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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.

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2 Answers 2

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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 $\theta=1.$

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    $\begingroup$ Where did that first inequality come from? Is it a theorem? $\endgroup$
    – Paprika
    Sep 14, 2018 at 21:17
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    $\begingroup$ 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. $\endgroup$
    – user587377
    Sep 14, 2018 at 21:32
  • $\begingroup$ 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? $\endgroup$
    – Paprika
    Sep 14, 2018 at 23:55
  • $\begingroup$ What are you allowed to use? What were you studying when you came across with this problem? $\endgroup$
    – user587377
    Sep 15, 2018 at 0:14
  • $\begingroup$ 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 $L_1$? $\endgroup$
    – Paprika
    Sep 15, 2018 at 22:05
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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.

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