Relation between metric spaces, normed vector spaces, and inner product space.

Question

I am wondering what exactly is the relationship between the three aforementioned spaced. All of them seem to show up many times in: Linear Algebra, Topology, and Analysis. However, I feel like I'm missing the bigger picture of how these spaces relate to each other. For example, in my course in multi-dimensional analysis, we started out talking about metric spaces, but later suddenly switched to normed vector spaces, without any explicit mention of this transition. In linear algebra we usually talked about inner product spaces, and in topology we talked about metric spaces and topological spaces.

The bigger picture of the relation between these three is still unclear to me. Which is used where, for what reason, and how do they relate?

I do know the definitions of all three of them:

A metric space is a pair $(S,d)$ with $S$ a set and $d: S \times S \to \mathbb{R}_{\geq 0}$ a metric:

• $d(x,x) = 0$ for all $x \in S$ and $d(x,y) >0$ for $x \neq y$,
• $d(x,y) = d(y,x)$,
• $d(x,z) \leq d(x,y) + d(y,z)$.

A (real) inner product space is a pair $(V,\langle \cdot \rangle)$ where $V$ is a (real) vector space and $\langle \cdot \rangle: V \times V \to \mathbb{R}$ is an inner product:

• $\langle v,w \rangle = \langle w,v \rangle$,
• $\langle a_1 v_1 + a_2v_2,w \rangle = a_1\langle v_1,w \rangle + a_2\langle v_2,w \rangle$ for all $a_1,a_2 \in \mathbb{R}$,
• $v \neq 0 \Longrightarrow \langle v,v \rangle > 0$.

A (real) normed vector space is a pair $(V,\|\cdot\|)$ where $V$ is a (real) vector space and $\|\cdot\|: V \to \mathbb{R}_{\geq 0}: v \mapsto \|v\|$ is a norm on $V$:

• $\|v\| \geq 0$ and $\|v\| = 0 \ \Longleftrightarrow \ v = 0$.
• For $t \in \mathbb{R}$ and $v \in V$ we have $\|tv\| = |t|\|v\|$
• $\|v+w\| \leq \|v\| + \|w\|$.

I also know that an inner product gives rise to a norm by taking $\|v\| = \sqrt{\langle v,v \rangle}$, for example the Euclidean norm derives from the standard inner product on $\mathbb{R}^n$ in this way. And Cauchy-Schwarz: $|\langle x,y \rangle| \leq \|x\|\|y\|$.

I'm not interested in details about the definitions but in the intuition and bigger picture of these three spaces, and how they show up in Analysis.

Answer 1

You have the following inclusions:

$$\{ \textrm{inner product vector spaces} \} \subsetneq \{ \textrm{normed vector spaces} \} \subsetneq \{ \textrm{metric spaces} \} \subsetneq \{ \textrm{topological spaces} \}.$$

Going from the left to the right in the above chain of inclusions, each "category of spaces" carries less structure. In inner product spaces, you can use the inner product to talk about both the length and the angle of vectors (because the inner product induces a norm). In a normed vector space, you can only talk about the length of vectors and use it to define a special metric on your space which will measure the distance between two vectors. In a metric space, the elements of the space don't even have to be vectors (and even if they are, the metric itself doesn't have to come from a norm) but you can still talk about the distance between two points in the space, open balls, etc. In a topological space, you can't talk about the distance between two points but you can talk about open neighborhoods.

Because of this inclusion, everything that works for general topological spaces will work in particular for all other spaces, but there are some things you can do in (say) normed vector spaces which don't make sense in a general topological space. For example, if you have a function $f \colon V \rightarrow \mathbb{R}$ on a normed vector space, you can define the directional derivative of $f$ at $p \in V$ in the direction $v \in V$ by the limit

$$ \lim_{t \to 0} \frac{f(p + tv) - f(p)}{t}. $$

In the definition, you are using the fact that you can add the vector $tv$ to the point $p$. If you try to mimick this definition in a topological space, then since the set itself doesn't have the structure of a vector space, you can't add two elements so this definition doesn't make sense. That's why during your studies you sometimes restrict your attention to a smaller category of spaces which has more structure so you can do more things in it.

You can discuss the notions of continuity, compactness only in the category (context) of topological spaces (but for reasons of simplicity it is often done in the beginning of one's studies in the category of metric spaces). However, once you want to discuss differentiability, then (in first approximation, before moving to manifolds) you need to restrict your category and work with normed vector spaces. If you also want to discuss the angle that two curves make, you will need to further restrict your category and work with inner product vector spaces in which the notion of angle makes sense, etc.

Answer 2

1. Every inner product space is (can be naturally made) a normed space by defining $$\|x\|:=\sqrt{\langle x, x\rangle} $$ (following the leading example $\Bbb R^n$)

2. Every normed space is, by definition, a linear space, and at the same time can be naturally equipped with a metric: $$d(x, y) :=\|y-x\|$$

You can check that the respective axioms are indeed satisfied.

Metric spaces provide a general framework for continuity and uniform continuity.
We can define differentiation on normed spaces.

Noticing that the class of (nice-in-a-way) real or complex valued functions themselves form a linear space, we can investigate several norms for them, even inner products, which is the study of functional analysis.

Answer 3

An inner product space $\left(V,\langle\cdot,\cdot\rangle\right) $ is also a normed vector space $\left(V,\lVert\cdot\rVert\right) $, taking $$ \lVert v\rVert \colon= \sqrt{\langle v,v\rangle} \qquad\forall v\in V$$

A normed vector space $\left(V,\lVert\cdot\rVert\right) $ is also a metric vector space $(V,d) $, taking $$ d(x,y)\colon=\lVert x-y\rVert \qquad\forall x,y\in V$$

In short $$ \text{Inner product}\implies \text{Norm}\implies \text{Metric} $$