MHB Range of Uniform convergence of dirchlet series

[Euge]

While I follow your solution, I can't quite explain to myself why you chose "converges uniformly on [p,∞), where p > 1", instead of (p,∞), where p > 1 ?

Hmm, I'm not sure if this question has do with a weak knowledge about intervals. The statement would be true if you put $(p,\infty)$ instead of $[p,\infty)$, but what I proved is the stronger result, since $(p,\infty)$ is a proper subset of $[p,\infty)$ ($p\in [p,\infty)$ but $p\notin (p,\infty)$).

 

[ognik]

Is it because you chose p earlier in your analysis, so it would be wrong to then use (p, inf) because p is not in (p, inf)?

 

[Euge]

I didn't say it would be wrong to use $(p,\infty)$. The point is that my argument shows uniform convergence is valid on $[p,\infty)$ for all $p> 1$. In particular, uniform convergence is valid on $(p,\infty)$ for all $p > 1$.

 

[ognik]

sorry, I should have said something like weaker ...

But I still can't grasp why one is different from the other, OR why one is better than the other.

My lecturer finally responded to my nagging and said "What this statement means is that you have two parameters s and x. However, the values of s are restricted by the condition indicated. When x is varied s should be fixed."

While that seems to make sense, I still am not getting it ...I keep hoping there is just a piece of math theory that I don't know, that will make this clear to me?

 

[Euge]

When you say that you can't grasp why one is different from the other, do you mean you don't understand why $(p,\infty)$ is different from $[p,\infty)$?

 

[ognik]

They are different stated like that because (p...) excludes p, and [p ...] includes p. But once we have that p > 0, they then both exclude p?

 

[Euge]

No, that's not true. The open interval $(a, b)$ is defined as the set of real numbers $x$ such that $a < x < b$; the half-open interval $[a, b)$ is defined as the set of real numbers $x$ such that $a \le x < b$. Here, we may allow $b = \infty$. Thus $(a,\infty)$ consists of those $x$ which satisfy $a < x < \infty$; similarly for $[a,\infty)$. For all real numbers $a$, $a\in [a,\infty)$ but $a\notin (a,\infty)$.

 

[ognik]

After some more fiddling with trying to grasp this, (from what you have explained, but also from what my prof said - he will be setting the end of year exam), I came up with the following:

If I was asked to write $ p \lt x \lt \infty $ as an interval, by definition I would write $ x \epsilon (p, \infty) $

If I was then told that $ p \gt 1 $ is a fixed parameter and that we had previously found that p is included in the interval, then I would write $ x \epsilon [p, \infty), p > 1 $ - I hope this is correct and well reasoned?

Then, if I wanted to try and write this semi-closed interval as an algebraic inequality, I would write $ 1 \lt p \lt x \lt \infty $, with p 'fixed' as P > 1

Have I got it?

 

[Euge]

You're so close! If you remove the "$x\in$" then you would be absolutely correct. [emoji2]

 

[ognik]

Thanks, Euge - but why remove the $ x \epsilon $ ... and do you mean from both places I have used it? ? x is the variable that we are trying to find the range for?

 

[Euge]

The notation $x\in (a, b)$ is not interval notation, but it express membership; it's $(a, b)$ itself that's interval notation. That's what I meant.

 

[ognik]

Hi Euge, I thought you might like to know, I found that the radius of convergence with the Weierstrauss test MUST be a closed interval, that helped. But what finally put my mind at rest, was to check convergence using the integral test - that showed me very clearly where the parameter (s) comes from.

A sincere thanks for your patience, it all finally came together for me - joy!