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I think I can prove that a harmonic function $u$ on $\mathbf{R}^n$ which satisfies $|u(x)|\leqslant C \ln(|x|+1)$ is constant. But what can we say about $u$ when the absolute value sign of $u$ is canceled? Can we still say that $u$ is constant? Any comments or references will be appreciated.

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You might want to read through "Slow Growth for Universal Harmonic Functions" at journalofinequalitiesandapplications.com/content/2010/1/253690 . –  A Walker Nov 8 '11 at 5:15
    
Thank you for this paper. But actually I do not figure out the relation between this paper and my question. –  Y.Z Nov 8 '11 at 7:19
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3 Answers 3

up vote 4 down vote accepted

We have the following theorem (which is a slight generalisation of the classical Liouville theorem for positive harmonic functions [see, for example, chapter 3 of Axler, Bourdon and Ramey's Harmonic Function Theory]; it may help to read that proof first to get an idea of the basic approach):

Theorem Let $f:[0,\infty)\to[0,\infty)$ be a (not necessarily strictly) increasing continuous function such that $\lim_{r\to\infty} f(r)/r = 0$. Let $u:\mathbf{R}^n\to\mathbf{R}$ be harmonic, such that $u(x) \geq - f(|x|)$, then $u$ is constant.

Proof: Observe that $u(x) + f(|x|)$ is a continuous, non-negative function.

Consider $u(x) - u(z)$ for some fixed $x,z$. Using the mean value property for harmonic functions, we write

$$ |B_R|(u(x) - u(z)) = \int_{B_R(x)} u(y) dy - \int_{B_R(z)} u(y) dy $$

The right hand side we rewrite

$$ = \int_{B_R(x)} u(y) + f(y) - f(y) dy - \int_{B_R(z)} u(y) + f(y) - f(y) dy $$

which is

$$ \leq \int_{B_R(x)\setminus B_R(z)} u(y) + f(y) dy + \int_{B_r(z) \setminus B_r(x)} f(y) dy $$

Writing $A\delta B$ for the symmetric set difference $(A\setminus B)\cup (B\setminus A)$, we get

$$ \leq \int_{B_R(x) \delta B_R(z)} u(y) + 2f(y) dy $$

Define $w = \max(|x|,|z|)$. Now using that $B_R(x) \delta B_R(z) \subset B_{R+ w}(0) \setminus B_{R-w}(0)$, we have

$$ \leq \int_{B_{R+ w}(0) \setminus B_{R-w}(0)} u(y) + 2f(y) dy $$

For the $u$ term, we use the mean value property again. For $f$, estimate by a supremum, you get

$$ \leq (u(0)+2f(R+w)) \left( |B_{R+ w}| - |B_{R- w}|\right) $$

Observe that $|B_{R+ w}| - |B_{R- w}| = O(R^{n-1})$, by our assumption on $f$ we have

$$\lim_{R\to\infty} (u(0) + 2f(R+w))\frac{|B_{R+ w}| - |B_{R- w}|}{|B_R|} = 0$$

and hence

$$ u(x) - u(z) \leq 0 $$

Since the derivation is symmetric in $x$ and $z$, this implies that $u(x) = u(z)$. q.e.d.

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One should note that the above-stated theorem uses strongly (twice, in both directions) the mean-value theorem for harmonic functions, which is tied to the exact form of the Green's identity on Euclidean space and the form of the Green's function of the Laplacian. In the case of noncompact Riemannian manifolds with nonnegative Ricci curvature, there is a theorem of S.Y. Cheng that states the same thing as the above, but requiring both upper and lower bounds $|u(x)| \leq f(|x|)$. The only one-sided bound result I know is Yau's theorem for nonnegative harmonic functions. –  Willie Wong Nov 8 '11 at 16:18
    
Thank you very much! This is more than enough for my question. Would you please tell me where this theorem appears? –  Y.Z Nov 9 '11 at 5:37
    
@Y.Z. if I knew where to find it, I would not have written it up right here. :) It is presumably known, but none of the potential theory textbooks or harmonic function theory textbooks I have on hand contains it. Since the entire proof is given above, if you need a reference you can just give the above answer as a citation. (For citations, click on the "link" button at the bottom of the answer, and in the pop-up window click on "cite" which is on the lower left hand corner; this will give you BibTeX or AMSRefs citations.) –  Willie Wong Nov 9 '11 at 8:56
    
If you really must give a more credible citation, you can additionally mention that the above Theorem is a "direct generalisation of the Liouville Theorem for positive harmonic functions" and refer to Axler, Bourdon, and Ramey for the Liouville Theorem. –  Willie Wong Nov 9 '11 at 8:58
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In two dimensions such a function still has to be constant. Let $v(z)$ be the harmonic conjugate of $u(z)$. Then ${\displaystyle f(z) = e^{u(z) + i v(z)}}$ is an entire function, and $$|f(z)| = e^{u(z)} \leq e^{C\ln(|z| + 1)}$$ $$\leq K(1 + |z|)^C$$ Here $K$ is some constant. It's a standard exercise to show that entire functions that grow polynomially are themselves polynomials, so $f(z)$ must be a polynomial.

If $z_0$ is any zero of $f(z)$, then $\lim_{z \rightarrow z_0} e^{u(z)} = |f(z)| = 0$, so $\lim_{z \rightarrow z_0} u(z) = -\infty$. Hence $u(z)$ can't be continuous at $z_0$, a contradiction.

Thus the only possibility is that $f(z)$ is constant, so the same is true for $u(z)$.

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As a general rule, if we control the behavior of a harmonic function on one side then we have the same control on the other side by the Harnack inequality. More rigorously:

In your particular case, we have $u$ harmonic and $u \geq -\log(1+|x|)$. Assume $u(0) = a$. Suppose at a point $x$ in $B_R$ we have $u(x) = C$. Sliding $u$ up by $\log(1+2R) + 1$, we obtain a positive harmonic function in $B_{2R}$ with the value $a + 1 + \log(1+2R)$ at $0$. By the Harnack inequality, we have $$C + \log(1+2R) + 1 \leq K(a + 1 + \log(1+2R))$$ which tells us that $u$ grows at worst logarithmically, reducing the problem to the case $|u| \leq C(1+\log(1+2R))$. The standard proof of Liouville theorem applies once we have the two-sided bound.

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