# Prove that $V$ is a subspace of $F$

How do I prove that $V$ is a subspace of $F$ and whether $f_1 \in \operatorname{span}\{f_1,f_2,f_3\}$ ?

We let F denote the vector space of all functions mapping real numbers to real numbers. We define $T\colon F\to \mathbb R$ by $T(f) = \int_0^1 f(x)dx$ for $f \in F$. We let $f_1(x) = 2x -1$, $f_2(x) = x^2 -1$, and $f_3(x) = e^x + 1$. We let $V = \{f \in F : T(f)= 0\}$.

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With your definition of $F$, the operator $T$ isn't well-defined. Not all functions are integrable. – David Mitra Sep 24 '12 at 20:54
@DavidMitra I got this question from my practice test and I wrote it exactly as it was stated. – diimension Sep 24 '12 at 21:02
@diimension , it never mind where did you get this question from: the definition of T still makes no sense as lots of elements of V aren't integrable. – DonAntonio Sep 24 '12 at 22:19
@DonAntonio So the question is misinterpreted ? – diimension Sep 24 '12 at 22:31
The question is sloppy. It would be better if it began, We let $F$ denote the vector space of all integrable functions mapping real numbers to real numbers. – Gerry Myerson Sep 25 '12 at 0:01

## 1 Answer

You can prove $V$ is a subspace of $F$ the same way you prove any subspace is a subspace: show it is closed under addition and under scaling. Explicitly, if $f,g$ are in $V$, show $f+g$ is in $V$, and $\alpha f$ is in $V$. If you don't know the defintions of addition or scaling of functions, you can ask, but I'm betting you've seen it.

The second half of the question is garbled. You wrote "f(1)" which as written is in $\mathbb{R}$, so it's not in $F$ at all. If you meant $"f1"$ instead, then the statement is trivial, because $x$ is surely in the span of $\{x,y,z,\dots\}$... can you see how to arrange coefficients on the three functions so that the result is f1?

Note The OP improved the typesetting by a lot since this was posted, but fortunately the answers still apply.

I'll add that as long as you show $T$ is linear, it isn't important how $T$ is defined: $\{x\in F\mid Tx=0\}$ is always going to be a subspace (because it's the kernel of a linear operator.)

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I edited my post and I am having trouble making the distinction of what is F and V in this problem? I know the axioms of a subspace but what is V and F particularly in this problem? – diimension Sep 24 '12 at 20:48
@rschwieb, I don't see any interpretation under which the domain of $T$ is the reals, rather than some function space. – Gerry Myerson Sep 25 '12 at 0:03
@GerryMyerson oh right! yeah, I messed that up. Lemme fix that. Lot of issues with this question. – rschwieb Sep 25 '12 at 0:07
@diimension Apparently, $F$ is the space of integrable functions from $\mathbb{R}$ into $\mathbb{R}$, and $V$ is a subset of $F$. Your map $T:F\rightarrow \mathbb{R}$ is going to turn out to be a linear functional. Check that it's a linear map (that's elementary rules of integration!) and then use that to show that your set $V$ is a subspace, verifying the axioms we talked about. – rschwieb Sep 25 '12 at 0:27
Ah, and now that I noticed the limits of integration again, $F$ can be all functions which are just integrable on that interval $[0,1]$. And somehow I deleted my thank-you to @GerryMyerson for alerting me I went off the rails. – rschwieb Sep 25 '12 at 11:54