If $u$ is harmonic, $\exists \alpha,\beta \in \mathbb{R},\forall x\in \mathbb{R}^d,u(x) \leq \alpha |x|+\beta,$ then $u$ is affine

Question

We consider a harmonic function $u:\mathbb{R}^d \to \mathbb{R}$ $(\Delta u=0).$ Suppose that $$\exists \alpha,\beta \in \mathbb{R},\forall x\in \mathbb{R}^d,u(x)\leq \alpha |x|+\beta.$$

Therefore $u-u(0)$ is linear: $$\exists\lambda\in \mathbb{R}^d,\forall x\in \mathbb{R}^d,u(x)=\langle\lambda,x\rangle+u(0)$$

I am looking for a probabilistic proof of this fact

We note that Liouville's theorem for bounded harmonic functions is a particular case, where we can use Ito's formula, martingale convergence and Blumenthal's $0$-$1$ law to prove it.

How to use probabilistic techniques to prove the above claim? (To be noted that it's sufficient to prove that partial derivatives of $u$ are constant)

Other proofs are also appreciated.

Answer 1

This is a standard exercise eg. see "Harmonic Function bounded by a linear function" where a similar proof works here too. It is immediate using the Cauchy-estimate.

In terms of probabilistic proof, at least we get to use Liouville theorem which has probabilistic proof. But I doubt one can do be better since one needs to deal with derivatives and the Cauchy-estimate.

As mentioned in the comments we will follow the hint of the exercise 4 4.8 in Brownian Motion and Classical Potential Theory by M.Rao (1977) via the mean value property.

To be clear the author made a typo that he corrected in his picture i.e. $A_{t}=B_{R-t}(x+th)$ and not $A_{t}=B_{R}(x+th)$. We again let $A:=B_{R}(x)$. Therefore

$$\frac{u(x+th)-u(x)}{t}=\frac{1}{t}\left(\frac{1}{|A_{t}|}\int_{A_{t}}u(y)dy-\frac{1}{|A|}\int_{A}u(y)dy\right)$$

$$=\frac{1}{t}\left(\frac{1}{|A_{t}|}-\frac{1}{|A|}\right)\int_{A_{t}}u(y)dy-\frac{1}{t|A|}\int_{A\setminus A_{t}}u(y)dy.$$

We have $|A_{t}|=c(R-t)^{2}$ and $|A|=cR^{2}$. We use the bound on the second term and mean-value on the first

$$\geq \frac{1}{t}\left(\frac{1}{|A_{t}|}-\frac{1}{|A|}\right)|A_{t}|u(x+th)-\frac{1}{t|A|}\int_{A\setminus A_{t}}\alpha |y|+\beta dy.$$

The first term we just a derivative in $t$. For the second term, we use $\int_{A}|y|=c_{n}R^{3}=|A|R$. After taking $t\to 0$ we are left with $$\geq c_{1}\frac{u(x)}{R}-c_{2}\alpha.$$

By taking $R\to +\infty$, we get that the partial derivative is lower bounded. Since it is also harmonic, we can apply Liouville ("An entire function whose real part is bounded above must be constant." for $Ref=-\partial_{h} u$).