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Vector subspace F is closed in E

  • Thread starter Demon117
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  • #1
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Let E be the vector space of bounded functions f:N --> R, with the norm(g) = sup|f|. Assume without proof that the norm holds, so that the function d(f,g)=norm(f - g) is a metric.

Prove that the vector subspace F={f in F | f(n) -->0 as n --> infinity} is closed in E.

Here is what I have come up with:

Consider the sequence (f_k) in F, such that f_k --> f. It must be shown that f is in F also.

For such (f_k) in F, it follows that (f_k)(n) -->0 as n --> infinity. Since f_k --> f (by a problem done earlier), it follows that f(n) -->0 as n --> infinity. Since the latter holds, by definition of F it follows that f is contained within. Hence F is closed in E.

This doesn't look right at all. I just wondered what kinds of advice anyone has.
 

Answers and Replies

  • #2
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Let E be the vector space of bounded functions f:N --> R, with the norm(g) = sup|f|. Assume without proof that the norm holds, so that the function d(f,g)=norm(f - g) is a metric.

Prove that the vector subspace F={f in F | f(n) -->0 as n --> infinity} is closed in E.
Are you sure you wrote this correctly? It would make more sense to me to say {f in E | f(n) --> 0 as n --> infinity}
Here is what I have come up with:

Consider the sequence (f_k) in F, such that f_k --> f. It must be shown that f is in F also.

For such (f_k) in F, it follows that (f_k)(n) -->0 as n --> infinity. Since f_k --> f (by a problem done earlier), it follows that f(n) -->0 as n --> infinity. Since the latter holds, by definition of F it follows that f is contained within. Hence F is closed in E.

This doesn't look right at all. I just wondered what kinds of advice anyone has.
Consider two elements of F, f1 and f2, and a scalar c.
1) Show that f1 + f2 is in F. This shows that F is closed under addition.
2) Show that cf1 is in F. This shows that F is closed under scalar multiplication.
 
  • #3
165
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Are you sure you wrote this correctly? It would make more sense to me to say {f in E | f(n) --> 0 as n --> infinity}


Consider two elements of F, f1 and f2, and a scalar c.
1) Show that f1 + f2 is in F. This shows that F is closed under addition.
2) Show that cf1 is in F. This shows that F is closed under scalar multiplication.
Wow I had a few typos. Thanks for clarifying that. I am not sure that this is the idea behind the proof. I need to get some added direction on this one.
 
  • #4
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What I laid out in post #2 is the standard way of showing that a subset U of a vector space V is in fact a subspace of V. When you say that a subset (F) of a vector space (E) is closed in E, you are saying that the subset is closed under vector addition and closed under scalar multiplication.
 
  • #5
Dick
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What I laid out in post #2 is the standard way of showing that a subset U of a vector space V is in fact a subspace of V. When you say that a subset (F) of a vector space (E) is closed in E, you are saying that the subset is closed under vector addition and closed under scalar multiplication.
Judging by the attempt at a proof, I really think they mean 'closed' here to mean a closed set in the sup norm topology. You want to show if lim as k->infinity |fk-f|=0 and fk are in F then f is in F. I would think about throwing some epsilons around.
 
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  • #6
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Sounds reasonable to me. The bit about vector space threw me off so that I wasn't thinking about "closed" in terms of accumulation points and such.
 

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