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Determine of the set of all functions $f:\mathbb{R}\to\mathbb{R}$ is a vector space with addition defined as $(f+g)(x)=\max\{f(x),g(x)\}$

Usual scalar multiplication

Clearly when we add or scalar multiply, our function $f$ is still mapping $\mathbb{R}\to\mathbb{R}$, and so it is closed under addition and scalar mult.

However I have trouble coming up with counter example as to why it is not a vector space

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  • $\begingroup$ Why are you taking a max when considering the scalar multiplication? That was defined as usual. $\endgroup$ Sep 10, 2017 at 18:21
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    $\begingroup$ What's the $0$ element? $\endgroup$
    – lulu
    Sep 10, 2017 at 18:29
  • $\begingroup$ Just the constant $y=0$ function $\endgroup$
    – K Split X
    Sep 10, 2017 at 18:50

2 Answers 2

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I suppose the field you are considering is $\mathbb{R}$.

In this structure of your example the operation "multiplication of a given scalar by an element of the underlying set" does not distribute over the operation $\operatorname{max}$

For instance: $$-1\times(f_1+f_2)\ne-f_1-f_2$$

because $\exists x\in\mathbb{R}$ such that $$-1\times\operatorname{max}\{f_1(x),f_2(x)\}\ne\operatorname{max}\{-f_1(x),-f_2(x)\}$$

That means that this structure is not a vector space over $\mathbb{R}$.

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Your "addition" (I'll denote it $\oplus$) does not have a neutral element. Suppose it had, there would be some $e: \mathbb{R} \to \mathbb{R}$ such that $e \oplus f = f$ for all $f$.

This means (by the definition of $\oplus$) that $\max(f(x), e(x)) = f(x)$ for all $x$. But define $f$ by $f(x) = e(x) - 1$ (usual addition) then $\max(f(x) , e(x)) = e(x) \neq f(x)$ for all $x$, (so we have $e \oplus f = e$) contradiction.

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