[proofplan]
We apply the interior derivative estimate from [Theorem 37](/theorems/37) to bound $|\nabla u(x_0)|$ for an arbitrary point $x_0 \in \mathbb{R}^n$. The $L^1$ norm of $u$ over a ball $B(x_0, r)$ is controlled by the $L^\infty$ norm times the volume of the ball, yielding a bound that decays as $1/r$. Since $u$ is bounded on all of $\mathbb{R}^n$, the ball radius $r$ can be taken arbitrarily large, forcing $\nabla u(x_0) = 0$. As $x_0$ is arbitrary, $u$ is constant.
[/proofplan]
[step:Bound $|\nabla u(x_0)|$ using the interior derivative estimate and the volume of $B(x_0, r)$]
Fix $x_0 \in \mathbb{R}^n$ and $r > 0$. Since $u$ is harmonic in $\mathbb{R}^n$, the ball $B(x_0, r) \subset \mathbb{R}^n$ and [Theorem 37](/theorems/37) (interior [derivative](/page/Derivative) estimates for harmonic functions) at order $k = 1$ gives
\begin{align*}
|\nabla u(x_0)| \le \frac{C_1}{r^{n+1}}\, \|u\|_{L^1(B(x_0, r))}.
\end{align*}
Since $u$ is bounded, $|u(y)| \le \|u\|_{L^\infty(\mathbb{R}^n)}$ for all $y \in \mathbb{R}^n$. Replacing $|u|$ by its supremum in the $L^1$ norm:
\begin{align*}
\|u\|_{L^1(B(x_0, r))} = \int_{B(x_0, r)} |u(y)|\, d\mathcal{L}^n(y) \le \|u\|_{L^\infty(\mathbb{R}^n)} \cdot \mathcal{L}^n(B(x_0, r)) = \|u\|_{L^\infty(\mathbb{R}^n)} \cdot \omega_n\, r^n,
\end{align*}
where $\omega_n = \pi^{n/2}/\Gamma(1 + n/2)$ is the volume of the unit ball. Substituting:
\begin{align*}
|\nabla u(x_0)| \le \frac{C_1\, \omega_n}{r}\, \|u\|_{L^\infty(\mathbb{R}^n)}.
\end{align*}
[guided]
The strategy is to exploit the fact that $u$ is defined on all of $\mathbb{R}^n$, so we can take arbitrarily large balls centred at any point. The interior derivative estimate gives control that improves as the ball grows.
Fix $x_0 \in \mathbb{R}^n$ and $r > 0$. Since $u$ is harmonic on $\mathbb{R}^n$, every ball $B(x_0, r) \subset \mathbb{R}^n$ is admissible for the [interior derivative estimate](/theorems/37) at order $k = 1$:
\begin{align*}
|\nabla u(x_0)| \le \frac{C_1}{r^{n+1}}\, \|u\|_{L^1(B(x_0, r))},
\end{align*}
where $C_1$ depends only on $n$. The $L^1$ norm involves an integral over a ball whose volume grows as $r^n$, while the prefactor decays as $r^{-(n+1)}$. What happens when we bound the $L^1$ norm using the $L^\infty$ norm? Since $|u(y)| \le \|u\|_{L^\infty(\mathbb{R}^n)}$ for all $y$:
\begin{align*}
\|u\|_{L^1(B(x_0, r))} \le \|u\|_{L^\infty(\mathbb{R}^n)} \cdot \mathcal{L}^n(B(x_0, r)) = \|u\|_{L^\infty(\mathbb{R}^n)} \cdot \omega_n\, r^n.
\end{align*}
The volume $\omega_n r^n$ grows with $r$, but it cancels all but one power of $r$ in the denominator:
\begin{align*}
|\nabla u(x_0)| \le \frac{C_1}{r^{n+1}} \cdot \omega_n\, r^n \cdot \|u\|_{L^\infty(\mathbb{R}^n)} = \frac{C_1\, \omega_n}{r}\, \|u\|_{L^\infty(\mathbb{R}^n)}.
\end{align*}
The net decay is $1/r$ -- exactly one power survives after the cancellation.
[/guided]
[/step]
[step:Send $r \to \infty$ to conclude $\nabla u \equiv 0$ and hence $u$ is constant]
The bound $|\nabla u(x_0)| \le C_1 \omega_n\, r^{-1}\, \|u\|_{L^\infty(\mathbb{R}^n)}$ holds for every $r > 0$. The right-hand side tends to $0$ as $r \to \infty$ (here $\|u\|_{L^\infty(\mathbb{R}^n)} < \infty$ by hypothesis). Therefore $\nabla u(x_0) = 0$.
Since $x_0 \in \mathbb{R}^n$ was arbitrary, $\nabla u \equiv 0$ on $\mathbb{R}^n$. Because $\mathbb{R}^n$ is [connected](/page/Connectedness), $u$ is constant on $\mathbb{R}^n$.
[/step]
