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Question Number 89530 by mathocean1 last updated on 17/Apr/20

F which is the set of funtions from R to R is a vectorial space and G(a part of F) is the set of odd functions such as G={ f ∈ F/∀ x∈ R, f(x)=−f(−x)} 1) Show that G is sub vector space of F in R.

$$\mathrm{F}\:\mathrm{which}\:\mathrm{is}\:\:\mathrm{the}\:\mathrm{set}\:\mathrm{of}\:\mathrm{funtions}\:\mathrm{from}\:\mathbb{R}\:\mathrm{to}\:\mathbb{R}\: \\ $$$$\mathrm{is}\:\mathrm{a}\:\mathrm{vectorial}\:\mathrm{space}\:\mathrm{and}\:\mathrm{G}\left(\mathrm{a}\:\mathrm{part}\:\mathrm{of}\:\mathrm{F}\right)\:\mathrm{is} \\ $$$$\mathrm{the}\:\mathrm{set}\:\mathrm{of}\:\mathrm{odd}\:\mathrm{functions}\:\mathrm{such}\:\mathrm{as} \\ $$$$\mathrm{G}=\left\{\:{f}\:\in\:\mathrm{F}/\forall\:\mathrm{x}\in\:\mathbb{R},\:{f}\left({x}\right)=−{f}\left(−{x}\right)\right\} \\ $$$$\left.\mathrm{1}\right)\:{S}\mathrm{how}\:\mathrm{that}\:\mathrm{G}\:\mathrm{is}\:\mathrm{sub}\:\mathrm{vector}\:\mathrm{space}\:\mathrm{of}\:\mathrm{F} \\ $$$$\mathrm{in}\:\mathbb{R}. \\ $$

Commented by arcana last updated on 18/Apr/20

The function zero f(x)=0, ∀x∈R, so f(x)=0=f(−x)⇒0∈G. G≠φ. G⊆F. let be f, g ∈G (f+g)(x)=f(x)+g(x) =−f(−x)+[−g(−x)] =−(f+g)(−x), ∀x∈R f+g ∈G. for α∈R, (αf)(x)=α(−f(−x))=−(αf)(−x) ⇒αf ∈G.

$$\mathrm{The}\:\mathrm{function}\:\mathrm{zero}\:{f}\left({x}\right)=\mathrm{0},\:\forall{x}\in\mathbb{R},\:\mathrm{so} \\ $$$${f}\left({x}\right)=\mathrm{0}={f}\left(−{x}\right)\Rightarrow\mathrm{0}\in\mathrm{G}.\:\mathrm{G}\neq\phi. \\ $$$$\mathrm{G}\subseteq\mathrm{F}. \\ $$$$\mathrm{let}\:\mathrm{be}\:{f},\:{g}\:\in\mathrm{G} \\ $$$$\left({f}+{g}\right)\left({x}\right)={f}\left({x}\right)+{g}\left({x}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=−{f}\left(−{x}\right)+\left[−{g}\left(−{x}\right)\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=−\left({f}+{g}\right)\left(−{x}\right),\:\forall{x}\in\mathbb{R} \\ $$$${f}+{g}\:\in\mathrm{G}. \\ $$$$\mathrm{for}\:\alpha\in\mathbb{R},\:\left(\alpha{f}\right)\left({x}\right)=\alpha\left(−{f}\left(−{x}\right)\right)=−\left(\alpha{f}\right)\left(−{x}\right) \\ $$$$\Rightarrow\alpha{f}\:\in\mathrm{G}. \\ $$$$ \\ $$

Commented by mathocean1 last updated on 18/Apr/20

thank you sir!

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}! \\ $$

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