Let $u:\mathbb R^d\longrightarrow \mathbb R$ a harmonic function. Show that,
1) $\int_{\mathbb R^d}|u|^2<\infty \implies u=0$
2) $\int_{\mathbb R^d}|\nabla u|^2<\infty \implies u$ constant.
My attempts
Let consider the ball $B(x,r)$. Then, $$|u(x)|=\left|\frac{1}{|B(x,r)|}\int_{B(x,r)}u(y) d y\right|\leq \frac{1}{|B(x,r)|}\int_{B(x,r)}|u|\underset{CS}{\leq}\frac{1}{\sqrt{|B(x,r)|}}\sqrt{\int_{B(x,r)}|u|^2}\leq \frac{C}{\sqrt{|B(x,r)|}}.$$ Letting $r\to \infty $, we get $|u(x)|=0$ and thus $u=0$.
Is it correct ? (In fact, $u\in L^1(\mathbb R^d)$ would be enough, no ?)
2) We have that $|\nabla u|$ is harmonic, and thus the claim follow for the previous exercise.
Is it correct ?
Your solution of (1) is correct. And you are right that $u\in L^1$ would lead to the same conclusion, but as Thomas noted, neither $L^1$ nor $L^2$ contains the other.
For (2), consider individual derivatives $u_{x_i}$ instead of the norm of the gradient. Since $\Delta u_{x_i} = (\Delta u)_{x_i}=0$, part 1) applies to $u_{x_i}$. Hence all the partial derivatives are zero.