4 Normed Linear Spaces
So far distance has had no relation to the linear structure. A norm ties the two together: it measures the size of a vector, and it must respect addition and scalar multiplication. Every norm induces a metric, but not every metric comes from a norm.
A Banach space is a normed space that is complete in that metric, and most of the spaces worth naming are Banach. The section then turns to the maps between them. For linear transformations, boundedness and continuity turn out to be the same condition — a genuine simplification, since one is far easier to check than the other.
The section ends with the dual space, the space of bounded linear functionals, and two results about it: the Riesz representation theorem, which identifies the dual of \(l_p\) concretely, and the Hahn–Banach theorem, which guarantees there are enough functionals to be worth studying.
Definition 4.1. A normed linear space is a vector space \(V\) over a field \(\mathbb {F}\) with a function \(||.||:V\rightarrow \mathbb {F}\) that satisfies the following conditions for all \(x,y\in V\) and \(\alpha \in \mathbb {F}\):
- 1.
- \(||x||\geq 0\)
- 2.
- \(||x||=0\) if and only if \(x=0\)
- 3.
- \(||\alpha x||=|\alpha |||x||\)
- 4.
- \(||x+y||\leq ||x||+ ||y||\)
The function \(||.||\) is called a norm on \(V\). \((V,||.||)\) this pair is called a normed space.
Example 4.2. Let \(\mathbb {R}^n\) be the set of all n-tuples of real numbers and \(p\geq 1\) for \(x=(x_1,x_2,\dots ,x_n)\) in \(\mathbb {R}^n\), define \(\displaystyle {||x||=\Bigg (\sum ^n_{i=1}|x_i|^p\Bigg )^\frac {1}{p}}\hspace {0.2cm}\) then \((\mathbb {R}^n,||.||_p)\) is a normed linear space.
Proof. Let \(x=(x_1,x_2,\dots ,x_n), y=(y_1,y_2,\dots ,y_n)\in \mathbb {R}^n\) and \(\alpha \in \mathbb {R}\).
- i
- \(\hspace {0.5cm}\displaystyle {||x||_p=\Bigg (\sum ^n_{i=1}|x_i|^p\Bigg )^{\frac {1}{p}}\geq 0}\hspace {0.2cm},\hspace {0.3cm}\) since \(|xi|\geq 0\) for all \(i\).
- ii
- If \(x=0=(0,\dots ,0)\) then \begin {align*} ||x||^p_p &=||x_1||+||x_2||+\cdots +||x_n||\\ &=|0|+|0|+\cdots +|0|\\ &=0 \end {align*}
\(\implies ||x||_p=0\). Conversely, if \(||x||_p=0\), then \(\displaystyle {\Bigg (\sum ^n_{i=1}|x_i|^p\Bigg )^{\frac {1}{p}}=0}\hspace {0.2cm},\hspace {0.3cm}\) Since \(|x_i|\geq 0\hspace {0.2cm} \forall i\implies |x_i|^p=0\hspace {0.2cm} \forall i\) or \(|x_i|=0\) or \(x_i=0\).
\(\implies x=(x_1,x_2,\dots ,x_n)=(0,0,\dots ,0)=\underline {0}\)
- iii
- \begin {align*} ||\alpha x||_p &=\Bigg (\sum ^n_{i=1}|\alpha x_i|^p\Bigg )^{\frac {1}{p}} =\Bigg (|\alpha |^p\sum ^n_{i=1}|x_i|^p\Bigg )^{\frac {1}{p}}\\ &=|\alpha |\Bigg (\sum ^n_{i=1}|x_i|^p\Bigg )^{\frac {1}{p}}\\ &=|\alpha |||x||_p\\ \end {align*}
- iv
- For \(k>n\), let \(x_k=0,y_k=0\). Then \begin {align*} ||x+y||_p &=\Bigg (\sum ^n_{i=1}|x_i+y_i|^p\Bigg )^{\frac {1}{p}} =\Bigg (\sum ^{\infty }_{i=1}|x_i+y_i|^p\Bigg )^{\frac {1}{p}}\\ &\leq \Bigg (\sum ^{\infty }_{i=1}|x_i|^p\Bigg )^{\frac {1}{p}}+\Bigg (\sum ^{\infty }_{i=1}|y_i|^p\Bigg )^{\frac {1}{p}}\hspace {0.3cm}\text {Minkowski's inequality}\\ &=\Bigg (\sum ^n_{i=1}|x_i|^p\Bigg )^{\frac {1}{p}}+\Bigg (\sum ^n_{i=1}|y_i|^p\Bigg )^{\frac {1}{p}}\\ &=||x||_p+||y||_p.\\ \end {align*}
Example 4.3. Let \(C[0,1]\) be the set of real continuous functions on \([0,1]\), for \(f\in C[0,1]\), define \(\displaystyle {||f||=\max _{0\leq x\leq 1}|f(x)|}.\) Then \((C[0,1],||.||)\) is a normed linear space.
Proof. Let \(f,g\in C[0,1]\) and \(\alpha \in \mathbb {R}\)
- i
- \(\hspace {0.3cm}\displaystyle {||f||=\max _{0\leq x\leq 1}|f(x)|\geq 0}\hspace {0.3cm}\) since \(|f(x)|\geq 0\) \(\forall x\)
- ii
- If \(f(x)=0\hspace {0.3cm} \forall x\in [0,1]\) then \(\displaystyle {||f||=\max _{0\leq x\leq 1}|f(x)|=\max _{0\leq x\leq 1}|0|=0}.\) Conversely, if \(||f||=0\) then \(\displaystyle {\max _{0\leq x\leq 1}|f(x)|=0}\hspace {0.2cm}\), since \(|f(x)|\geq 0\hspace {0.2cm}\forall x\implies |f(x)|=0\) or \(f(x)=0\hspace {0.2cm} \forall x\in [0,1]\). That is \(f=0\), a zero function.
- iii
- \begin {align*} ||\alpha x|| &=\max _{0\leq x\leq 1}\Big |(\alpha f)(x)\Big | =\max _{0\leq x\leq 1}\Big |\alpha f(x)\Big |\\ &=\max _{0\leq x\leq 1}|\alpha ||f(x)|\\ &=|\alpha |\max _{0\leq x\leq 1}|f(x)|\\ &=|\alpha |||f|| \end {align*}
- iv
- \begin {align*} ||f+g|| &=\max _{0\leq x\leq 1}\Big |(f+g)(x)\Big | =\max _{0\leq x\leq 1}\Big |f(x)+g(x)\Big |\\ &\leq \max _{0\leq x\leq 1}\Big (|f(x)|+|g(x)|\Big )\\ &\leq \max _{0\leq x\leq 1}|f(x)|+\max _{0\leq x\leq 1}|g(x)|\\ &=||f||+||g||.\\ \end {align*}
Lemma 4.4. If \((X,||.||)\) is a normed linear space, then the function \(d:X\times X\rightarrow \mathbb {R}\) defined by \(d(x,y)=||x-y||\hspace {0.2cm}\) is a metric on \(X\).
Proof. Let \(x,y,z\in X\). Then
- i
- \(d(x,y)=||x-y||\geq 0\) (since \(||x-y||\) is a norm)
- ii
- \(||x-y||=0\) iff \(x-y=0\iff x=y\). i.e \(d(x,y)=0\) iff \(x=y\).
- iii
- \(d(x,y)=||x-y||=|-1|||y-x||=||y-x||=d(y,x)\).
- iv
- \begin {align*} d(x,z) &=||x-z|| =||x-y+y-z||\\ &\leq ||x-y||+||y-z||\\ &=d(x,y)+d(y,z). \end {align*}
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