Determine if the function is a linear space?

[Cassi]

Homework Statement

For the real-valued function All f with 2f(0)=f(1) with domain [0, 1], determine whether the given set is a real linear space, if addition and multiplication by real scalars are defined in the usual way. For those that are not, tell which axioms fail to hold.

Homework Equations

Supposed to use the Closure axioms, axioms for additions, and axioms for multiplication for a proof.

The Attempt at a Solution

To me these axioms are trivial and it is obvious that the function is a real linear space; however, the question wants me to prove them. Here are my attempts:

In the following situations, let f, g, h be functions that satisfy the condition 2f(0) = f(1).
Axiom 1 (Closure under addition):Then 2f(0) + 2g(0) = f(1) + g(1).
Axiom 2 (Closure under multiplication): Let a be a real number. Therefore af(0) = af(1).
Axiom 3 (Commutative law): 2f(0) + 2g(0) = f(1) + g(1)
Axiom 4 (Assosciative law): 2f(0) +) 2g(0) + 2h(0)) = (f(1) + g(1)) +h(1)
My other arguments follow in this way. Is this enough of a proof for each statement? Is there a better way to work through this?

 

[mfb]

I would introduce "let k=f+g." and prove "2k(0)=2f(0) + 2g(0) = f(1) + g(1)=k(1)" for the first axiom, and similar for the others.
Yes, all those steps are very basic.

 

[gopher_p]

Homework Statement

For the real-valued function All f with 2f(0)=f(1) with domain [0, 1], determine whether the given set is a real linear space, if addition and multiplication by real scalars are defined in the usual way. For those that are not, tell which axioms fail to hold.

Homework Equations

Supposed to use the Closure axioms, axioms for additions, and axioms for multiplication for a proof.

The Attempt at a Solution

To me these axioms are trivial and it is obvious that the function is a real linear space; however, the question wants me to prove them. Here are my attempts:

In the following situations, let f, g, h be functions that satisfy the condition 2f(0) = f(1).
Axiom 1 (Closure under addition):Then 2f(0) + 2g(0) = f(1) + g(1).
Axiom 2 (Closure under multiplication): Let a be a real number. Therefore af(0) = af(1).
Axiom 3 (Commutative law): 2f(0) + 2g(0) = f(1) + g(1)
Axiom 4 (Assosciative law): 2f(0) +) 2g(0) + 2h(0)) = (f(1) + g(1)) +h(1)
My other arguments follow in this way. Is this enough of a proof for each statement? Is there a better way to work through this?

Ok. So you're trying to show that the set of real-valued functions $V=\{f: f\text{ is real-valued with domain }[0,1]\text{ and }2f(0)=f(1)\}$ is a linear/vector space over $\mathbb{R}$. And in order to do that, you're meant to use the axioms for vector spaces. From the work that you've shown, your list of axioms is a bit more extensive than the usual list.

Your "proof" that your Axiom 1 holds is almost complete. You want to show that $f+g$ is in $V$ whenever both $f$ and $g$ are. You know from precalc how $f+g$ is defined and that given $f,g\in V$, $f+g$ has the right domain and range for $V$. You just need to show that that $2(f+g)(0)=(f+g)(1)$, which you basically did. You just want to be a little more explicit about it. Your assertion that this is "trivial" is mostly correct -straightforward is probably a better way to put it - but that doesn't mean there isn't something to say here. You want to recognize exactly what is going on here and show your teacher that you, in fact, really do know what is going here.

It looks like you missed the boat (or made a typo) on Axiom 2. You want to show that $af\in V$ whenever $f\in V$ and $a\in\mathbb{R}$. Again from precalc you can say some things about the function $af$, and you might want to briefly mention those things in order to show that you know that it's relevant. The key part here is showing that $2(af)(0)=(af)(1)$. Again, this is straightforward if you know what's going on.

Axioms 3 and 4 look to be way off the mark. You're meant to show that $f+g=g+f$ and $(f+g)+h=f+(g+h)$ in $V$. That has nothing to do with the $2f(0)=f(1)$ part of the definition of $V$ and everything to do with how $+$ is defined for real-valued functions. The proof here is precalc stuff that you may have brushed off as "trivial" back in the day. The same thing will be true for your proofs of the "multiplication axioms"; all about what $af$ means and not at all about the $2f(0)=f(1)$ part of this specific example. To put it another way, your proofs here for the addition and multiplication axioms should be no different than your proofs of those axioms would be in the "show that the set of real-valued functions with shared domain $D$ form a real-linear space under the usual 'precalc' definitions of addition and scalar multiplication" version of the problem.

 

[Fredrik]

There's a slightly different way to do this. First show that the set of all functions from [0,1] into $\mathbb R$ is a vector space over $\mathbb R$. Then show that your set is a subspace of that vector space. But I recommend that you finish your current approach first, and then look at this alternative approach.

axioms for multiplication

The operation you're talking about is called "scalar multiplication", not "multiplication". Their definitions are similar, but if we denote the vector space of functions from [0,1] to $\mathbb R$ by V, then scalar multiplication is a map from $\mathbb R\times V$ into $V$ and multiplication is a map from $V\times V$ into $V$.

Scalar multiplication: For each $f\in V$ and each $a\in\mathbb R$, define $af:[0,1]\to\mathbb R$ by $(af)(x)=a(f(x))$ for all $x\in[0,1]$.

Multiplication: For each $f,g\in V$, define $fg:[0,1]\to\mathbb R$ by $(fg)(x)=f(x)g(x)$ for all $x\in[0,1]$.