Here is a more thorough write up of my comment.

Fix a non-negative smooth function $\phi$ which is identically $1$ on $B_1$ and vanishes identically outside $B_2$. Pick $M$ so $|\Delta \phi| \leq M$. Set $\phi_R(x)=\phi(x/R)$. We have $|\Delta \phi_R(x)|\leq M/R^2$.

By the Bochner identity $$ \Delta \frac{1}{2} |\nabla u|^2= \nabla u \cdot \nabla \Delta u + |\nabla^2 u|^2=|\nabla^2 u|^2 $$

We have $$ \int_{B_R} |\nabla^2 u|^2\leq \int_{\mathbb{R}^n} \phi_R |\nabla^2 u|^2=\frac{1}{2} \int_{\mathbb{R}^n} \phi_R \Delta |\nabla u|^2\leq \frac{M}{2R^2} \int_{B_{2R}} |\nabla u^2| \leq \frac{CM}{2 R^2}. $$ Sending $R\to \infty$ implies $\int_{\mathbb{R}^n}|\nabla^2 u|^2=0$ so $\nabla^2 u$ vanishes identically.

This means $u(x)=\mathbf{a}\cdot x +b$ but finite energy forces $\mathbf{a}=0$.

Here is a more thorough write up of my comment.

Fix a non-negative smooth function $\phi$ which is identically $1$ on $B_1$ and vanishes identically outside $B_2$. Pick $M$ so $|\Delta \phi| \leq M$. Set $\phi_R(x)=\phi(x/R)$. We have $|\Delta \phi_R(x)|\leq M/R^2$.

By the Bochner identity $$ \Delta \frac{1}{2} |\nabla u|^2= \nabla u \cdot \nabla \Delta u + |\nabla^2 u|^2=|\nabla^2 u|^2 $$

We have $$ \int_{B_R} |\nabla^2 u|^2\leq \int_{\mathbb{R}^n} \phi_R |\nabla^2 u|^2=\frac{1}{2} \int_{\mathbb{R}^n} \phi_R \Delta |\nabla u|^2\leq \frac{M}{2R^2} \int_{B_{2R}} |\nabla u|^2 \leq \frac{CM}{2 R^2}. $$ Sending $R\to \infty$ implies $\int_{\mathbb{R}^n}|\nabla^2 u|^2=0$ so $\nabla^2 u$ vanishes identically.

This means $u(x)=\mathbf{a}\cdot x +b$ but finite energy forces $\mathbf{a}=0$.

Here is a more thorough write up of my comment.

Fix a non-negative smooth function $\phi$ which is identically $1$ on $B_1$ and vanishes identically outside $B_2$. Pick $M$ so $|\Delta \phi| \leq M$. Set $\phi_R(x)=\phi(x/R)$. We have $|\Delta \phi_R(x)|\leq M/R^2$.

By the Bochner identity $$ \Delta \frac{1}{2} |\nabla u|^2= \nabla u \cdot \nabla \Delta u + |\nabla^2 u|^2=|\nabla^2 u|^2 $$

We have $$ \int_{B_R} |\nabla^2 u|^2\leq \int_{\mathbb{R}^n} \phi_R |\nabla^2 u|^2=\frac{1}{2} \int_{\mathbb{R}^n} \phi_R \Delta |\nabla u^2|\leq \frac{M}{2R^2} \int_{B_{2R}} |\nabla u^2| \leq \frac{CM}{2 R^2}. $$ Sending $R\to \infty$ implies $\int_{\mathbb{R}^n}|\nabla^2 u|^2=0$ so $\nabla^2 u$ vanishes identically.

This means $u(x)=\mathbf{a}\cdot x +b$ but finite energy forces $\mathbf{a}=0$.

Here is a more thorough write up of my comment.

Fix a non-negative smooth function $\phi$ which is identically $1$ on $B_1$ and vanishes identically outside $B_2$. Pick $M$ so $|\Delta \phi| \leq M$. Set $\phi_R(x)=\phi(x/R)$. We have $|\Delta \phi_R(x)|\leq M/R^2$.

By the Bochner identity $$ \Delta \frac{1}{2} |\nabla u|^2= \nabla u \cdot \nabla \Delta u + |\nabla^2 u|^2=|\nabla^2 u|^2 $$

We have $$ \int_{B_R} |\nabla^2 u|^2\leq \int_{\mathbb{R}^n} \phi_R |\nabla^2 u|^2=\frac{1}{2} \int_{\mathbb{R}^n} \phi_R \Delta |\nabla u|^2\leq \frac{M}{2R^2} \int_{B_{2R}} |\nabla u^2| \leq \frac{CM}{2 R^2}. $$ Sending $R\to \infty$ implies $\int_{\mathbb{R}^n}|\nabla^2 u|^2=0$ so $\nabla^2 u$ vanishes identically.

This means $u(x)=\mathbf{a}\cdot x +b$ but finite energy forces $\mathbf{a}=0$.