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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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    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.2012-09-25

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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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    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.2012-09-25