# $u$ harmonic, $|u(x)|\le C(1+|x|^\theta), x\in \mathbb{R}^N$ then $u$ is constant

Let $u$ be an harmonic function in $\mathbb{R}^N$. Suppose that there exists constants $C>0$ and $0\le\theta<1$ such that $$|u(x)|\le C(1+|x|^\theta), x\in \mathbb{R}^N$$ Show that $u$ is contant. Show also that this conclusion is not valid for $\theta=1$

I believe this has somthing to do with the maximum principle:

if $f$ is a harmonic function, then $f$ cannot exhibit a true local maximum within the domain of definition of $f$. In other words, either $f$ is a constant function, or, for any point $x_0$, $x_{0}$ inside the domain of $f$, there exist other points arbitrarily close to $x_0$ at which $f$ takes larger values.

However, since $u$ is defined in all of $\mathbb{R}^N$, then I think $u$ should be constant or unbounded with no local maximum points (I can only picture a function strictly increasing or decreasing).

I don't know how to proceed here. This inequality condition makes no sense to me.

If $$f$$ is harmonic in $$\mathbb{R}^n$$, then given $$r>0$$, there is $$M>0$$ such that $$| Df(x_0)|\leq \frac{M}{r^{n+1}}\|f\|_{L_1(B(x_0,r))}.$$ That is $$| Df(x_0)|\leq \frac{C\cdot M}{r^{n+1}}\|1+|x|^\theta\|_{L_1(B(x_0,r))}\leq \frac{C\cdot M}{r^{n+1}}\|1+|r|^\theta\|_{L_1(B(x_0,r))}$$ Hence $$| Df(x_0)|\leq \frac{C\cdot M}{r^{n+1}}\cdot(1+|r|^\theta)\cdot \|1\|_{L_1(B(x_0,r))} = \frac{C\cdot M\cdot r^n}{r^{n+1}}\cdot(1+|r|^\theta)\cdot | B(x_0,1)|.$$

So $$| Df(x_0)|\leq \frac{C\cdot M}{r}\cdot(1+|r|^\theta)\cdot | B(x_0,1)|.$$

Since $$0\leq\theta<1$$ and $$f$$ is harmonic in $$\mathbb{R}^n$$, taking $$r$$ arbitrarily big, the righ side goes to zero. Then, for all $$x_0$$, we have that $$|Df(x_0)|=0,$$ and hence $$f$$ is constant since $$\mathbb{R}^n$$ is connected.

Notice that this proof doesn't work for $$\theta=1.$$

• Where did that first inequality come from? Is it a theorem? Sep 14, 2018 at 21:17
• Yes. In Evan's Book on Partial Differential Equations it's called 'Estimates on derivatives'. This theorem is in the section about Mean-value Formulas in the chapter about Laplace Equations.
– user587377
Sep 14, 2018 at 21:32
• Thanks, I just found the proof. However I don't know if I'm allowed to use it. Do you have an idea on how it would be done without this theorem? Sep 14, 2018 at 23:55
• What are you allowed to use? What were you studying when you came across with this problem?
– user587377
Sep 15, 2018 at 0:14
• I've already seen green functions, the representation formula, weak and strong maximum principles, mean value properties and soe other minor things. I didn't see this theorem so it's kinda strange that I should come up with it before solving the exercise. Also, what is this $||||_{L_1}$? It's a norm? What is $L_1$? Sep 15, 2018 at 22:05

Cauchy-type estimate: There is a constant $$A$$ such that for all $$u$$ harmonic on $$B(0,1)$$ and continuous on $$\overline {B(0,1)},$$

$$|\nabla u(0)| \le A\max_{|x|=1}|u(x)|.$$

This follows from the Poisson integral representation; just differentiate through the integral sign to see each partial derivative at $$0$$ satisfies the above, hence so does $$\nabla u(0).$$

In our problem, let $$u_r(x) = u(rx)$$ for $$r>0.$$ Then by the above

$$|\nabla u_r(0)| = r|\nabla u(0)| \le A\max_{|x|=1}|u_r(x)|$$ $$= A\max_{|x|=r}|u(x)| \le AC(1+r^\theta).$$

It follows that $$|\nabla u(0)|\le AC(1+r^\theta)/r.$$ Letting $$r\to \infty$$ gives $$\nabla u(0)=0.$$

To finish, we translate $$u.$$ Let $$x_0\in \mathbb R^N$$ and define $$v(x) = u(x+x_0).$$ Then for $$|x|\ge |x_0|,$$ we have

$$|v(x)|=|u(x+x_0)|\le C(1+|x+x_0|^\theta) \le C(1+2|x|^\theta).$$

This is enough to make the previous argument work, giving $$|\nabla v(0)| = |\nabla u(x_0)|=0.$$ It follows that $$\nabla u(x)=0$$ everywhere, proving $$u$$ is constant.

If $$\theta =1,$$ then $$u(x)=x_1$$ is a non constant harmonic function that satisfies the inequality.