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Thread: function vector space

  1. #1
    Senior Member Sampras's Avatar
    May 2009

    function vector space

    Do functions that vanish at the endpoints of  [0,L] form a vector space? How about periodic functions such that  f(0) = f(L) ?

    Let  f(x) = \sin x and  g(x) = \tan x . Then  h(x) = f(x)+g(x) seems to be an element of this "space" if  L = 0+ 2 n \pi . E.g. closure of addition. The same goes for the periodic functions.

    The null function seems to be  f(0) = 0 . The inverse function is  f(x) - g(x) .

    Is this correct?
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  2. #2
    Jul 2008
    To demonstrate a vector space, we have to show that the vectors (in this case, the functions) with addition form an abelian group and that the scalars (in this case, real numbers) distribute over sums of vectors and vice versa.

    Let L,a\in\mathbb{R} be given and define S_{L,a}=\{f:\mathbb{R}\rightarrow\mathbb{R}: f(0)=f(L)=a\}. Addition of functions f,g\in S_{L,a} is defined as (f+g)(x)=f(x)+g(x) for all x\in\mathbb{R}, and scaling of a function f by c\in\mathbb{R} is defined as (cf)(x)=c(f(x)) for all x\in\mathbb{R}. We want to decide if S_{L,a} is a vector space.

    It is fairly straightforward to check that S_{L,0} is an abelian group (in fact a subgroup of real-valued functions on the real line under addition) since (f+g)(0)=f(0)+g(0)=0 and similarly for L. What does this tell us about S_{L,a} for nonzero a?

    Similarly, the distributive laws fall out of the field structure of \mathbb{R} but we must be careful that scaled functions are still vanishing at the endpoints. This is, of course, easy to show by a similar argument to the addition closure.
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