I saw $\|u\|_{L^1(B(x,r))}$ in my text. But I don't know what it means. I know what $L^p$ norm is, but with the ball, I am not sure. Does it mean the volume inside $L^p$ norm$=r$? There isn't integral sign though.
This is the text:
THEOREM 7 (Estimates on derivatives). Assume $u$ is harmonic in $U$. Then $$ \tag{18} |D^\alpha u(x_0)| \le \frac{C_k}{r^{n+k}} \| u\|_{L^1(B(x_0, r))}.$$
On
Definition: Let $(X,\mu)$ be a measure space. Then $L^p(X)$ is defined to be $$ \left\{ f : X \to \mathbb{C} \ \middle|\ \|f\|_{L^p(X)} < \infty \right\}, $$ where $$ \|f\|_{L^p(X)} := \left( \int_{X} |f|^p \,\mathrm{d}\mu \right)^{1/p}. $$
In your example, $B(x_0,r)$ is a measure space, with the measure $\mu$ being the restriction of the ambient measure to the $r$-ball (presumably, Lebesgue measure coming from $\mathbb{R}^n$ or $\mathbb{C}^n$, though it is hard to say from the small excerpt you provide—it doesn't really do any harm to ignore the measure theory here and regard the integrals as Riemann integrals). Hence $$ \| u \|_{L^1(B(x_0,r))} := \int_{B(x_0,r)} |u(x)| \, \mathrm{d}x. $$
It is the $\mathbf{L}^1$ norm of the restriction of $u$ to $B(x_0,r)$.