Intuition of a norm vector space/ infinite dimensional vector space

Question

I'm finding it terribly difficult to build an intuition of what a norm vector space and an infinite dimensional vector space is. There aren't any good notes online that builds the intuition-most notes are overwhelmingly detailed which doesn't quite serve any purpose for me at this point.

Would someone be helpful enough to give me an intuitive description of what those concepts are? Even better if someone could connect the above to orthonormal basis.

Answer 1

I think an easy example of an infinite dimensional vector space would be to consider a sequence space. Concretely consider the space $\mathcal{l}^2(\mathbb{R})$, which consists of all infinite sequences

$$ \mathbf{x}=(x_1, x_2, \ldots )$$

with elements $x_i \in \mathbb{R}$.

You could view this as vectors, just that you have infinitely many coordinates.

Endow this space with a norm that is derived from the usual Pythagorean theorem:

$$\|\mathbf{x}\|^2 := \sum_i x_i^2.$$

Then this gives you an infinite dimensional vector space with which you can work with almost in the same way as with the usual finite dimensional spaces.

In fact it also has an inner product given by

$$<\mathbf{x},\mathbf{y}> := \sum_i x_i y_i$$

and has an orthonormal basis with respect to this inner product.

Answer 2

One standard example the vector space of all real valued continuos functions defined on $[0,1]$, let's call it $V$. Convince yourself that $V$ indeed is a vector space if you define the sum $f+g$ for $f$ and $g$ in $V$ to be the function $x\mapsto f(x)+g(x)$ and $cf$ is defined by $x\mapsto cx$ for any real $c$.

One may define an inner product of $V$ by $$\langle f,g\rangle:=\int_0^1f(x)g(x)\,dx.$$ Verify that $\langle.,.\rangle$ is indeed an inner product. Then, e.g., all straight lines with zero $2/3$ are orthogonal to the identity function.

Define as usual $\|f\|:=\sqrt{\langle f,f\rangle}$. Then $B^{\|.\|}(0,1)$,the ball with radius $1$ centered at $0$ generated by $\|.\|$, is the set of all functions satisfying $\|f\|^2=\int_0^1 f^2(x)\,dx\leq1$.

But you may equip $V$ with a different norm $\|.\|_1$ defined by $$\|f\|_1:=\max_{[0,1]}f(x).$$ Imagine what the ball $B^{\|.\|_1}(0,1)$ is now. Furthermore, convince yourself that the set $B^{\|.\|}(0,1)$ is unbounded in respect to $\|.\|_1$.