7
$\begingroup$

I have recently started a undergraduate linear algebra course in which these definitions came up:

Let $V$ be the vector space $C[a, b]$ of all continuous functions on $[a, b]$. Then the inner product and norm are defined as:

\begin{align} \langle f, g \rangle &= \int_a^b f(t) g(t) \,\mathrm{d}t \\ \| f \| &= \sqrt{\langle f, f \rangle} = \sqrt{\int_a^b f^2(t) \,\mathrm{d}t} \end{align}

Concerns:

What does it mean if $\int_a^b f^2(t) \,\mathrm{d}t < 0$?

It is also, strangely, possible to calculate the angle between functions (non-linear), is this considered the average angle in $[a, b]$ or what is it’s geometrical representation?

$\endgroup$
4
  • 1
    $\begingroup$ $\int_{a}^{b}f^2(t)dt$ is always positive for $a<b$. $\endgroup$
    – Learner
    Apr 12, 2013 at 18:43
  • $\begingroup$ I did not know that. But is not $C[1,0]$ the same vector space? Or is it a necessity that a<b when vector spaces are defined? $\endgroup$
    – EricAm
    Apr 12, 2013 at 18:49
  • 1
    $\begingroup$ The standard interval notation $[a,b]$ means $a<b$. $\endgroup$
    – Learner
    Apr 12, 2013 at 18:50
  • $\begingroup$ @AdamYac : there are many inner products on $V$. Whoever wrote that definition was sloppy. $\endgroup$ Apr 12, 2013 at 22:46

1 Answer 1

6
$\begingroup$

It is important to note that the (inner-product) space you are working with is not the product of $[a,b]$ and the image of functions $f$. It is the infinite dimensional space of continuous functions defined on $[a,b]$. You should picture each $f$ in that space as an infinite vector.

$\endgroup$
3
  • $\begingroup$ Alright, I did not have that picture in mind but I do see your point. A question I should not ask but since we can deal with all kinds of C^0-functions; I know a vector have magnitude and direction but is it necessary a straight line? $\endgroup$
    – EricAm
    Apr 12, 2013 at 18:56
  • $\begingroup$ Or should I imagine all kinds of elementary functions as infinite straight vectors? In that case the angle makes more sense.. $\endgroup$
    – EricAm
    Apr 12, 2013 at 18:59
  • 1
    $\begingroup$ Yes, linearity here comes the view that you treat each function as an infinite vector. The area of mathematics that studies this (among other things) is "functional analysis". $\endgroup$
    – Learner
    Apr 12, 2013 at 19:03

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .