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I am having trouble understanding where to start with the following question:

Let $F$ be the set of all differentiable functions on $[a,b]$. Show $F$ is a vector space with the standard operations.

So I know that I need to prove the axioms, however I don't have a function to start with. How do I represent $F$ as a set of differentiable functions so I can prove the axioms?

Thanks.

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    $\begingroup$ your vectors are the functions, your scalars are elements of the field you're working on. So, take arbitrary vectors (functions) and scalars, and show the vector space properties hold $\endgroup$ – Alan Aug 5 '15 at 12:02
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Hint: Suppose $f$ and $g$ are differentiable functions, and $a$ is a Real number. What is $f+g$? What about $af+g$? You don't need to be given any particular functions, you can imagine any two functions in $F$ and show how $F$ is a vector space using two arbitrary functions.

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  • $\begingroup$ I hope everything comes out ok. $\endgroup$ – Todd Wilcox Aug 5 '15 at 12:41
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HINT: $(af(x)+bg(x))'=af'(x)+bg'(x)$ and your vectors are $f$ and $g$. You can omit verifying all axioms, because $F$ is a subspace of $C[a,b]$ (continuous functions).

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  • $\begingroup$ The way I understand it is that you prove VA and SM for subspaces, but when proving a vector space you need to prove all axioms. The question states that F is the vector space, however you've said that F is a subspace of C[a,b]. Could you please explain the relevance of a subspace in this question? Thanks. $\endgroup$ – Angie01 Aug 5 '15 at 23:18
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    $\begingroup$ @Angie01 If you know, that $F$ is a subset of $C[a,b]$, you must only verify that multiplying a vector by a scalar is in $F$, and sum of vectors is in $F$, which is verified by the first equation in my answer. I am assuming that you are not expected to verify all conditions of the linear space, but you may use this characterization of a linear subspace. $\endgroup$ – Przemysław Scherwentke Aug 6 '15 at 18:01
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This is my answer now:

Take two functions f(x), g(x) $\in$ F, which are differentiable.

$(f(x)+g(x))' = f'(x) +g'(x) \in [a,b]$ as for a function to be differentiable, it must be continuous over a given interval.

let c $\in$ $R$,

$(cf(x))'=cf'(x) \in [a,b]$ as f(x) is continous on $[a,b]$

I don't really see how this is proving much - it's just rehashing the axioms with different notation.

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