Banach Space

Normed Linear Space

We have already learn the definition of linear space with its number domain \(K\). Now we introduce a topology in it.

Definition of Norm

Assume \(E\) is a linear space. If \(\forall x\in E\), there exists a corresponding real number, denoted as \(\|x\|\), which satisfies

(i) Positive & Definite. \(\|x\|\geq 0\). And \(\|x\|=0\), iff \(x=\theta\).

(ii) Homogeneous. \(\forall \alpha\in K\), \(\|\alpha x\|=|\alpha| \|x\|\).

(iii) Triangle inequation. \(\forall x,y\in E\),

\[ \|x+y\|\leq \|x\|+\|y\|. \]

then we call \(\|\cdot\|\) norm and \(E\) is normed linear space.

Readers could check norm of finite-dimensional linear space in Numerical Analysis.

If we define

\[ \rho(x,y)=\|x-y\| \]

then it is easy to see the above defines a metric in \(E\). With metric, all the conclusions in previous chapter Metric Space could apply here. The following is about convergence.

Properties of Norm

(i) As a functional \(\|x\|\) is continuous.

(ii) If \(\{x_n\}\subset E\) converges to \(x\in E\), then \(\{\|x_n\|\}\) is bounded.

The linear operations are continuous:

(iii) Assume \(\{x_n\}, \{y_n\}, x,y\in E\), and \(x_n\rightarrow x\) and \(y_n\rightarrow y\), respectively, then

\[ x_n+y_n\rightarrow x+y. \]

(iv) Assume \(\{\alpha_n\}, \alpha\in K\), and \(\alpha_n\rightarrow \alpha\). And \(\{x_n\}, x\in E\), and \(x_n\rightarrow x\). Then we have

\[ \alpha_n x_n\rightarrow \alpha x. \]

Proof

(i) Since \(\{x_n\}\rightarrow x\), and by triangle inequation

\[ |\|x_n\|-\|x\||\leq \|x_n-x\|. \]

(ii) From above we have

\[ \|x_n\|\leq \|x_n-x\|+\|x\|<M. \]

(iii) By

\[ \|x_n+y_n-x-y\|\leq \|x_n-x\|+\|y_n-y\|. \]

(iv) By \(|\alpha_n|\) is bounded

\[ \|\alpha_n x_n-\alpha x\|\leq \|\alpha_n x_n-\alpha_n x\|+\|\alpha_n x-\alpha x\|\leq |\alpha_n| \|x_n-x\|+|\alpha_n-\alpha|\|x\|. \]

\(\square\)

A normed linear space is called Banach Space if it is complete.

Example. Assume space \(C^k[a,b]\) is composed by functions with \(k\)-th continuous derivatives. Define norm

\[ \|x\|=\sum_{i=0}^k \max_{t\in [a,b]}|x^{(i)}(t)|. \]

Then in this norm the above space is Banach space.

Proof

It is easy to show that norm satisfies triangle inequation.

We now show it is complete. For all Cauchy sequence \(\{x_n\}\subset C^k[a,b]\), \(\forall \varepsilon\), \(\exists N>0\), \(\forall n,m>N\),

\[ \begin{align} \|x_n-x_m\|=\sum_{i=0}^k \max_{t\in [a,b]}|x_n^{(i)}(t)-x_n^{(i)}(t)|<\varepsilon.\label{Cauchy-continuous-k} \end{align} \]

So for each \(i\), we have

\[ |x_n^{(i)}(t)-x_m^{(i)}(t)|<\varepsilon,\quad \forall t\in [a,b]. \]

So \(x_n^{(i)}\) is a Cauchy sequence, by mathematical analysis, we have a convergent function \(x_i(t), (i=0,\cdots,k)\), such that

\[ x_n^{(i)}(t)\rightrightarrows x_i(t),\quad i=0,1,\cdots,k \]

and satisfy \(x^{(i)}(t)=x_i(t)\) for each \(i\). So \(x(t)\in C^k[a,b]\). We show that \(x_n(t)\) converges to \(x(t)\) in this norm.

In equation \(\ref{Cauchy-continuous-k}\), fix \(n\), let \(m\rightarrow \infty\), and we have the result.

\(\square\)

Example. \(L^p(E)\) is a Banach space.

Proof

• From a Cauchy sequence in this metric to get a Cauchy sequence in measure.

For all Cauchy sequence \(\{x_n\}\), \(\forall \varepsilon, \sigma>0\), \(\exists N>0\), \(\forall n,m>N\),

\[ \begin{align*} \varepsilon>&\|x_n,x_m\|\\ &=\left(\int_E |x_n(t)-x_m(t)|^p dt\right)^{\frac{1}{p}}\\ &\geq \left(\int_{\{|x_n(t)-x_m(t)|\geq \sigma\}} |x_n(t)-x_m(t)|^p dt\right)^{\frac{1}{p}}\\ &\geq \left(\int_{\{|x_n(t)-x_m(t)|\geq \sigma\}} \sigma^p dt\right)^{\frac{1}{p}}\\ &\geq \sigma m(\{|x_n(t)-x_m(t)|\geq \sigma\})^{\frac{1}{p}} \end{align*} \]

which means \(x_n\) is a Cauchy sequence in measure.

• From a Cauchy sequence in measure to get a convergent function in usual metric \(||\).

One method is, by what we have learned, there exists a finite-valued function \(x\) a.e. on \(E\) such that \(x_n\) converges to \(x\) in measure. By Riesz Theorem, there exists a subsequence \(\{x_{n_k}\}\) such that \(x_{n_k}\) converges to \(x\) a.e. on \(E\).

Or we take another version inspired by Lemma for summation in \(L^p\) space. Since \(x_n\) is a Cauchy sequence in \(L^p(E)\), \(\forall k>0\), there exists a subsequence \(x_{n_k}\) such that

\[ \left(\int_E|x_{n_k}-x_{n_{k-1}}|^p dt\right)^\frac{1}{p}=\|x_{n_k}-x_{n_{k-1}}\|<\frac{1}{2^k} \]

So by Hölder inequation, \(1\in L^q\), \(q=\frac{p}{p-1}\),

\[ \int_E |x_{n_k}-x_{n_{k-1}}| \cdot 1 dt\leq (\int_E |x_{n_k}-x_{n_{k-1}}|^p dt)^\frac{1}{p}m(E)^\frac{1}{q}<\frac{1}{2^k} m(E)^\frac{1}{q} \]

So by Levy Theorem, \(f_n=\sum_{k=1}^n |x_{n_k}-x_{n_{k-1}}|\) converges to \(x\).

\[ \begin{align*} \int_E \sum_{k=1}^\infty |x_{n_k}-x_{n_{k-1}}| dt &= \lim_{k\rightarrow \infty}\int_E \sum_{k=1}^K |x_{n_k}-x_{n_{k-1}}| dt \\ &\leq \lim_{k\rightarrow \infty} \sum_{k=1}^K \int_E |x_{n_k}-x_{n_{k-1}}| dt<\infty. \end{align*} \]

So \(x_n=\sum_{k=1}^\infty x_{n_k}- x_{n_{k-1}}+x_i\) converges to \(x\).

• Use Fatau theorem to change the order of limit & integral.

By Fatau theorem, \(\forall n_k>N\), we still have \(\left(\int_E |x_n(t)-x_{n_k}(t)|^p dt\right)^{\frac{1}{p}}<\varepsilon\), so

\[ \begin{align*} \int_E |x_n-x|^pdt &= \int_E \lim_{k\rightarrow \infty}|x_n-x_{n_k}|^pdt\\ &\leq \underline{\lim}\limits_{k\rightarrow \infty} \int_R |x_n-x_{n_k}|^pdt\\ &\leq \varepsilon^p. \end{align*} \]

So \(x_n\) converges to \(x\) in the above metric. Then by Minkowski inequation, we have \(\|x\|\leq \|x_n-x\|+\|x_n\|<\infty\), so \(x\in L^p(E)\).

Quotient Space

Equivalence of two points

Assume \(X\) is a linear space, \(F\subset X\) is a subspace. Two points \(x_1, x_2\in X\) are called equivalent modulo \(F\), denoted as \(x_1\equiv x_2 (\text{ mod } F)\), or simply \(x_1\sim x_2\), if \(x_1-x_2\in F\).

Properties of Equivalence

(i) Reflexive. \(x\sim x\).

(ii) Symmetric. If \(x\sim y\), then \(y\sim x\).

(iii) Tansitive. If \(x\sim y\), \(y\sim z\), then \(x\sim z\).

The above property tells us equivalence mod \(F\) is an equivalence relation, meaning we could divide \(X\) into distinct equivalence classes mod \(F\), denoted as

\[ \xi:=x+F=\{x+y: y\in F\},\text{ where } x\in \xi. \]

The whole equivalence relation compose a set \(X / F\). If we define proper linear operations on \(X/F\), we could show that it is also a linear space.

Linear operations on Space \(X/F\)

(i) \(\forall \xi,\eta\in X/F\), choose \(x\in \xi, y\in \eta\), define

\[ \xi+\eta:=\{x+y+z: z\in F\}=x+y+F. \]

(ii) \(\forall \xi\in X/F\), \(\forall \alpha\in K\), define

\[ \alpha \xi:=\{\alpha x + y: y\in F\}= \alpha x+F. \]

we have to show that the above definition is irrelevant with the choice of \(x\).

Proof

\(\forall x'\in \xi,y'\in \eta\),

\[ \begin{align*} x'+y'+F&=x+(x'-x)+y+(y'-y)+F\\ &=x+y+(x'-x)+(y'-y)+F\\ &=x+y+F. \end{align*} \]

Similar to \(\alpha x'\).

\(\square\)

Easy to see that \(X/F\) is a linear space based on the above linear operations. And its zero element is \(F\).

Provided \(F\) is closed, quotient space \(X/F\) could be a normed linear space.

Normed linear space for \(X/F\)

Assume \(F\) is a closed subspace of a normed linear space \(X\). For any equivalence class \(\xi\in X/F\), we define

\[ \|\xi\|=\inf_{x\in \xi}\|x\|. \]

then the above definition is a norm and \(X/F\) is a normed linear space.

Proof

(i) Positive and definive.

Positive is apparent. We prove definitive. Let \(\|\xi\|=0\), if \(\xi=x_0+F\), \(x_0\not\in F\). Then by definition of infimum, \(\forall \varepsilon_n=\frac{1}{n}\), \(\exists \{y_n\}\subset F\) such that

\[ \|x_0+y_n\|<\frac{1}{n} \]

which means \(y_n\rightarrow -x_0\). Since \(F\) is closed, \(-x_0\in F\), so by linear operation, \(x_0\in F\), which contradicts! So \(\xi=F\).

(ii) Homogeneous. Easy to show.

(iii) Triangle inequation. \(\forall \{x_j\}, \{y_j\}\in X/F\), \(\forall \varepsilon>0\), we have

\[ \|x_j\|\leq \|\{x_j\}\|+\varepsilon, \quad |y_j\|\leq \|\{y_j\}\|+\varepsilon. \]

So

\[ \|\{x_j\}+ \{y_j\}\|\leq \|x_j+y_j\|\leq \|x_j\|+\|y_J\|\leq \|\{x_j\}\|+ \|\{y_j\}\|+2\varepsilon. \]

Let \(\varepsilon\rightarrow 0\), and we are done.

\(\square\)