converges.
Definition: A Banach space is a complete, normed vector space.
We have the following important result about $L^p$ spaces:
Theorem: If $p > 1$, then the pair $( L^p(X), \Vert \cdot \Vert_p )$ is a complete normed vector space. That is, $L^p(X)$ is a Banach space.
Proof. Let $\{ f_n \}_{n \in \mathbb{Z}_{> 0}} \subseteq L^p(X)$ be a Cauchy sequence.