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

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 $\theta=1.$

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.