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 March 8th, 2017, 12:07 PM #1 Senior Member   Joined: Jan 2015 From: usa Posts: 104 Thanks: 1 locally uniformly convergence I want to answer the following problem; please help me: Suppose that $G\subset \mathbb{C}$ is a Jordan domain. Consider an increasing sequence of continuous functions $f_n:\partial G\to \mathbb{R}, n\in\mathbb{N},$ that is uniformly bounded above. Let $u_n$, $n\in\mathbb{N},$ be the solution of the Dirichlet problem for $f_n$ in $G$. Show that the corresponding sequence of functions $u_n$, $n\in\mathbb{N},$ converges locally uniformly on $G$ to a function $u$ harmonic in $G$ Thanks in advance. Last edited by skipjack; March 8th, 2017 at 01:26 PM.
March 8th, 2017, 02:38 PM   #2
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This is above my level. I wonder if you would mind explaining some of the technical terms. I'd be interested to understand the question. And maybe if you explain your problem in detail you'll see the answer. That's a great trick at work. When someone comes to you with a complicated problem you just ask them to explain it to you in detail. Halfway through their explanation they figure out the answer and think you're a genius.

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 Originally Posted by mona123 Suppose that $G\subset \mathbb{C}$ is a Jordan domain.
I Googled that and did not see a definition. Can you say what is a Jordan domain?

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 Originally Posted by mona123 Consider an increasing sequence of continuous functions $f_n:\partial G\to \mathbb{R}, n\in\mathbb{N},$ that is uniformly bounded above.
Ok $\partial G$ is the boundary of $G$, that's the topological usage and has nothing to do with partial derivatives. So we have a sequence of continuous functions from the boundary of $G$ to the reals. Say a function from the unit circle to the reals, where the unit circle is the boundary of the unit disk. Something like that, depending on exactly what a Jordan domain is.

Quote:
 Originally Posted by mona123 Let $u_n$, $n\in\mathbb{N},$ be the solution of the Dirichlet problem for $f_n$ in $G$.
Even though I don't know what a Dirichlet problem is, I think that notation is a little ambiguous. Given a function $f_n$, when you say $u_n$, $n\in\mathbb{N}$ that indicates that there is a whole collection of $u_n$'s for each $f_n$. But you're using the same $n$. What I think you mean is that for each $n\in\mathbb{N}$, $u_n$ is a solution of the Dirichlet problem for $f_n$. There is one solution per function, and not a sequence of solutions for each $f_n$. Is that right?

From Wiki: "... a Dirichlet problem is the problem of finding a function which solves a specified partial differential equation (PDE) in the interior of a given region that takes prescribed values on the boundary of the region."

I wonder if you would be willing to tell us a little about this, what it means in the scheme of things. Honestly this won't help me much, I'm way too ignorant of PDE's. But someone else might be interested.

Quote:
 Originally Posted by mona123 . Show that the corresponding sequence of functions $u_n$, $n\in\mathbb{N},$ converges locally uniformly on $G$ to a function $u$ harmonic in $G$
I know what uniform convergence is. What is local uniform convergence? I found an SE thread but I didn't read it yet.

Well I hope you don't mind a response that doesn't answer your question But if you want to talk about what this problem means, I'd be interested to learn.

Last edited by Maschke; March 8th, 2017 at 02:44 PM.

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