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I have a question about proving that something is a vector space. For example, I am trying to prove Axiom 6: $$ (A + B)(f(x)) = A(f(x)) + B(f(x)), $$ where $A$ and $B$ are constants.

Now, to me, this seems like common sense. You just distribute. Is there a more formal way to prove this? I find myself getting stuck a lot of times when trying to prove that something is a vector space because it seems too obvious.

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What's your definition of continuity? What if you define $g(x) = A f(x) + B f(x)$ and use the definition for $f$? Then what happens with $g$? – Pragabhava Oct 2 '12 at 1:21
g is still continuous. A continuous function multiplied by a constant is always continuous. And similarly, the sum of two continuous functions is continuous. – mh234 Oct 2 '12 at 1:23
Meaning $C[a,b]$ is closed under scalar multiplication. – Pragabhava Oct 2 '12 at 1:27
up vote 3 down vote accepted

I totally agree that "proving" the space as defined complies with this axiom of the definition of a vector space seems trivial. Common sense does tell us we can "just distribute", and that's the problem. We have to be wary of our common sense in proofs, because it is the exact thing that will let us leave out an important step, or even worse, get an invalid result.

I see this proof as being about notation. We are using the same notation, namely the parenthesis in something like $(x)y$ to mean two different operations: the multiplication of a real number times a continuous function on $[a,b]$, and the scalar multiplication operation in our candidate vector space. What we have to prove is that multiplying a continuous function by a real number fulfills the necessary attributes to be a valid scalar multiplication operation.

Specifically, this is how I write a "proof" (it's so simple I hesistate to call it that) for this axiom:$$(A+B)(f(x))=((A+B)f)(x)=(Af+Bf)(x)=(Af)(x)+(Bf)(x)=A(f(x))+B(f(x))$$ The expressions before the first equal sign and after the last equal sign both use parentheses to denote scalar multiplication in the vector space. The second expression uses the parentheses around $A+B$ to denote multiplying a function by a real number, which is where the meat of this proof lies.

It absolutely looks like an almost insultingly simple thing to prove, and those are the exact areas where I remind myself to be most careful about how I'm thinking.

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If $f$ and $g$ are real functions with a common domain $D$, we define $(f + g)(x) = f(x) + g(x)$ for $x\in D$ and $cf(x) = c\cdot f(x)$ for $x \in D$ and for a constant $c$. This gives a vector space structure. The continuous functions will be a subspace if $D$ is a topological space.

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