# If vector space of all real functions is modified to have $cf(x) = f(cx)$, which axiom of vector space is broken?

There is vector space $F$ of all real functions. $f(x) \in F$.

Now suppose that scalar multiplication of vector space $F$ is modified so that it is now defined as $cf(x) = f(cx)$.

In this case, which axiom of vector space is broken?

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Hi! Have you tried just writing down the axioms and seeing whether you can contradict them one at a time? In general, it's best to tell us what you've tried and exactly where you're stuck. – Sharkos Feb 10 '14 at 12:35

## 3 Answers

Hint: What would happen with something like $(c+d)f(x)$?

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Given two field elements $c$ and $d$, and given a vector $f$, we must have: $$(c+d)\,f=c\,f+d\,f.$$ But, instead, we get the vector $$x\mapsto f((c+d)\,x)\qquad\text{as our (c+d)\,f}$$ and the vector $$x\mapsto f(c\,x)+f(d\,x)\qquad\text{as our c\,f+d\,f}$$ The two are not necessarily equal.

For instance consider the square function $x\mapsto x^2$. $$((c+d)\,x)^2\quad\text{is not the same as}\quad (cx)^2+(dx)^2\quad\text{for all x.}$$

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When one multiplies a vector by$~0$, the result should be the zero vector; however, this definition gives $0f:x\mapsto f(0)$, a constant function, but not necessarily the zero function (neutral element for addition).

Now $\def\vv{\vec v}0\vv=\vec0$ is not a vector space axiom, at least not one in the usual list. However it can be proved using the fact that $\vec0$ is the unique solution to the vector equation $\vec x+\vec x=\vec x$, together with the identity $0\vv+0\vv=(0+0)\vv=0\vv$. The former property only involves vector addition, so it certainly (still) holds in the example. The latter part uses only the distributive law $a\vv+b\vv=(a+b)\vv$ (the simplification $0+0=0$ does not involce the vector space structure), so it is this law that must fail. Indeed if $\vv$ is for instance the constant function $1$, then $0\vv=\vv$, and $0\vv+0\vv\neq0\vv$ (LHS is constant$~2$, RHS is constant$~1$).

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