Functions forming a vector space

Homework Statement

1.1.3
1) Do functions that vanish at the endpoints x=0 and L=0 form a vector space?
2) How about periodic functions? obeying f(0)=f(L) ?
3) How about functions that obey f(0)=4 ?

If the functions do not qualify, list what go wrong.

The Attempt at a Solution

1) Considering a set of functions which vanish at the end points x = {0,L}. Let's say f,g,h belong to this set of functions. Then all the properties in the above image are verified. So, this set of functions form a vector space.
2) Similarly, for 2 also all properties get verified except existence of a null vector. Can a function h(x) = 0 for all x, be a periodic function?
3) For 3, this set of functions do not have closure feature. So, it does not form a vector space.

Is this correct?

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PeroK
Homework Helper
Gold Member
2020 Award
Technically, the zero function is periodic for any given period. In particular, it has the same value at the end points.

Pushoam
Ray Vickson
Homework Helper
Dearly Missed

Homework Statement

1.1.3
1) Do functions that vanish at the endpoints x=0 and L=0 form a vector space?
2) How about periodic functions? obeying f(0)=f(L) ?
3) How about functions that obey f(0)=4 ?

If the functions do not qualify, list what go wrong.

Homework Equations

View attachment 229213

The Attempt at a Solution

1) Considering a set of functions which vanish at the end points x = {0,L}. Let's say f,g,h belong to this set of functions. Then all the properties in the above image are verified. So, this set of functions form a vector space.
2) Similarly, for 2 also all properties get verified except existence of a null vector. Can a function h(x) = 0 for all x, be a periodic function?
3) For 3, this set of functions do not have closure feature. So, it does not form a vector space.

Is this correct?

You need to have confidence in your own reasoning. (i) Have you verified all the defining properties of vector spaces in (1)-(2)? (ii) Have you shown that some vector-space property fails in (3)?

Pushoam
Mark44
Mentor

Homework Statement

1.1.3
1) Do functions that vanish at the endpoints x=0 and L=0 form a vector space?
2) How about periodic functions? obeying f(0)=f(L) ?
3) How about functions that obey f(0)=4 ?

If the functions do not qualify, list what go wrong.

Homework Equations

View attachment 229213

The Attempt at a Solution

1) Considering a set of functions which vanish at the end points x = {0,L}. Let's say f,g,h belong to this set of functions. Then all the properties in the above image are verified. So, this set of functions form a vector space.
2) Similarly, for 2 also all properties get verified except existence of a null vector. Can a function h(x) = 0 for all x, be a periodic function?
What's the definition of a periodic function?
Pushoam said:
3) For 3, this set of functions do not have closure feature. So, it does not form a vector space.

Is this correct?

Pushoam
fresh_42
Mentor
1) Considering a set of functions which vanish at the end points x = {0,L}. Let's say f,g,h belong to this set of functions. Then all the properties in the above image are verified. So, this set of functions form a vector space.
The notion of the corresponding equation would have been almost shorter: ##(\alpha f + \beta g)(0)=\alpha f(0)+\beta g(0)=\alpha f(L)+ \beta g(L)=(\alpha f + \beta g)(L)##
2) Similarly, for 2 also all properties get verified except existence of a null vector.
Why this? ##0(0)=0=0(L)##
Can a function h(x) = 0 for all x, be a periodic function?
This is the extreme version of a periodic function, namely with period ##0##. There is no reason to exclude it.
3) For 3, this set of functions do not have closure feature. So, it does not form a vector space.
Yes, and again ##0(0)=0\neq 4##

Is this correct?
Yes, except that I couldn't see your reasoning why the null vector in section 2 shouldn't work.

Edit: Please do the following little exercise: A set which is closed under addition and scalar multiplication automatically contains zero.

Last edited:
Pushoam
You need to have confidence in your own reasoning. (i) Have you verified all the defining properties of vector spaces in (1)-(2)? (ii) Have you shown that some vector-space property fails in (3)?
Yes, I verified it in mind. I didn't write the verification here as it was easy.

Mark44
Mentor
This is the extreme version of a periodic function, namely with period 0.
Or any period.
If f(x) = 0 for all x, then f(x) = f(x + L), for any L.

The vector space properties in the attached image in post 1 is a bit unusual. The usual definition includes statements about closure under vector addition and closure under scalar multiplication.

Minor point: The usual definitions are of the properties these operations. I've never seen them called features.

Pushoam
vela
Staff Emeritus
Homework Helper
This is the extreme version of a periodic function, namely with period ##0##. There is no reason to exclude it.
Typically the period is required to be positive to exclude possibilities like claiming f(x)=x is periodic since f(x)=f(x+0) for all x.

Pushoam
fresh_42
Mentor
Typically the period is required to be positive to exclude possibilities like claiming f(x)=x is periodic since f(x)=f(x+0) for all x.
Makes sense. I thought the zero might be necessary for the set of all periods, or to get vector spaces for periodic functions, but "any" sounds better than "0" although "any" includes it.

Pushoam
The notion of the corresponding equation would have been almost shorter: ##(\alpha f + \beta g)(0)=\alpha f(0)+\beta g(0)=\alpha f(L)+ \beta g(L)=(\alpha f + \beta g)(L)##

Edit: Please do the following little exercise: A set which is closed under addition and scalar multiplication automatically contains zero.
Let's say that ## \alpha ## and ## \beta ## belongs to this set and a, b belong to the scalar field. Then, the set being closed under addition and scalar multiplication means that ## a \alpha + b \beta ## must belong to the set.
Taking ## b \beta = - a \alpha ## implies that 0 belongs to the set.

Thus, a set which is closed under addition and scalar multiplication automatically contains zero.

fresh_42