anonymous
  • anonymous
In each of the following cases, either prove that the given subset W is a subspace of V , or show why it is not a subspace of V . a) \[V= c ^\infty (\mathbb{R})\] , W is the set of functions f⊆V for which lim f(x) = 0 \[x \rightarrow \infty\].
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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KingGeorge
  • KingGeorge
Can you explain what \(e^{\infty} (\mathbb{R})\) is?
anonymous
  • anonymous
its not e, it's C . like, constant i believe?
anonymous
  • anonymous
its a vector space.

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KingGeorge
  • KingGeorge
I'm still confused as to what \(c^{\infty}\) means. I've never seen notation like that.
anonymous
  • anonymous
i think it's like all of constants .. like constant to infinity vector space or something .. for all real numbers? heres to guessing hahaha.
KingGeorge
  • KingGeorge
In any case however, I think W is still a subspace because it contains the function \(f(x)=0\), W is closed under addition, and it's closed under scalar multiplication.
KingGeorge
  • KingGeorge
See http://en.wikipedia.org/wiki/Linear_subspace for more information.
anonymous
  • anonymous
ok thanks! one more question if you dont mind same question applied .. but to V=c∞(R) , W is the set of functions f⊆V for which f(4) = 1
KingGeorge
  • KingGeorge
Definitely not. It doesn't contain the function \(f(x)=0\), it isn't closed under addition, and it isn't closed under scalar multiplication.
anonymous
  • anonymous
ok, are you able to explain to me sort of what i'm looking for here, because there are 2 other parts to this same question but i'm unsure how to define them
KingGeorge
  • KingGeorge
First, see if the function \(f(x)=0\) is in the subset. Then, take two functions \(f, g\) in the subset. Find if the function \(h=f+g\) is in the subset as well. Finally, take a function \(f\) in the subset, and some constant \(c\). Then find if \(c\cdot f\) is in the subspace. If all three of these conditions are true, you have a subspace.
anonymous
  • anonymous
ok thank you so much!!!!!
KingGeorge
  • KingGeorge
You're welcome.

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