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Let $U$ be the set of all continuous functions $f: [a,b] \rightarrow \mathbb{R}$ such that $\int_a^b f(x) dx=1$. With the usual operations of pointwise addition and scalar multiplication, is $U$ a vector space over $\mathbb{R}$?

My first thought is no, since letting $a=0, b=1$ and letting $f(x)=1$, and taking $\int_a^b [f(x)+f(x)] dx=2$ so this is not closed under addition. Is this reasoning correct?

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Your reasoning is correct. –  Alex Becker Feb 3 '13 at 19:36
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Almost correct. Who said that the constant function $f(x)\equiv 1$ lies in $U$? –  Giuseppe Negro Feb 3 '13 at 19:36
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Also, $\,\int_a^b2f\,dx=2\,$... –  DonAntonio Feb 3 '13 at 19:37
    
Re to edit: ok with $a=0, b=1$ but what about the general case? You can follow DonAntonio advice above. –  Giuseppe Negro Feb 3 '13 at 19:40

1 Answer 1

"With the usual operations of pointwise addition and scalar multiplication..." These are not operations on the set $U$, as you noticed. Far from it: for all $f,g\in U$, $f+g\not\in U$, and for all $f\in U$, $\lambda\in\mathbb R\setminus\{1\}$, $\lambda f\not\in U$. (There is also no additive identity, but that is a little strange to consider when there is no addition in the first place.)

One concrete example suffices as in your solution, but technically the example should apply to the general $[a,b]$, not just to $[0,1]$. You could adapt your example slightly to accommodate the general case.

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