# Functions forming a vector space (1 Viewer)

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#### Pushoam

1. The problem statement, all variables and given/known data
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.

2. Relevant equations 3. 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
2018 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
1. The problem statement, all variables and given/known data
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.

2. Relevant equations
View attachment 229213

3. 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
1. The problem statement, all variables and given/known data
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.

2. Relevant equations
View attachment 229213

3. 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
2018 Award
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

#### 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
2018 Award
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

#### 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

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