Modes of convergence Let \((f_n)_{n \in \mathbb{N}}\) be a sequence of complex-valued functions on a space \(X\).
\(f_n \ _{\overrightarrow{n\to \infty}} \ f\) uniformly on \(X\) iff \(\sup_{x \in X}|f_n(x)-f(x)| \ _{\overrightarrow{n\to \infty}}\ 0\). In other words, for any \(\varepsilon>0\) there is \(N_{\varepsilon} \in \mathbb{N}\) such that \[n \geq N_{\varepsilon} \quad \text{ implies } \quad |f_n(x)-f(x)|<\varepsilon \quad \text{ for all } \quad x\in X.\]
\(f_n \ _{\overrightarrow{n\to \infty}} \ f\) pointwise iff \(|f(x)-f_n(x)| \ _{\overrightarrow{n\to \infty}} \ 0\) for any \(x \in X\). In other words, for any \(x \in X\) and \(\varepsilon>0\) there is \(N_{\varepsilon, x} \in \mathbb{N}\) such that \[n \geq N_{\varepsilon, x} \quad \text{ implies } \quad |f_n(x)-f(x)|<\varepsilon.\]
Uniform convergence implies pointwise convergence.
Pointwise convergence implies pointwise convergence a.e.
Let \((X,\mathcal M,\mu)\) be a measure space and let \(L^0(X)\) be the space of all complex-valued measurable functions on \(X\).
We say that a sequence of measurable functions \((f_n)_{n \in \mathbb{N}}\subseteq L^0(X)\) converges in measure to a measurable function \(f\in L^0(X)\) if for every \(\varepsilon>0\) we have \[\mu \left(\{x \in X: |f_n(x)-f(x)|\geq \varepsilon\}\right) \ _{\overrightarrow{n\to \infty}} \ 0.\]
Cauchy sequence We say that a sequence of measurable functions \((f_n)_{n \in \mathbb{N}}\subseteq L^0(X)\) is Cauchy in measure if for every \(\varepsilon>0\) we have \[\mu \left(\{x \in X: |f_m(x)-f_n(x)|\geq \varepsilon\}\right) \ _{\overrightarrow{m, n\to \infty}} \ 0.\]
Proposition. Let \((X,\mathcal M,\mu)\) be a measure space, \((f_n)_{n \in \mathbb{N}}\subseteq L^1(X)\) and \(f\in L^1(X)\). If \(f_n \ _{\overrightarrow{n\to \infty}} \ f\) in \(L^1(X)\), which means that \(\lim_{n\to \infty}\int_X|f_n-f|d\mu=0\), then \(f_n \ _{\overrightarrow{n\to \infty}} \ f\) in measure.
Proof. For any \(\lambda>0\), recall Chebyshev’s inequality \[\mu\left(\{x \in X:|h(x)| \geq \lambda\}\right) \leq \frac{1}{\lambda}\int_E |h(x)|d\mu(x),\] where \(E=\{x \in X:|h(x)| \geq \lambda\}\). Let \(E_{n,\varepsilon}=\{x \in X:|f_n(x)-f(x)| \geq \lambda\}\). Then by Chebyshev’s inequality we have \[\qquad \varepsilon \mu(E_{n,\varepsilon}) \le \frac{1}{\varepsilon}\int_{E_{n,\varepsilon}}|f_n-f|d\mu \le \frac{1}{\varepsilon}\int_X |f_n-f|d\mu \ _{\overrightarrow{n\to \infty}} \ 0. \qquad \tag*{$\blacksquare$}\]
Pointwise convergence does not imply \(L^1\) convergence, it suffices to consider \(f_n(x)=n^{-1}\mathbf{1}_{{(0, n)}}(x)\), which converges to \(0\) uniformly, but \(L^1\) convergence fails, since \(\int_{\mathbb R}|f_n(x)|dx=1\).
\(L^1\) convergence does not imply pointwise convergence. To see this consider \(f_n(x)=\mathbf{1}_{{[j2^{-k}, (j+1)2^{-k}]}}\) for \(n=2^k+j\) with \(0\le j<2^k\). Then we have \(\int_{0}^1|f_n(x)|dx=2^{-k}\) for any \(2^k\le n< 2^{k+1}\), but \(f_n\) does not converge for any \(x\in [0, 1]\), since there are infinitely many \(n\in\mathbb N\) for which \(f_n(x)=0\) and infinitely many for which \(f_n(x)=1\).
On the other hand, if \(f_n \ _{\overrightarrow{n\to \infty}} \ f\) a.e. on \(X\) and \(|f_n| \le g \in L^1(X)\) for all \(n \in \mathbb{N}\) then \(f_n \ _{\overrightarrow{n\to \infty}} \ f\) in \(L^1(X)\), which means \(\int_X |f_n-f|d\mu \ _{\overrightarrow{n\to \infty}} \ 0.\)
If \(f_n \ _{\overrightarrow{n\to \infty}} \ f\) in \(L^1(X)\) then there exists \((n_k)_{k \in \mathbb{N}}\subseteq \mathbb N\) such that \(\lim_{k \to \infty}f_{n_k}=f\) a.e. on \(X\). This will follow from Riesz’s theorem.
Let \((X,\mathcal M,\mu)\) be a measure space and let \((f_n)_{n \in \mathbb{N}}\subseteq L^0(X)\) be a Cauchy sequence in measure. Then there is a subsequence \((n_k)_{k \in \mathbb{N}}\subseteq \mathbb N\) and a measurable function \(f\in L^0(X)\) such that
\(f_{n_k} \ _{\overrightarrow{k\to \infty}} \ f\) a.e. on \(X\).
\(f_n \ _{\overrightarrow{n\to \infty}} \ f\) in measure.
Moreover, if \(f_n \ _{\overrightarrow{n\to \infty}} \ g\) in measure, then \(f=g\) a.e. on \(X\).
In particular, if \(f_n \ _{\overrightarrow{n\to \infty}} \ f\) in \(L^1(X)\) then there is \((n_k)_{k \in \mathbb{N}}\subseteq \mathbb N\) such that \(f_{n_k} \ _{\overrightarrow{k\to \infty}} \ f\) a.e. on \(X\).
Proof. \((f_n)_{n \in \mathbb{N}}\) is Cauchy in measure if for every \(\varepsilon>0\), we have \[\mu \left(\{x \in X: |f_m(x)-f_n(x)|\geq \varepsilon\}\right) \ _{\overrightarrow{m, n\to \infty}} \ 0.\]
We choose a sequence \((g_j)_{j \in \mathbb{N}}=(f_{n_j})_{j \in \mathbb{N}}\) such that \[\text{ if } \ E_j=\{x \in X: |g_j(x)-g_{j+1}(x)| \geq 2^{-j}\}, \quad \text{ then } \quad \mu(E_j) \leq {2^{-j}}.\]
If \(F_k=\bigcup_{j \geq k}E_j\), then \[\mu(F_k) \leq \sum_{j = k}^{\infty}\mu(E_j)=\sum_{j = k}^{\infty}{2^{-j}}=2^{1-k} \ _{\overrightarrow{k\to \infty}} \ 0.\]
If \(x \not\in F_k\), then for any \(n \geq m \geq k\) one has \[|g_m(x)-g_n(x)| \leq \sum_{j=m}^{n-1}|g_{j+1}(x)-g_j(x)| \leq \sum_{j=m}^{n-1}{2^{-j}} \leq 2^{1-m} \ _{\overrightarrow{m\to \infty}} \ 0. \quad {\color{purple}(*)}\]
We have proved that \((g_n)_{n \in \mathbb{N}}\) is Cauchy in \(F_k^c\). Let \[F=\bigcap_{k=1}^{\infty}F_k=\limsup_{n\to \infty} E_n.\]
Since \(\mu(F_k) \leq 2^{1-k}\) we deduce that \[\mu(F) \leq \mu(F_k) \leq 2^{1-k}, \quad \text{ which implies } \quad \mu(F)=0.\]
If we set \[f(x)= \begin{cases} \lim_{n \to \infty}g_n(x) & \text{ if } x \not\in F,\\ 0 & \text{ if } x \in F, \end{cases}\] then we see that \(f\) is measurable on \(X\) and thus \(g_n \ _{\overrightarrow{n\to \infty}} \ f\) a.e.
If \(x \not\in F_m\) and \(m\ge k\), then by (*) we obtain \[|g_m(x)-f(x)| \leq 2^{1-m}.\]
Thus we pick \(m\in \mathbb N\) so that \(2^{1-m}<\varepsilon\) and note that \[\begin{aligned} &\mu \left(\{x \in X:|g_m(x)-f(x)| \geq \varepsilon\}\right) \\ &\qquad \leq \mu\left(\{x \in F_m^c:|g_m(x)-f(x)| \geq \varepsilon\}\right)+\mu(F_m) \ _{\overrightarrow{m\to \infty}} \ 0. \end{aligned}\]
We have shown that \(g_k=f_{n_k} \ _{\overrightarrow{k\to \infty}} \ f\) in measure, but then \(f_n \ _{\overrightarrow{n\to \infty}} \ f\) in measure, since \[\begin{gathered} \{x \in X:|f_n(x)-f(x)| \geq \varepsilon\} \subseteq\\ \{x \in X: |f_n(x)-f_{n_k}(x)| \geq {\varepsilon}/{2}\} \cup \{x \in X: |f_{n_k}(x)-f(x)| \geq {\varepsilon}/{2}\}. \end{gathered}\]
If \(f_n \ _{\overrightarrow{n\to \infty}} \ g\) in measure, then \[\begin{gathered} \{x \in X: |f(x)-g(x)| \geq \varepsilon\}\subseteq \\ \{x \in X: |f(x)-f_n(x)|\geq {\varepsilon}/{2}\} \cup \{x \in X: |f_n(x)-g(x)|\geq {\varepsilon}/{2}\}. \end{gathered}\]
Since \(\varepsilon>0\) is arbitrary we deduce that \[\mu(\{x \in X: |f(x)-g(x)| \geq \varepsilon\})=0,\] showing that \(f=g\) a.e. on \(X\).$$\tag*{$\blacksquare$}$$
Although the notions of measurable sets and measurable functions represent new tools, we should not overlook their relation to the older concepts they replaced. Littlewood aptly summarized these connections in the form of three principles:
Three Littlewood principles
Every set is nearly a finite union of intervals.
Every convergent sequence is nearly uniformly convergent.
Every function is nearly continuous.
If \(E\in \mathcal L(\mathbb R^d)\) and \(\lambda_d(E)<\infty\), then for every \(\varepsilon>0\) there is a set \(A\) that is a finite union of open rectangles such that \(\lambda_d(E\triangle A)<\varepsilon.\)
The same result is true if \(\mu\) is a Lebesgue–Stieltjes measure on \(\mathbb R\).
Let \((X,\mathcal M,\mu)\) be a finite measure space \(\mu(X)<\infty\). Let \((f_n)_{n \in \mathbb{N}}\subseteq L^0(X)\) and \(f_{n} \ _{\overrightarrow{n\to \infty}} \ f \in L^0(X)\) a.e. on \(X\). Then for every \(\varepsilon>0\) there exists a set \(E \in \mathcal M\) such that \(\mu(E)<\varepsilon\) and \(f_{n} \ _{\overrightarrow{n\to \infty}} \ f\) uniformly on \(E^c\).
Proof. We may assume (wlog) that \(f_n(x) \ _{\overrightarrow{n\to \infty}} \ f(x)\) for every \(x \in X\). Let \[E_n(k)=\bigcup_{m \geq n}\{x \in X: |f_m(x)-f(x)| \geq k^{-1}\}.\]
For a fixed \(k \in \mathbb{N}\), observe that \(E_n(k) \supseteq E_{n+1}(k)\) and \[\bigcap_{n \in \mathbb{N}}E_n(k) =\varnothing.\] Since \(\mu(X)<\infty\) by the continuity of the measure \[\lim_{n \to \infty}\mu(E_n(k))=0.\]
Given \(\varepsilon>0\) and \(k \in \mathbb{N}\) we choose \(n_k\) so large that \[\mu(E_{n_k}(k))<{\varepsilon}{2^{-k}}.\] We now consider \[E=\bigcup_{k \in \mathbb{N}}E_{n_k}(k).\] Then \[\mu(E) \leq \sum_{k \in \mathbb{N}}\mu(E_{n_k}(k)) \leq \sum_{k \in \mathbb{N}}{\varepsilon}{2^{-k}}=\varepsilon,\] and for \(x \in E^c\) we have that \[|f_n(x)-f(x)|<{k}^{-1}\quad \text{ for } \quad n \geq n_k.\] Thus \(f_n \ _{\overrightarrow{n\to \infty}} \ f\) uniformly on \(E^c\).$$\tag*{$\blacksquare$}$$
A step function on \(\mathbb R^d\) is a finite linear combination, with complex coefficients, of characteristic functions of rectangles in \(\mathbb R^d\), i.e. \[\varphi(x)=\sum_{j=1}^na_j\mathbf{1}_{{R_j}},\] where \(R_j\subseteq \mathbb R^d\) is a rectangle and \(a_j\in \mathbb C\). A family of step functions on \(\mathbb R^d\) is a subfamily of simple functions on \(\mathbb R^d\).
Let \(f: \mathbb R^d\to \mathbb C\) be a measurable function. Then there exists a sequence of step functions \((\varphi_n)_{n\in \mathbb N}\) such that \[\lim_{n\to \infty}\varphi(x)=f(x) \quad \text{ for almost every } \quad x\in \mathbb R^d.\]
We know that if \(f:\mathbb R^d \to \mathbb{C}\) is measurable then there is a sequence \((f_n)_{n \in \mathbb{N}}\) of simple functions such that \(0 \leq |f_1| \leq |f_2| \leq \ldots \leq |f|\) and \(f_n \ _{\overrightarrow{n\to\infty}}\ f\) pointwise everywhere on \(\mathbb R^d\).
Thus it suffices to show that if \(E\) is a measurable set with finite measure, then \(f = \mathbf{1}_{{E}}\) can be approximated by step functions.
We now use the first Littlewood’s principle, asserting that for every \(\varepsilon>0\) there are rectangles \(Q_1,\ldots, Q_k\) so that \(\lambda_d(E\triangle \bigcup_{j=1}^kQ_j)<\varepsilon\).
By considering the grid formed by extending the sides of these cubes, we find almost disjoint rectangles \(P_1, \ldots, P_m\) such that \[ \bigcup_{j=1}^kQ_j=\bigcup_{j=1}^mP_j \quad \text{ and } \quad \lambda_d(E\triangle \bigcup_{j=1}^mP_j)<\varepsilon.\]
By taking rectangles \(R_j\) contained in \(P_j\), and slightly smaller in size, we find a collection of disjoint rectangles that satisfy \[\lambda_d(E\triangle \bigcup_{j=1}^mR_j)<2\varepsilon.\]
Therefore \[f(x)=\mathbf{1}_{{E}}(x)=\sum_{j=1}^m\mathbf{1}_{{R_j}}(x)\] except possibly on a set of measure \(<2\varepsilon\).
The same remains true if \(f\) is a simple function.
Consequently, for every \(n\in \mathbb N\), there exists a step function \(\varphi_n\) such that if \[E_n=\{x\in \mathbb R^d: f(x)\neq\varphi_n(x)\},\] then \(\lambda_d(E_n)<2^{-n}\).
If we let \(F_m=\bigcup_{j=m+1}^\infty E_j\) and \(F=\bigcap_{m\in\mathbb N} F_m\), then \(\lambda_d(F)=0\), since \(\lambda_d(F_m)<2^{-m}\) and \[\lim_{n\to\infty}\varphi_n(x)=f(x)\] on the complement of \(F\), which gives the desired result. $$\tag*{$\blacksquare$}$$
Given a measurable set \(E \subseteq \mathbb R^d\) and given \(f : E \to \mathbb C\), the following statements are equivalent:
\(f\) is measurable,
for every \(\varepsilon>0\) there exists a closed set \(F\subseteq E\) such that \(\lambda_d(E\setminus F)<\varepsilon\) and \(f\) restricted to the set \(F\) is continuous.
Proof of (a) \(\Longrightarrow\) (b). Assume that \(\lambda_d(E)<\infty\) and fix \(\varepsilon>0\).
Let \((f_n)_{m\in\mathbb N}\) be a sequence of step functions such that \(f_{n} \ _{\overrightarrow{n\to \infty}} \ f\) a.e. on \(\mathbb R^d\). Then we find measurable sets \(E_n\) so that \(\lambda_d(E_n) < \varepsilon2^{-n-1}\) and \(f_n\) is continuous outside \(E_n\).
By Egoroff’s theorem we find a measurable set \(U\) such that \(f_{n} \ _{\overrightarrow{n\to \infty}} \ f\) uniformly on \(U\) and \(\lambda_d(E\setminus U)<\varepsilon/4\).
Define a new set \[D=U\setminus \bigcup_{n\in\mathbb N}E_n,\] then we see that \(f_{n} \ _{\overrightarrow{n\to \infty}} \ f\) uniformly on \(D\) and \((f_n)_{m\in\mathbb N}\) is a sequence of continuous functions on \(D\), thus \(f\) is a continuous function on \(D\).
Now we find a closed set \(F\subseteq D\) such that \(\lambda_d(D\setminus F)<\varepsilon/4\). Then it is easy to see that \(\lambda_d(E\setminus F)<\varepsilon\).
If \(\lambda_d(E)=\infty\) then we may apply the previous result to the sets \(A_k=E\cap\{x\in \mathbb R^d: k\le |x|<k+1\}\) and then use regularity of \(\lambda_d\).
Proof of (b) \(\Longrightarrow\) (a). We may assume that \(f\) is real-valued. Then for \(n\in \mathbb N\) we can find a closed set \(F_n\subseteq E\) such that \(f\) is continuous on \(F_n\) and \(\lambda_d(E\setminus F_n)<n^{-1}\). Define \(F=\bigcup_{n\in\mathbb N}F_n\) and note that \(\lambda_d(E\setminus F)=0\). We see that for every \(a>0\) the following set \[\{x\in E: f(x)>a\}=\bigcup_{n\in \mathbb N}\{x\in F_n: f(x)>a\}\cup \{E\setminus F: f(x)>a\}\] is measurable since \(\lambda_d\) is complete. $$\tag*{$\blacksquare$}$$
Let \(\mathbb K\) denote either \(\mathbb{R}\) or \(\mathbb{C}\) and let \(X\) be a vector space over \(\mathbb K\). We denote the zero element of \(X\) simply by \(0\).
By a subspace we shall always mean a vector subspace.
If \(x \in X\) we denote by \(\mathbb Kx=\{\lambda x: \lambda \in \mathbb K\}\) the one-dimensional space spanned by \(x\).
Definition. A seminorm on \(X\) is a function \(X\ni x \mapsto \|x\|\in [0,\infty)\) such that
\(\|x+y\| \leq \|x\|+\|y\|\) for all \(x,y \in X\),
\(\|\lambda x\|=|\lambda|\|x\|\) for all \(x \in X\) and \(\lambda \in \mathbb K\).
Definition. A seminorm such that \(\|x\|_X=0\) only when \(x=0\) is called a norm. A vector space equipped with a norm is called a normed vector space.
Norms and metrics. If \((X, \|\cdot\|)\) is a normed vector space the function \(\rho(x,y)=\|x-y\|\) is a metric on \(X\). The topology it defines is called the norm topology on \(X\).
Definition. A normed vector space \((X, \|\cdot\|)\) that is complete (Cauchy complete) with respect to the norm metric is called a Banach space.
Remark. Every normed vector space can be embedded in a Banach space as a dense space. One way to do this is to mimic the construction of \(\mathbb{R}\) from \(\mathbb{Q}\) via Cauchy sequences.
Definition. Two norms \(\|\cdot\|_1\) and \(\|\cdot\|_2\) on \(X\) are equivalent if there exists \(C \geq 1\) such that \[C^{-1}\|x\|_1 \leq \|x\|_2 \leq C\|x\|_1 \quad \text{ for any }\quad x \in X.\]
Equivalent norms define equivalent metrics and hence the same topology and the same Cauchy sequences.
Definition. If \((x_n)_{n \in \mathbb{N}}\) is a sequence in \(X\), the series \(\sum_{n=1}^{\infty}x_n\) is said to converge to \(x\in X\) if \[\sum_{n=1}^{N}x_n \ _{\overrightarrow{N\to \infty}} \ x\quad \iff \quad \lim_{N \to \infty}\bigg\|\sum_{n=1}^N x_n-x\bigg\|=0,\] and it is called absolutely convergent if \(\sum_{n \in \mathbb{N}}\|x_n\|<\infty\).
A normed vector space \((X,\|\cdot\|)\) is complete iff every absolutely convergent series in \(X\) converges.
Proof \((\Longrightarrow)\). Assume that \(X\) is complete and \(\sum_{n \in \mathbb{N}}\|x_n\|<\infty\). Define \[S_n=\sum_{m=1}^{n}x_m \quad \text{ for every }\quad n\in\mathbb N.\] Then for \(n >k\) we have \[\|S_n-S_k\|=\bigg\|\sum_{m=k+1}^n x_m\bigg\| \leq \sum_{m=k+1}^n\|x_m\| \leq \sum_{m=k+1}^\infty\|x_m\| \ _{\overrightarrow{k\to \infty}} \ 0.\] So \((S_n)_{n \in \mathbb{N}}\) converges since \(X\) is complete.
Proof \((\Longleftarrow)\). Let \((x_n)_{n \in \mathbb{N}}\) be a Cauchy sequence in \(X\). For any \(k \in \mathbb{N}\) there is \(n_k \in \mathbb{N}\) such that for every \(n,m \geq n_k\) we have \[\|x_n-x_m\| \leq 2^{-k}.\] Thus we find \(n_1<n_2<\ldots\) such that for every \(k \in \mathbb{N}\) one has \[\|x_{n_{k+1}}-x_{n_k}\| \leq 2^{-k}.\] Define \({\color{blue}y_0=x_{n_1}}\) and \({\color{blue}y_k=x_{n_{k+1}}-x_{n_k}}\) for \(k \geq 1\). Then the sequence \[\sum_{k \in \mathbb{N}}\|y_k\|=\|x_{n+1}\|+\sum_{k \in \mathbb{N}}\|x_{n_{k+1}}-x_{n_k}\| \leq \|x_{n_1}\|+\sum_{k \in \mathbb{N}}2^{-k}<\infty,\] thus \(\sum_{k \in \mathbb{N}}y_k\) converges.
But \[x_{n_k}=\sum_{j=0}^{k-1}y_j=\sum_{j=0}^{k-1}(x_{n_{j+1}}-x_{n_j}),\] hence the sequence \((x_{n_k})_{k\in \mathbb N}\) converges as well. Let \(\lim_{k \to \infty}x_{n_k}=x\).
We show that \[\lim_{n \to \infty}x_n=x.\] Indeed, \(\|x_n-x_m\|<\frac{\varepsilon}{2}\) for all \(n,m \geq N_1\), and \(\|x_{n_k}-x\|<\frac{\varepsilon}{2}\) for all \(k\ge N_2\).
Let \(n, k \geq N=\max\{N_1,N_2\}\), then \(n_{k} \geq k \geq N_2\) and \(n \geq N_1\) yielding \[\qquad\qquad \|x_n-x\|\le \|x_n-x_{n_{k}}\|+\|x_{n_{k}}-x\|<\varepsilon. \qquad\qquad \tag*{$\blacksquare$}\]
Bounded and bounded continuous functions are Banach spaces If \(X\) is a topological space then \[B(X)=\{f:X \to \mathbb{C}: \sup_{x \in X}|f(x)|=\|f\|_{\infty}<\infty\}\] and \[BC(X)=\{f:X \to \mathbb{C}\::\:f\text{ is continuous and } \sup_{x \in X}|f(x)|=\|f\|_{\infty}<\infty\}\] are Banach spaces with the uniform norm \[{\color{red}\|f\|_{\infty}=\sup_{x \in X}|f(x)|}.\] It is straightforward to show that \(f\mapsto\|f\|_{\infty}\) defines a norm on \(B(X)\) and \(BC(X)\) and both spaces are complete with respect to this norm.
Let \((X, \mathcal M, \mu)\) be a measure space. For every \(0<p<\infty\) and a measurable function \(f:X\to \mathbb C\) we define the quantity \[\|f\|_{L^p}=\|f\|_{L^p(X)}=\left(\int_X |f(x)|^pd\mu(x)\right)^{1/p},\] and for \(p=\infty\) we set \[\|f\|_{L^{\infty}}=\|f\|_{L^{\infty}(X)}={\rm esssup}_{x \in X}|f(x)|,\] where \[{\rm esssup}_{x \in X}|f(x)|=\inf\{C \geq 0:\mu(\{x \in X:|f(x)| \geq C\})=0\}.\]
For every \(0<p\le \infty\) we define \[L^p(X,\mathcal M, \mu)=\{f:X \to \mathbb{C}: f \text{ is measurable and }\|f\|_{L^p(X)}<\infty\}.\]
Remarks.
We abbreviate \(L^p(X,\mathcal M, \mu)\) by \(L^p(X,\mu)\) or \(L^p(X)\) or simply \(L^p\) when this will cause no confusions.
As we have done with \(L^1(X)\), we consider two functions to define the same element of \(L^p(X)\) when they are equal almost everywhere.
\(L^p(X)\) is a vector space. For \(p=\infty\) there is nothing to prove. For \(0<p<\infty\) if follows from a simple inequality \[|f+g|^p\le \big(2\max\{|f|, |g|\}\big)^p\le 2^p(|f|^p+|g|^p) \quad \text{ for } \quad f, g\in L^p(X).\]
If \(X\) is a countable set we define \(\ell^p(X)\) to be \(L^p(X, \mathcal P(X), \mu)\), where \(\mu\) is counting measure on \((X, \mathcal P(X))\), then \[\|f\|_{\ell^p(X)}=\Big(\sum_{x\in X}|f(x)|^p\Big)^{1/p} \quad \text{ for } \quad 0<p<\infty.\]
Proposition (Hölder’s inequality). Let \((X,\mathcal M, \mu)\) be a measure space and let \(1 \leq p,q \leq \infty\) with \(\frac{1}{p} + \frac{1}{q} = 1\). Then, for any \(f \in L^p(X)\) and \(g \in L^q(X)\), we have \[\|fg\|_{L^1(X)} \leq \|f\|_{L^p(X)} \|g\|_{L^q(X)}.\] The parameter \(p'=q=\frac{p}{p-1}\) is called the conjugate exponent to \(p\). We will use convention that \(1'=\infty\) and \(\infty'=1\).
Proof. By homogeneity, we may assume that \(\|f\|_{L^p(X)} = \|g\|_{L^q(X)} = 1\). Note that \[\begin{gathered} \|fg\|_{L^1(X)} = \int_X |fg| d\mu = \int_X (|f|^p)^{1/p}(|g|^q)^{1/q} d\mu \leq \int_X \max(|f|^p,|g|^q) d\mu \\ \leq \int_X |f|^p d\mu + \int_X |g|^q d\mu = 2 = 2 \|f\|_{L^p(X)} \|g\|_{L^q(X)}. \end{gathered}\]
Thus, we have proved that \(\|fg\|_{L^1(X)} \leq 2 \|f\|_{L^p(X)} \|g\|_{L^q(X)}\). That is, Hölder’s inequality holds, but with a constant of \(2\) instead of \(1\).
We use the best constant argument to complete the proof. Let \(C > 0\) be the smallest constant such that, for all measure spaces \((X, \mathcal M, \mu)\), we have \[\|fg\|_{L^1(X)} \leq C \|f\|_{L^p(X)} \|g\|_{L^q(X)}.\]
We know that \(C \leq 2\), and we will show that \(C \leq 1\) using the “tensor trick.” Let \(F(x,y) = f(x)f(y)\) and \(G(x,y) = g(x)g(y)\). Then \[\begin{gathered} \|fg\|_{L^1(X)}^2 = \|FG\|_{L^1(X \times X)} \leq C \|F\|_{L^p(X \times X)} \|G\|_{L^q(X \times X)}\\ = C \|f\|_{L^p(X)}^2 \|g\|_{L^q(X)}^2. \end{gathered}\]
Taking square roots, we have \(\|fg\|_{L^1(X)} \leq C^{1/2} \|f\|_{L^p(X)} \|g\|_{L^q(X)}\), so \(C \leq C^{1/2}\), (since \(C\) is the best constant), which implies \(C \leq 1\). $$\tag*{$\blacksquare$}$$
Proposition (Minkowski’s inequality). Let \((X,\mathcal M, \mu)\) be a measure space and let \(1 \leq p\leq \infty\). Then, for any \(f, g \in L^p(X)\), we have \[\|f+g\|_{L^p(X)} \leq \|f\|_{L^p(X)}+ \|g\|_{L^p(X)}.\]
Proof. For \(f, g \in L^p(X)\) note that \[|f+g|^p\le |f||f+g|^{p-1}+|g||f+g|^{p-1}.\] Then by Hölder’s inequality, with \(q=\frac{p}{p-1}\), we immediately obtain \[\begin{aligned} \|f+g\|_{L^p(X)}^p&\le \big(\|f\|_{L^p(X)}+ \|g\|_{L^p(X)}\big)\||f+g|^{p-1}\|_{L^q(X)}\\ &\le \big(\|f\|_{L^p(X)}+ \|g\|_{L^p(X)}\big)\|f+g\|_{L^p(X)}^{p/q}. \end{aligned}\] Dividing both side by \(\|f+g\|_{L^p(X)}^{p/q}=\|f+g\|_{L^p(X)}^{p-1}\) we obtain Minkowski’s inequality. $$\tag*{$\blacksquare$}$$
Let \((X,\mathcal M, \mu)\) be a measure space. For every \(1 \leq p\leq \infty\) the set \(L^p(X)\) is a norm vector space with the norm \(\|\cdot\|_{L^p(X)}\), where \[\|f\|_{L^p(X)}=\left(\int_X |f(x)|^pd\mu(x)\right)^{1/p}, \quad \text{ for } \quad 1<p<\infty,\] and \[\|f\|_{L^{\infty}(X)}={\rm esssup}_{x \in X}|f(x)|, \quad \text{ for } \quad p=\infty.\]
Proof. Let \(f, g \in L^p(X)\) be given.
It is obvious that \(\|f\|_{L^p(X)}=0\) if and only if \(f=0\) a.e. on \(X\).
It is also easy to see that \(\|\lambda f\|_{L^p(X)}=|\lambda|\|f\|_{L^p(X)}\) for any \(\lambda\in \mathbb C\).
By Minkowski’s inequality we also have the triangle inequality \[\qquad\qquad \|f+g\|_{L^p(X)} \leq \|f\|_{L^p(X)}+ \|g\|_{L^p(X)}. \qquad\qquad \tag*{$\blacksquare$}\]
Let \((X,\mathcal M, \mu)\) be a measure space. For every \(1 \leq p\leq \infty\) the set \(L^p(X)\) with the norm \(\|\cdot\|_{L^p(X)}\) is a Banach space.
Proof. To prove completeness if suffices to show that every absolutely convergent series in \(L^p(X)\) converges.
Suppose that \((f_k)_{k\in\mathbb N}\subseteq L^p(X)\) and \(\sum_{k\in\mathbb N}\|f_k\|_{L^p(X)}=B<\infty\).
Let \(G_n=\sum_{k=1}^n|f_k|\) and \(G=\sum_{k\in \mathbb N}|f_k|\). Then \(G_n\in {L^p(X)}\), and \(G_n\le G_{n+1}\) and \(\|G\|_{L^p(X)}^p=\lim_{n\to \infty}\|G_n\|_{L^p(X)}^p\le B^p\) by the (MCT).
Hence \(G\in L^p(X)\) and \(G(x)<\infty\) a.e. on \(X\), which ensures that \(F=\sum_{k\in \mathbb N}f_k<\infty\) a.e. on \(X\). Since \(|F|\le G\) we obtain \(F\in L^p(X)\).
Then \(|F-\sum_{k=1}^nf_k|^p\le (2G)^p\in L^1(X)\), so by the (DCT) we have \[\Big\|F-\sum_{k=1}^nf_k\Big\|_{L^p(X)} \ _{\overrightarrow{n\to \infty}} \ 0.\]
Thus the series \(\sum_{k=1}^nf_k\) converges in the \(L^p(X)\) norm as desired. $$\tag*{$\blacksquare$}$$