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well, $a$ is true I guess but not sure , $b$ is true as $\max\{f,g\}=\frac{1}{2}(|f-g|-|f+g|\}$ $c$ is also true as $f\in V\Rightarrow f^2\in V$ so any polynomial expression will also be in $V$, could any one tell me my logic are correct or not? and help me to solve fully correctly?

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3 Answers 3

You are right. For (a) note that $fg=(\frac{f+g}2)^2-(\frac{f-g}2)^2$.

For (b) your expression for $\max$ seems to be wrong, however, try again, the correct one is not much different. Observe that $\max\{t,0\}=\frac12(t+|t|)$.

For (c) you should at least remark that $f^n\in V$ uses induction.

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ahhhhhhhhhhhhhhhhhhhhh :( thank you – La Belle Noiseuse Oct 3 '12 at 13:18

Relate $fg$ to $(f+g)^2$ and $(f-g)^2$.

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a is indeed true, but you should give an argument for it (hint: $fg$ occurs in what kind of expression involving $f$, $g$ and $x \mapsto x^2$? With a bit of cleverness, you can make the excess $f^2$ and $g^2$ bits go away)

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why I am not clever :'( :( :( :'( – La Belle Noiseuse Oct 3 '12 at 13:19

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