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My book states that the set of all continuous functions defined on a closed interval $[a, b]$ where $a ≠ b$ satisfies the axioms of a vector space.

I am understanding that this means that for the axiom of closure under multiplication, any function within that interval that we would multiply by any scalar would still be a function within that interval. I am a little confused about this. For example how does $C[2, 4]$ with scalar $0$ satisfy this? If we would multiply any function by scalar $0$, how would it still be a function within $C[2, 4]$?

Just as a comparison, I have also understood that the set of all quadratic functions is not a vector function because if we would multiply by scalar $0$, it would cease to be a quadratic function. Am I wrong about this as well? If not, what is the difference?

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    $\begingroup$ If you multiply a function with the scalar $0$, then the result is the zero function that has function value zero for any $x$ that you plug in. What do you mean by "quadratic function"? If you mean polynomials of order two, the set they consitute is of course a vector space. $\endgroup$ – amsmath Mar 24 '19 at 18:46
  • $\begingroup$ My book says polynomials of degree two is not a vector space, only polynomials of degree two or less. $\endgroup$ – agblt Mar 24 '19 at 18:50
  • $\begingroup$ @agblt Is $x^2 + -x^2$ a second order polynomial? $\endgroup$ – John Douma Mar 24 '19 at 18:52
  • $\begingroup$ agblt Please look again at what you wrote last. You are contradicting yourself. $\endgroup$ – amsmath Mar 24 '19 at 18:53
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    $\begingroup$ @agblt Ok, the set of polynomials of degree exactly 2 is clearly not a vector space. $\endgroup$ – amsmath Mar 24 '19 at 20:01
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Take, for instance,$$\begin{array}{rccc}f\colon&[2,4]&\longrightarrow&\mathbb R\\&x&\mapsto&x.\end{array}$$Then $f\in\mathcal C[a,b]$, right? And $0\times f$ is the null function, which is also an element of $\mathcal C[a,b]$.

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