I found the following statement:

"Example of a linear - vector - space: The set $C^{(k)}[a,b]$ of all (real-valued) continuous functions on a finite interval $a ≤ t ≤ b$ with addition and real number multiplication

1) $( f + g)(t) = f (t)+ g(t )$

2) $(α f )(t )= \alpha f (t )$, $t∈[a,b]$ forms a linear space. The zero vector $Θ: f (t )= 0$ for all $t∈[a,b]$."

I am a bit confused with 2).

Take $f(x) = x^2$. Let $\alpha = 2$, then:

$2*x^2 = 2*f(2) = 2*4 = 8 \neq 2^2*x^2 = f(2*2) = 4*4 = 16 $

Hence $(α f )(t ) \neq \alpha f (t )$

So how is it that degree two polynomials form a linear space?

Added: I realize I had a typo and that my question makes little sense. Thanks Race Bannon for pointing this out.

  • $\begingroup$ Just a comment on notation: don't use the asterisk to denote a product, because in certain fields the asterisk is used to denote the convolution of two functions. Juxtaposition is sufficient between two letters or a number and a letter; between two numbers you can use the symbol $\times$. $\endgroup$ – Massimo Ortolano Aug 30 '15 at 19:18

You calculated $f(\alpha x)$ instead of $(\alpha f)(x)$ in your example.

  • $\begingroup$ You are right, I read it wrong. So what would $(\alpha f)(x)$ look like in this case? $\endgroup$ – Gustavo Aug 30 '15 at 17:10
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    $\begingroup$ Well, it would look like what you'd expect. If $\alpha = 2$ and $f(x) = x^2$, then $(\alpha f)(x)$ would just be $2x^2$. The whole point is, it is giving you the rule of how to treat $(\alpha f)$. If you treat it the way they tell you, the functions will act like a vector space. $\endgroup$ – Race Bannon Aug 30 '15 at 17:13
  • $\begingroup$ Do you have an example of a function where $(α f )(t ) \neq \alpha f (t )$? $\endgroup$ – Gustavo Aug 30 '15 at 17:17
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    $\begingroup$ It depends on how you define $(\alpha f)$ in the first place. You can define $(\alpha f)(t) = f(\alpha t)$ or $f(t) ^ {\alpha}$ or $\frac{f(t)}{\alpha}$ or any other which way you like. They are just telling you how to interpret the symbol $(\alpha f)$ which on its own, and before the definition given, has little meaning when applied to things like specific polynomials. So for example, if you define $(\alpha f)(t) = \frac{f(t)}{\alpha}$, then $(\alpha f)(t) \neq \alpha f(t)$ as long as $\alpha \neq 1$ or $-1$. $\endgroup$ – Race Bannon Aug 30 '15 at 17:20
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    $\begingroup$ @Gustavo: It's worth noting, though, that the way addition and scalar multiplication are defined in your exercise is, in fact, the standard way to define those operations for functions. If someone writes $g = \alpha f$ without explicitly stating what they mean by that, the usual assumption is indeed that $g(x) = \alpha f(x)\ \forall x$. $\endgroup$ – Ilmari Karonen Aug 30 '15 at 17:48

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