Undergrad Proof of Harmonic Function Infinitely Differentiable

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A harmonic function is defined as infinitely differentiable within a simply-connected region G, as demonstrated by the theorem linking harmonic functions to holomorphic functions. The proof begins by establishing that for any point z_0 in G, a disk can be found around z_0 that remains within G. However, concerns arise when considering boundary points, such as z_0 = 0 in the region G = C \ R_<0, where a disk cannot be formed. This leads to the conclusion that the discussion assumes G is an open set, as boundary points complicate the conditions necessary for harmonicity. Clarification is sought on the nature of harmonic functions at boundary points, highlighting the need for specific conditions in such cases.
Silviu
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Hello! I have this Proposition: "A harmonic function is infinitely differentiable". The book gives a proof that uses this theorem: "Suppose u is harmonic on a simply-connected region G. Then there exists a harmonic function v in G such that ##f = u + iv## is holomorphic in G. ". In the proof they present in the book they begin with: "Suppose u is harmonic in G and ##z_0 ∈ G##. Let ##r > 0## such that the disk ##D[z_0, r]## is contained in G. " and as a disk is simply connected the conclusion follows from the theorem. My question is, how can you make sure that for any ##z_0 \in G## you can have a disk around ##z_0##? (For example if ##G=\mathbb{C}##\##\mathbb{R}_{<0}## and ##z_0=0##, you can't find such a disk. What am I missing?
 
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Typically, these things are only considered in the interior of the set G. Statements about points on the boundary of G would require a lot of conditions and restrictions.
 
they evidently assume ##G## to be an open set:
Silviu said:
Let r>0r > 0 such that the disk D[z0,r]D[z_0, r] is contained in G.
 
Region usually means open and connected.
 
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Silviu said:
For example if G=CG=\mathbb{C}\R<0\mathbb{R}_{z0=0z_0=0, you can't find such a disk.
by the way, what is a harmonic function at such a point ##z_0##?
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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