Let $u:\mathbb{R}^n \to \mathbb{R}$ be harmonic on $B_1(0)$, the ball of radius one centered at $0$. I saw this result that
$$ \sup_{B_{1/2}(0)} |D^k u| \leq C \| u \|_{L^2(B_1(0))}$$
where $k \geq 0$ is an integer, and $C = C(n,k)$.
I think induction works.
Attempt for $k=1$: Let $x_0 \in B_{1/2}(0)$. Then we have for $r=1/8$
$$ |u_{x_i}(x_0)| = \Big| C \int_{B_r(x_0)} u_{x_i} dx \Big| = \Big| \frac{C}{r^n} \int_{\partial B_r(x_0)} u \nu_i ds \Big| \leq C \|u\|_{L^\infty(\partial B_r(x_0))} $$
where we have used the Mean Value Property, and divergence theorem, and $C= C(n)$.
For $x \in \partial B_r(x_0)$ we have by MVP,
$$ u(x) = \frac{C}{r^n} \int_{B_r(x)} u(x) dx \leq \frac{C}{r^n} \int_{B_1(x)} u(x) dx = C \| u \|_{L^1(B_1(0))}$$
So $$ |u_{x_i}(x_0)| \leq C \| u \|_{L^1(B_1(0))} \leq C \| u \|_{L^2(B_2(0))} $$
Is this method correct?
Question: I understand we can bound the $L^1$ norm in terms of the $L^2$ norm, but why not leave the result in terms of $L^1$? And why stop at $L^2$ why not go higher?