MHB Solving Attached Problem, Plus MHF Update

[Cairo]

Could anybody help me with the attached problem?

I've managed to prove part (ii) and (iv) of the question, but have no idea where to begin with part (i). I'm assuming that the result follows by using the definitions, but cannot make any progress with it.

For part (iii) I have thought about looking at the ration of the two norms and then using the squeeze rule but again, have yet to make any progress with this.

Any help would be much appreciated here.

Do we also know when MHF will be up and running again?

 

[tarantella]

Where is the attached problem?

 

[Jameson]

Do we also know when MHF will be up and running again?

I don't see any attachment and please read http://www.mathhelpboards.com/showthread.php?119-is-the-mhf-dead in regards to MHF. This is a new site that will remain for a long time. Shoot me a PM if you can't work out the attachment issue.

 

[tarantella]

For question i), if $a=0$ it's clear, and otherwise, for $r<\infty$ use the fact that the sequence $\{a_n\}$ converges to $0$. For $r=\infty$, it's not true (take $a_n=1-\frac 1n$).

 

[Cairo]

For question i), if $a=0$ it's clear, and otherwise, for $r<\infty$ use the fact that the sequence $\{a_n\}$ converges to $0$. For $r=\infty$, it's not true (take $a_n=1-\frac 1n$).

Thanks for replying so quickly, Girdav. However, I'm still not sure how to translate this into a proof as such.

 

[tarantella]

That's your job, so I will only give a sketch of proof. If $a=0$ then the result is obvious, and if $a\neq 0$, since $\lim_{n\to \infty}a_n=0$ we can find an integer $N$ such that $|a_n|\leq \frac{\lVert a\rVert_{\infty}}2$ if $n\geq N$. So the index we are looking for is between $0$ and $N-1$.

 

[Cairo]

I've tried and tried and tried and still don't get this!

 

[tarantella]

What do you get?

 

[Ackbach]

Do we also know when MHF will be up and running again?

Nobody knows. What I do know is that MHB is not going away any time soon. It also does not suffer from the same problem MHF did: a site owner who had orphaned the site.

 

[Cairo]

What do you get?

It's just the first part I am unsure of. Am I to show that a convergent sequence attains its supremum?

 

[tarantella]

Yes, and here it's a little more easy since it converges to $0$. Did you try with my hints?

 

[Cairo]

That's your job, so I will only give a sketch of proof. If $a=0$ then the result is obvious, and if $a\neq 0$, since $\lim_{n\to \infty}a_n=0$ we can find an integer $N$ such that $|a_n|\leq \frac{\lVert a\rVert_{\infty}}2$ if $n\geq N$. So the index we are looking for is between $0$ and $N-1$.

OK - I give up with this problem! I just cannot see how you make the jump from converges to zero on to attaining its supremum.

Where did the 2 in the denominator come from in your inequality?

I'm at least partially happy that I have managed to prove the remaining parts of the problem.