Inner products¶

Inner product

Let \(V\) be a vector space over \(F\), where \(F\) is either \(\mathbb{R}\) or \(\mathbb{C}\). An inner product on \(V\) is a map \(\langle\cdot,\cdot\rangle:V\times V\to V\) that satisfies

\[ \forall y\in V,\langle\cdot,y\rangle\in V^*, \]

\[ \langle x,y\rangle=\overline{\langle y,x\rangle}, \]

\[ \langle x,x\rangle\ge0,\langle x,x\rangle=0\iff x=0. \]

\(V\) together with an inner product is called an inner product space.

Example

Let \(V=\mathsf{C}\left(\left[0,1\right],\mathbb{C}\right)\) be the vector space of all continuous functions defined on \([0,1]\) with codomain \(\mathbb{C}\). Let

\[ \langle f(x),g(x)\rangle:=\int_{0}^{1}f(x)\overline{ g(x) }dx, \]

then it is an inner product on \(V\).

Norm¶

Norm

Let \(V\) be a vector space over \(F\), where \(F\) is either \(\mathbb{R}\) or \(\mathbb{C}\). A norm on \(V\) is a map \(\|\cdot\|:V\to\mathbb{R}_{\ge0}\) that satisfies

\[ \forall x\in V,\|x\|\ge0,\text{ and }\|x\|=0\iff x=0, \]

\[ \forall x,y\in V,\|x\|+\|y\|\ge\|x+y\|, \]

\[ \forall\alpha\in F, x\in V, \|\alpha x\|=|\alpha|\|x\|, \]

where \(|\alpha|\) is the absolute value on \(F\).

\(V\) together with a norm is called a normed vector space.

Metric¶

Metric

Let \(V\) be a vector space over \(F\), where \(F\) is either \(\mathbb{R}\) or \(\mathbb{C}\). A metric on \(V\) is a map \(d:V\times V\to\mathbb{R}\) that satisfies

\[ \forall x\in V d(x,y)\ge0,\,\text{and }d(x,y)=0\iff x=y \]

\[ \forall x,y\in V,d(x,y)=d(y,x), \]

\[ \forall x,y\in V,d(x,z)+d(z,y)\ge d(x,y). \]

Inner product induces norm

Let \(V\) be an inner product space over \(F\). Then the map \(\|\cdot\|:V\to\mathbb{R}\) defined as

\[ \|x\|:=\sqrt{\langle x,x\rangle} \]

is a norm on \(V\).

Norm induces metric

Let \(V\) be a normed vector space over \(F\). Then the map \(d:V\times V\to\mathbb{R}\) defined as

\[ d(x,y):=\|x-y\| \]

is a metric on \(V\).

Note

The converse fails for both of the theorems above.

For example, let \(V=\mathbb{R}^2\), \(F=\mathbb{R}\) and define \(\|(a,b)\|:=|a|+|b|\). Assume that there is an inner product such that

\[ \left\langle (a,b),(a,b)\right\rangle=\left(|a|+|b|\right)^2 \]

Then

\[ \begin{align*} 4=\left\langle(1,-1),(1,-1)\right\rangle&=\left\langle(1,0),(1,1)\right\rangle+\langle(0,-1),(1,1)\rangle\\ &=\langle(1,0),(1,0)\rangle+\langle(1,0),(0,1)\rangle+\langle(0,-1),(1,0)\rangle+\langle(0,-1),(0,1)\rangle\\ &=1+\langle(1,0),(0,1)\rangle-\langle(0,1),(1,0)\rangle-1\\ &=1+\langle(1,0),(0,1)\rangle-\langle(1,0),(0,1)\rangle-1\\ &=0\rightarrow\leftarrow \end{align*} \]

The discrete metric does not induce a norm on \(\mathbb{R}^2\). Since if \(x\neq0\), then \(d(x,0)=\|x\|=1\), and this causes \(1=d(2x,0)=\|2x\|=2\|x\|=2\).

Cauchy-Schwarz inequality¶

Cauchy-Schwarz inequality

Let \(V\) be an inner product space.

The inequality

\[ \|x\|\|y\|\ge|\langle x,y\rangle| \]

holds for all \(x,y\in V\)

Remarks¶

The other side of the triangle inequality

Let \(V\) be a normed vector space, then

\[ \|x-y\|\ge\left|\|x\|-\|y\|\right| \]

holds for all \(x,y\in V\).

The parallelogram law

Let \(V\) be an inner product space. Then the norm induced by the inner product satisfies

\[ 2(\|x\|^2+\|y\|^2)=\|x+y\|^2+\|x-y\|^2 \]

holds for all \(x,y\in V\).

Norm with parallelogram law induces inner product

Let \(V\) be a normed vector space. The norm satisfies the parallelogram law if and only if the norm is induced by some inner product on \(V\).