Relations among Inner Product Spaces, Normed Spaces, and Metric Spaces
Description
Let’s say an inner space $\left( X, \langle\cdot, \cdot\rangle \right)$ is given. Then, one can naturally define a norm as follows from the inner product.
$$ \begin{equation} \left\| x \right\| := \sqrt{ \langle x, x\rangle},\quad x\in X \end{equation} $$
Hence, if it is an inner space, then it’s a normed space. Subsequently, one can define a distance from the norm thus defined.
$$ \begin{equation} d(x,y):=\left\| x -y \right\| =\sqrt{ \langle x-y, x-y \rangle},\quad x,y\in X \end{equation} $$
Therefore, if it is an inner space, then it is both a normed space and a metric space. Some textbooks mention metric spaces upfront and then use the concepts of norm or inner product, and that’s precisely because of this reason. Although it is mentioned as a metric space, it is assumed to be given an inner space.
Conversely, saying ‘an inner space $X$ is given’ is synonymous with saying ‘a metric space $X$ is given’, ‘a normed space $X$ is given’. Additionally, the concept of completeness is defined in metric spaces, but the reason one can say a normed space or an inner space is complete is because distance can be defined through inner product and norm. The proof is not difficult through the definitions, so I will only introduce about $(1)$.
Theorem
If it is an inner space, then it is a normed space.
Proof
Let’s assume an inner space $X$ is given. And let’s say $x,y\in X$ and $c\in \mathbb{C}$. Then, by the definition of inner product,
$$ \left\| x \right\| \ge 0 $$
holds. Also, by the definition of inner product,
$$ \left\| x \right\| =0 \iff x=0 $$
holds. Similarly, by the definition of inner product,
$$ \begin{align*} \left\| cx \right\| =&\ \sqrt{ \langle cx,cx\rangle } \\ =&\ \sqrt{ \left| c \right| ^{2} \langle x,x \rangle} \\ =&\ \left| c \right| \sqrt{\langle x,x \rangle} \\ =&\ \left| c \right| \left\| x \right\| \end{align*} $$
holds. The last condition also holds by the definition of the inner product:
$$ \begin{align*} \left\| x + y \right\|^{2} =&\ \langle x+y,x+y \rangle \\ =&\ \langle x,x+y\rangle +\langle y,x+y \rangle \\ =&\ \langle x,x\rangle + \langle x,y\rangle + \langle y,x\rangle + \langle y,y\rangle \\ =&\ \left\| x \right\|^{2}+\langle x,y \rangle +\overline{ \langle x,y \rangle}+ \left\| y \right\| ^{2} \\ \le& \left\| x \right\| ^{2} + 2 \left| \langle x,y \rangle \right| + \left\| y \right\| ^{2} \\ \le& \left\| x \right\|^{2} +2\langle x,x \rangle ^{1/2}\langle y,y \rangle^{1/2} + \left\| y \right\|^{2} \\ =&\ \left\| x \right\|^{2}+2\left\| x \right\|\left\| y \right\| +\left\| y \right\|^{2} \\ =&\ \left( \left\| x \right\| + \left\| y \right\| \right)^{2} \end{align*} $$
The fifth line holds because for any complex number $c\in \mathbb{C}$, $c+\overline{c} \in \mathbb{R}$ holds. The sixth line is satisfied by the Cauchy-Schwarz inequality. Hence,
$$ \left\| x \right\| := \sqrt{\langle x,x \rangle} $$
$\left\| \cdot \right\|$ defined as above satisfies the definition of a norm.
■