Look at this doc it may be helpful for your topic. Good luck.

Dear Daniel,

May be next article will be useful for you

Ukrainian Mathematical Journal

August 1993, Volume 45, Issue 8, pp 1260–1271

On subharmonic extension and extension in the Hardy-Orlicz classes

Authors

J. Riihentaus P. M. Tamrazov

Thanks. It did mention if the exceptional set is a polar set,

then it is extendable.

For a geometer, "polar" set is too technical. Is there any

simply saying that a subharmonic (the general sense since

we are working on compact Riemannian manifolds) function can be extended over a condimension 2 subset?

Here is one of my argument, which reduces the problem to the harmonic one:

Whenever there is a bounded continuous nonnegative function f on M such that (1) f(N) = 0, (2) f is real analytic on M −N and (3) ∆f ≤ 0 on M−N, then f = 0. Here N could be just a codimension two subset. The reason is that if we define Ms = {x ∈ M|dist(x,N)≥s} and hs = ∂Ms, then the measure of hs is smaller than O(s) when s tends to zero. Therefore, 0 ≥ ln2\int _M2δ ∆fω^n ≥\int ^2δ _δ [\int _Ms ∆f ω^n]s^{−1} ds =\int ^2δ _δ [\int _hs ∂f/∂n dτ]s^{−1}ds.

But by applying an integration by parts to the single variable integral, the last term is about (δ)^{−1} \int _h2δ (f −g)dτ → 0 since f is bounded and f −g tends to 0 near N, where g is the f value of the corresponding point on hδ.

A set ${\displaystyle \mathbb {R} ^{n}} \mathbb {R} ^{n}$ (where${\displaystyle n\geq 2}$) is a polar set if there is a non-constant subharmonic function

${\displaystyle u}$ on ${\displaystyle \mathbb {R} ^{n}}$

such that

${\displaystyle Z\subseteq \{x:u(x)=-\infty \}.}$

Note that there are other (equivalent) ways in which polar sets may be defined, such as by replacing "subharmonic" by "superharmonic", and ${\displaystyle -\infty }$ with ${\displaystyle \infty }$ in the definition above.

The most important properties of polar sets are:

A singleton set in ${\displaystyle \mathbb {R} ^{n}}$ is polar.

A countable set in ${\displaystyle \mathbb {R} ^{n}}$ is polar.

The union of a countable collection of polar sets is polar.

A polar set has Lebesgue measure zero in ${\displaystyle \mathbb {R} ^{n}.}$

Ransford, Thomas (1995). Potential theory in the complex plane. London Mathematical Society Student Texts. 28. Cambridge: Cambridge University Press. ISBN 0-521-46654-7. Zbl 0828.31001

Thank you for your answers. My question should be more precisely as following:

If a continuous function f on a compact manifold M is superharmonic (with the given Riemannian metrics) on M-N with N a subset of codimension 2. Does it means that

f is constant?

This seemly was a general fact as early as 1981. In the paper:

"Y. T. Siu & P. Yang: Compact K¨ahler-Einstein surfaces of nonpositive bisectional curvature, Invent. Math. 64 (1981), 471–487."

it was mentioned as a fact without even mentioning a reference.

In my recent paper:

"On Bisectional Nonpositively Curved Compact K¨ahler Einstein Surfaces"

to appear in Pacific J. of Math.

I gave a short argument as above. Someone asked me about this.

Anyway, for our special case, my earlier post reduced the problem to the

harmonic case.

A similar argument proved the result we need for the special case as following:

"Therefore, ∆f = 0 on M−N. A similar arguments shows that \int _hd (f−g)dτ = 0, where g is the f value of the corresponding point on hs for any given s < d. Let s tends to zero, we get \int _hd fdτ = 0. By f ≥ 0 on hd we obtained that f = 0 near N. Therefore f extends over N as a harmonic function. This implies that f = 0 on M."

The last sentence came from the Hodge Theorem.

This solved the special case in our paper.

One more question is following:

Does this proof work, in some more general sense, for the general case as I mentioned

at the beginning of this post?