function vector space

**Sampras** · July 8th 2009, 01:49 PM

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?

**siclar** · July 8th 2009, 08:36 PM

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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