Vector subspace F is closed in E

[Demon117]

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.

 

[Mark44]

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.

 

[Demon117]

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.

 

[Mark44]

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.

 

[Dick]

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.

 

[Mark44]

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.