Question About Linear Subspaces of Functions

[bZNyQ7C2]

Is the set of functions $\displaystyle f\left( x \right)$ such that $\displaystyle f\left( 2 \right) = f\left( 5 \right)$ a linear subspace?

This question appears on lem.ma, a webpage that supports Pavel Grinfeld's online courses. Lem.ma says the answer is yes.

I do see that this is true for all polynomial functions which include the factors $\displaystyle \left( {x - 2} \right)\left( {x - 5} \right)$ no matter what the degree of the polynomial, even if it's 0 (a constant function).

But how do I know there aren't other functions, piece-wise functions perhaps, that have this property? If there are, how can I prove that they still belong to this linear subspace? I'm having trouble visualizing this linear subspace. Should I just assume that polynomial functions are meant?

 

[Romsek]

go back and look at the definition of a linear subspace

you should assume nothing about the functions other than f(2)=f(5)

 

[HallsofIvy]

1. Does this set include the "0" function, f(x)= 0 for all x? (What are f(2) and f(5)?)

2. If f(x) and g(x) are in this set (f(2)= f(5) and g(2)= g(5)) is f+ g also in this set? (Is [f+ g](2)= [f+g](5)?)

3. If f(x) is in this set (f(2)= f(5)) and a is a number, if af in this set? (is af(2)= af(5)?)

 

[bZNyQ7C2]

@HallsofIvy ... thanks for your questions, always helpful.

A few quick comments.

Dr. Grinfeld never mentions the 0 function. And he's not the only one. I've looked at several courses in LA both in print and video formats, and they seem to gloss over that, though it's simple once you think of it. I guess the people who teach LA will assume their students haven't taken Abstract Algebra yet, so they can gloss over that.

It helps me to write f(2) = f(5) as f(2) - f(5) =0. This makes the "yes" answers to your questions 2 and 3 easier to see.

 

[HallsofIvy]

In the definition of "Vector Space" or "Subspace of a Vector Space" some people include the requirement
"must include a (the) 0 vector" while others just say "must be non-empty". Those are equivalent. If a set includes the 0 vector then it is obviously non-empty. On the other hand, if it is non-empty then it includes some vector, v. Since another "axiom" says the set is "closed under scalar multiplication" the set must include -1 times v, -v. And then since still another "axiom" says that the set is "closed under vector addition" it must include v+ (-v)= 0.