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10.13 Inner Product, Topology

Inner product

Let \(V\) be a vector space over \(\R\), the inner product \(\lang ,\rang:V\times V\to \R\) where \(x,y\in V\Rightarrow\lang x,y\rang \in \R\)

Axioms

  1. Bilinearity

    1. \(\lang x_1+x_2,y\rang =\lang x_1,y\rang +\lang x_2,y\rang\quad\forall x_1,x_2,y\in V\)

      \(\lang \lambda x,y\rang =\lambda \lang x,y\rang \quad\forall x,y\in V,\lambda \in \R\) 2. \(\lang x,y_1+y_2\rang =\lang x,y_1\rang+\lang x,y_2\rang\quad \forall x,y_1,y_2\in V\)

      \(\lang x,\lambda y\rang=\lambda \lang x,y\rang\quad \forall x,y\in V,\lambda\in\R\)

  2. Symmetricity: \(\lang x,y\rang =\lang y,x\rang\quad\forall x,y\in V\)

  3. Positive definiteness: \(\forall x\neq 0,\lang x,x\rang >0\)

Examples

  1. \(V=\{(x_{1},...,x_{n}):x_{1},...,x_{n}\in \R\}=\R^n\) is the space of n-tuples

    \(\lang x,y\rang =x_1y_1+...+x_ny_n=xy\) (the dot-product) where \(x=(x_{1},...,x_{n}),y=(y_{1},...,y_{n})\)

  2. \(C^0[a,b]=\{f:[a,b]\to \R:f\text{ is continuous}\}\) where \(\lang f,g\rang=\displaystyle\int_{a}^{b}f(t)g(t)dt\)

Norm

\(|x|=\sqrt{\lang x,x\rang}\) is the length of \(x\in V\) (the Euclidean norm)

Properties

  1. \(|\lambda x|=|\lambda||x|,\forall \lambda \in \R,x\in V\)

  2. \(|\lang x,y\rang |\leq |x||y|,\forall x,y\in V\) this Cauchy Schwarz inequality holds\(\iff\)\(x,y\) are linear independent

    Proof

    1. \(y=\lambda x\) where \(\lambda \in \R\), then \(|\lang x,y\rang |=|\lang x,\lambda x\rang |=|\lambda|\lang x,x\rang\) and \(|x||y|=|x||\lambda x|=|\lambda ||x|^2\), then \(|\lang x,y\rang |= |x||y|,\forall x,y\in V\)

    2. \(y\neq \lambda x,\forall \lambda \in \R\), then \(\lang y-\lambda x,y-\lambda x\rang=\lambda^2\lang x,x\rang-2\lambda\lang x,y\rang +\lang y,y\rang^2>0\quad\forall \lambda\)

      This polynomial doesn't have a solution, then \(\frac{1}{4}\mathcal{D}=\lang x,y\rang^2-\lang x,x\rang\lang y,y\rang=\lang x,y\rang^2-|x|^2|y|^2<0\)

      Then \(\lang x,y\rang^2<|x|^2|y|^2\)

    Example: \(V=C^0[a,b]\), \(\displaystyle\left|\int^b_af(t)g(t)dt\right|^2\leq \int^b_af^2(t)dt\cdot \int^b_ag^2(t)dt\)

  3. \(|x+y|\leq |x|+|y|\) this triangle inequality holds\(\lang x,y\rang \geq 0\)

    Proof

    \(|x+y|^{2}=\lang x+y,x+y\rang=\lang x,x\rang +2\lang x,y\rang+\lang y,y\rang\leq |x|^{2}+2|x|\cdot|y|+|y|^{2}=(|x|+|y|)^{2}\)

    \(|z-x|=|(z-y)+(y-x)|\leq |z-y|+|y-x|\)

    image

  4. \(\lang x,y\rang =\frac{1}{4}(|x+y|^2-|x-y|^2)\) (The polarization identity)

Measuring angles

image

\(\alpha=\arccos\dfrac{\lang x,y\rang}{|x||y|}\) where \(\left|\dfrac{\lang x,y\rang}{|x||y|}\right|\leq 1\) since \(\lang x,y\rang =\cos\alpha\cdot |x||y|\)

Cos Theorem: \(|x-y|^2=\lang x-y,x-y\rang =\lang x,x\rang+\lang y,y\rang -2\lang x,y\rang=|x|^2+|y|^2-2\cos\alpha \cdot|x||y|\)

Euclidean spaces

Definition

A ES is a finite-dimensional space with a fixed inner product (\(E\) is our notation)

Example

Main example: \(E=\R^n+\text{dot-products}\) (\(\R^n\) with the dot-product of n-tuples)

Other cases: \(E\) is a \(ES\) where \(e_1,...,e_n\in E\) and \(\{e_1,...,e_n\}\) is an orthonormal basis: \(\lang e_i,e_j\rang =\begin{cases} 1&\text{if }i=j\\ 0&\text{if }i\neq j \end{cases}\)

\((\lang e_{i},e_{j}\rang)\Big|^{n}_{i,j=1}= \begin{pmatrix} 1 & & 0 \\ & \ddots & \\ 0 & & 1 \end{pmatrix}=I\)

Let \(x,y\in E:x=x_1e_1+...+x_ne_n,\,y=y_1e_1+...+y_ne_n\), then \(\lang x,y\rang =x_1y_1+...+x_ny_n\) is the dot-product of \((x_1,...,x_n)\) and \((y_1,...,y_n)\)

Topology in \(\R^n\)

Topological spaces

Let \(X\) be a set, then a fixed collection of subsets of \(X\) which is called open

Axiom

  1. \(\empty,X\) are open

  2. The union of any set of open subsets is open

  3. The intersection of a finite number of open subsets is open

Definition

A set with a fixed collection of open subsets satisfying the axioms is called a topological space where the collection is called a topology

Let \(W\subset X\) is closed\(\iff\)​\(X\setminus W\) is open where \(X\) is a topological space

Theorem

  1. \(\empty,X\) are closed

  2. The union of any set of closed subsets is closed

  3. The intersection of a finite number of closed subsets is closed

Proof

Let \(x,y\subset X\), then \(X\setminus (x\cap y)=(X\setminus x)\cup (X\setminus y)\) and \(X\setminus (x\cup y)=(X\setminus x)\cap (X\setminus y)\)

Open subsets in \(\mathbb{R}^n\)

Definition: \(U_{x}^{\varepsilon} \subset \mathbb{R}^{n}\) is the \(\varepsilon\)-neighborhood of \(x\) \((\varepsilon \in \mathbb{R}, \varepsilon>0,x \in \mathbb{R}^{n})\) if \(U_{x}^{\varepsilon}= \{y \in \mathbb{R}^{n}\mid |y-x| < \varepsilon \}\)

image

Definition: \(U\) is open\(\iff\)\(\forall x \in U,\exists U_{x}^{\varepsilon}\subset U\)

image

\(\mathbb{R}\): \((a,b)\) is open and \([a,b]\) is not open since image. Also \((c, +\infty)\) is open

The union of open subsets is open and a point is not open

The intersection of a infinite number of open subsets is not necessarily open

Example: \(\bigcap_{n=1}^{\infty} (a - \frac{1}{n}, a + \frac{1}{n}) = \{a\}\)

The intersection of a finite number of open subsets is open

Proof

Let \(U_{1}, \dots, U_{k}\) are open and \(x \in U_1 \cap \dots \cap U_k\), then \(\exists U_{x}^{\epsilon_i}\subset U_{i},\forall i=1,...,k\)

Let \(\epsilon = \min \{\epsilon_{1}, ..., \epsilon_{k}\}\), then \(\forall i=1,...,k \quad U_{x}^{\epsilon}\subset U_{x}^{\epsilon_i}\subset U_i\implies U_{x}^{\epsilon}\subset U_1\cap ... \cap U_k\)

Self Problem

In lecture we know a general definition

Let \(X\) be a set, then a fixed collection of subsets of \(X\) which is called open

Then if \(X=\R\) and let \(\R\) be a topology space, then I can define a collection which is \(\{\empty,(-\infty,1],(1,+\infty),\R\}\). This is a valid collection and those 4 subsets of \(\R\) are topology

By that general definition, we know \((-\infty,1]\) is open.

But in \(\R\), there is another definition says: \(U\) is open\(\iff\)\(\forall x \in U,\exists U_{x}^{\varepsilon}\subset U\)

Then by this definition, \((-\infty,1]\) is not open.

What happen?? why \((-\infty,1]\) is open and not open at same time?

Answer: For same topology space, there are many topologies.

In your custom topology, it is open. But in other topology(standard, Euclidean topology in \(\R\)), it is open by definition.

Thus, a subset can be open or closed according to the different topology.