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It says in the book that the set $F[a,b]$ of all functions on the interval $[a,b]$ is a vector space if pointwise addition and scalar multiplication are used.

I can't comprehend this statement very well.

If it means that the set of vectors form a space, then when I want to verify it through all the axioms, should I pick different functions and input some random number in the interval to be the arbitrary two different vectors or should I pick one function and input different numbers to be the arbitrary two different vectors?

I really hope that someone could understand what I'm asking here.

Please let me know if there is anywhere ambiguous in the question.

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    $\begingroup$ Choose arbitrary functions $[a,b] \to \mathbb R$ and arbitrary scalars in $\mathbb R$ $\endgroup$ – Simon S Nov 13 '15 at 1:19
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    $\begingroup$ The vectors are functions. For example, the sum $f+g$ of $f,g\in F[a,b]$ is the function whose values at every $x\in [a,b]$ is given by $(f+g)(x) = f(x)+g(x)$. This function $(f+g)\in F[a,b]$ (why?). Similarly for $\alpha f$ for any $\alpha\in \Bbb R$. $\endgroup$ – BrianO Nov 13 '15 at 1:30
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Define $\color{red}{+}\colon F[a,b]\times F[a,b]$, by $f\color{red}{+}g \in F[a,b] $, where $f\color{red}{+}g$ is defined by $(f\color{red}{+}g)(x) = f(x)+g(x)$, where $+$ is addition in $\Bbb R$. Do note that $\color{red}{+}$ and $+$ are different things.

Also define $\color{red}{\cdot}\colon \Bbb R \times F[a,b] \to F[a,b]$ by $\lambda \color{red}{\cdot} f$, where $(\lambda \color{red}{\cdot} f)(x) = \lambda \cdot f(x)$. Again, $\color{red}{\cdot}$ and $\cdot$ are different things.

You must check that $F[a,b]$ verify the axioms and such for $\color{red}{+}$ and $\color{red}{\cdot}$, and not for $+$ and $\cdot$. For example, I'll prove that $ f\color{red}{+}g = g \color{red}{+} f$. You have to prove that they're equal as functions, so you need to check equality pointwise. Let $x \in [a,b]$ be arbitrary. So: $$(f\color{red}{+}g)(x) \stackrel{\rm def}{=} f(x) + g(x) = g(x) + f(x) \stackrel{\rm def}{=}(g \color{red}{+} f)(x) \implies f \color{red}{+} g = g \color{red}{+} f,$$where the middle inequality is because $+$ is commutative in $\Bbb R$ and $f(x),g(x) \in \Bbb R$.

Now try to do the rest.

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    $\begingroup$ The green does not show up very well in my browser. Perhaps you could use red? $\endgroup$ – Ian Nov 13 '15 at 1:30
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    $\begingroup$ I just thought about that! $\endgroup$ – Ivo Terek Nov 13 '15 at 1:31
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    $\begingroup$ ah thank you very much. I just didn't know what "pointwise addition" means and thank you for a very specific explanation. $\endgroup$ – Ximing Nov 13 '15 at 2:06
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Given any vector space, first make sure that you completely understand the definition of the things that live in that space and the definition of addition and multiplication in this space. In this case, a vector is a function $f: [a,b] \rightarrow \mathbb{R}$. Do not think of numbers anymore, and do not assume that a vector is a string of numbers. "Point-wise" addition simply means that, given vectors $f_1$ and $f_2$, $f_1 + f_2$ is the vector $f_+$ that behaves like follows: $\forall x \in [a,b], f_+ (x) = f_1(x) + f_2(x)$

As an example of verifying the additive inverse axiom, pick any vector $f$ and let the additive inverse be $-f$ (which also lives in the space). Then, $f + (-f)$ is zero because the definition of pointwise addition implies $f(x) + (-f(x)) = y + (-y) = 0$, where $y$ is a real number

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For some field $F$ let $$ V = \{ f \mid f : [a,b] \to F \} $$ with $$ +: f + g: \forall {x \in [a, b]}: (f+g)(x) = f(x) + g(x) \quad (f, g \in V) \\ \cdot: \alpha \, f: \forall {x \in [a, b]}: (\alpha f)(x) = \alpha f(x) \quad (\alpha \in F, f \in V) $$ define a vector space over $F$. Then for the above definitions exercise your vector space criterion of choice, e.g. checking the axioms for $(V,+)$ and $(V, \cdot)$.

E.g. "C" (commutative property) for $(V,+)$:

Let $f, g \in V$, $x \in [a, b]$, then $$ (f + g)(x) = \underline{f(x) + g(x) = g(x) + f(x)} = (g + f)(x) \iff \\ f + g = g + f $$

Or "A" (associative property) for $(V,\cdot)$:

Let $f \in V$, $\alpha, \beta \in K$, $x \in [a,b]$: $$ ((\underbrace{\alpha \beta}_{K \cdot K}) f)(x) = \underline{(\alpha \beta) f(x) = \alpha (\beta f(x))} = \alpha ((\beta f) (x)) = (\underbrace{\alpha (\beta f)}_{K\cdot(K\cdot V)})(x) \iff \\ (\alpha \beta) f = \alpha (\beta f) $$

The operations on the vectors or scalar and vector are derived from the pointwise addition and multiplication of function values which are elements of the field $F$ and use its addition and multiplication and their properties (underlined parts above).

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    $\begingroup$ This wedge notation is unusual (to me at least) $\endgroup$ – Simon S Nov 13 '15 at 1:39
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    $\begingroup$ I guess it is more European, perhaps by Bourbaki. $\endgroup$ – mvw Nov 13 '15 at 1:48
  • $\begingroup$ What did that wedge mean? $\endgroup$ – YoTengoUnLCD Nov 13 '15 at 2:00
  • $\begingroup$ The all/universal quantifier, see: de.wikipedia.org/wiki/Quantor#Existenz-_und_Allquantor $\endgroup$ – mvw Nov 13 '15 at 2:01

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