Skip to content

Normed Space

Def'n. Norm

Let \(V\) be a vector space over \(\mathbb{R}\), a norm \(\|\cdot\|\) over \(V\) is a function \(V\rightarrow \mathbb{R}^+\) s.t.

  • positive definite: \(\|x\|\geq 0 \land \|x\| = 0 IFF x\equiv 0 \in V\)
  • homogeneous \(\|ax\| = |a|\|x\|. \forall a\in \mathbb{R}\)
  • triangular inequality \(\|x+y\| = \|x\| + \|y\|\)

Example: Euclidean space over \(\mathbb{R}^n\)

\(\|x\|_2 = \sqrt{\sum |x_k|^2}\) - positive definite
\(|x_k|^2 \geq 0 \Rightarrow \sqrt{\sum |x_k|^2} \geq 0\)
\(\|x\| = 0\Rightarrow x = 0\) - homogeneous \(\|ax\| = \sqrt{\sum |ax_k|^2} = \sqrt{|a|^2\sum |x_k|^2} = |a|\|x\|\) - Tri-ineq
By Cauchy Swartz Inequality
\(\|x+y\|^2 = \sum |x_k + y_k|^2 \leq \sum |x_k|^2 + |y_k|^2 = \sum |x_k|^2 + \sum|y_k|^2 = \|x\|^2 + \|y\|^2\)

Example: Some well-known norms

Some norms are

  • p-norm \(\|x\|_p:=(\sum |x_k|^p)^{p^{-1}}, p\geq 1\)
  • Lp-norm \(\|f\|_{L_p} := (\int_S f(x)^p dx)^{p^{-1}}, p\geq 1\) is a norm over \(C:=\) the set of all continuous functions
  • sup-norm \(\|f\|_\infty:= sup\{|f(x)|:x\in S\}\) is a norm over \(C_b(S):=\) the set of all continuous bounded functions or over \(C(S)\) if \(S\) is compact.

Example: sup-norm

Claim. sup-norm is a norm proof.

  1. \(|f(x)| \geq 0\), \(f(x):= 0 \Rightarrow \|f\|_\infty = 0, \|f\|_\infty=0\Rightarrow |f(x)|\leq 0\Rightarrow f(x)=0\)
  2. \(\|af\|_\infty = sup|af(S)| = |a|sup|f(S)|= |a|\|f\|_\infty\)
  3. \(\|f+g\|=sup(|f+g|) \leq sup(|f| + |g|) \leq sup|f| + sup|g| = \|f\| + \|g\|\)

\(C^k(S):=\) The set of all real number functions whose k-first derivative exists and continuous

Some norms are defined on \(C^k\), such as \(\|f\|':= \max\|f'\|_\infty; \|f\|_{C^k} = \sum \|f^{(n)}\|_\infty\)

Remark \(C^\infty\) a.k.a. smooth are normed space that obey completeness

Topology of normed-spaces

Thrm 1.

The set \(C:=\{f:[0,1]\rightarrow \mathbb{R}: f(0)=1\}\) is closed in \((C([0,1]), \|f\|_\infty:= sup_{x\in [0,1]}f(x))\).

proof. Take a sequence \(g_n \in C\) s.t. \(\|g_n - g\| \rightarrow 0\). Then, consider

\[\begin{align*} |g(0) - 1| &\leq |g(0) - g_n(0)| + |g_n(0) -1|\\ &\leq |g(0)-g_n(0)| \leq \|g_n-g\|=0\\ &\Rightarrow g(0)\rightarrow 1\Rightarrow g(0) = 1\Rightarrow g\in C \end{align*}\]

Thrm 2.

Let \(A:=\{f\in [0,1]: f(x) > 0, \|f'\|_\infty < 2\}\) is open in \(C^1([0,1])\) wrt \(\|f\|_{\infty, C^1}:=\|f\|_\infty + \|f'\|_\infty\)

proof. Take any \(g\in A\).
Since \(g(x) > 0\), by EVT on \([0,1]\), take \(\delta_1\) s.t. \(g(x)> \delta_1 > 0\).
Since \(g'(x) < 2\), by EVT on \([0,1]\), take \(\delta_2\) s.t. \(g'(x)< 2-\delta_2 < 2\).
Take \(\delta = \min(\delta_1, \delta_2)/ 2\).

\[\begin{align*} \|h-g\|_\infty < \delta &\implies |h(x)-g(x)|< \delta \\ &\implies h(x) > g(x)-\delta > \delta_1/2 > 0\\ \|h'-g'\|_\infty < \delta &\implies |h'(x) - g'(x)| < \delta\\ &\implies h'(x) < g'(x)+\delta < 2-\delta_2 + \delta_2/2 < 2 \end{align*}\]

Thrm 3.

\(C_c(\mathbb{R}):=\) space of compactly supported function on reals, \(C_c(\mathbb R), f\in C_c(\mathbb R)\) if \(\exists M > 0\) s.t. \(f(x)=0, \forall |x| > M\)

Claim \(C_c(\mathbb R)\) is not complete wrt \(\|f\|_\infty\)

proof. WTF a Cauchy sequence \(f_n \in C_c(\mathbb R)\) s.t. \(f_m \rightarrow f\not\in C_c(\mathbb R)\).
Take \(f_n(x) = \frac{1-(x/n)^2}{1+x^2}\mathbb I (|x|\leq n)\) and we can show that \(f_n\) is Cauchy wrt \(\|\cdot\|_\infty\)

wlog, assume \(n > m\),
Consider

\[\|f_n - f_m\| = \sup_{x\in [-n,n]} |\frac{1-(x/n)^2}{1+x^2} - f_m(x)|\]

Suppose that \(x\in [-m,m]\),

\[\sup_{x\in [-n,n]} |\frac{1-(x/n)^2}{1+x^2} - \frac{1-(x/m)^2}{1+x^2}| = |\frac{x^2}{1+x^2}||n^{-2} - m^{-2}| \rightarrow |1||0|=0\]

Suppose that \(x\in [-n, -m)\cup (m, n]\),

\[\sup_{x\in [-n,n]} |\frac{1-(x/n)^2}{1+x^2}|\leq |\frac{1}{1+x^2}|\leq |\frac{1}{1+m^2}|\rightarrow 0\]

Therefore, take \(N = \epsilon^{1/2}\) we can prove Cauchy.

However, \(f_n(x)=\frac{1-(x/n)^2}{1+x^2}\rightarrow \frac{1}{1+x^2}\rightarrow 0\) is not compactly supported

Therefore, only compact in metric space \(\rightarrow\) closed and bounded but not the converse.