Entire functions with finite $L^1$ norm must be identically $0$ Another question from Complex Variables: An Introduction by Berenstein and Gay.

Show that an entire function has finite $L^1$ norm on $\mathbb{C}$ iff $f\equiv0$. Does this also hold true for $L^2, L^{\infty}$?

So, the $L^{\infty}$ case I think just follows from Liouville's Theorem, but the others I'm not sure about. How should I approach these? My intuition would be to use a power series expansion at $0$ for $f$, but the work I've done on paper with this hasn't really gone anywhere fruitful.
Edit: As discussed in the comments, there are two cases to consider wrt the behavior at $\infty$: if $f$ has a pole there, then $f$ must be a polynomial so it must have infinite norm. So we are reduced to the case where $f$ has an essential singularity at $\infty$.
 A: If I remember my complex analysis, the real and imaginary parts of $f(z)$, denoted $u(x,y)$ and $v(x,y)$ are harmonic on $\mathbb{R}^{2}$. If $\|f\|_{L^{1}(\mathbb{R}^{2})}$ is finite, then $\max\{\|u\|_{L^{1}},\|v\|_{L^{1}}\}<\infty$. By the mean value property, we have for any fixed $(x,y)\in\mathbb{R}^{2}$,
$$|u(x,y)|\lesssim\dfrac{1}{r^{2}}\int_{B_{r}(x,y)}|u(s,t)|dsdt\leq\dfrac{\|u\|_{L^{1}}}{r^{2}},\quad\forall r>0$$
where the implied constant is independent of $r>0$ and $(x,y)$. Letting $r\rightarrow \infty$, we obtain that $|u(x,y)|=0$. By the same argument, one obtains $|v(x,y)|=0$.
If $\|f\|_{L^{2}}<\infty$, then by convexity, $\max\{\|u\|_{L^{2}},\|v\|_{L^{2}}\}<\infty$. By the above argument and Holder's inequality,
$$|u(x,y)|\lesssim\dfrac{1}{r^{2}}\int_{B_{r}(x,y)}|u(s,t)|dsdt\lesssim\left(\dfrac{1}{r^{2}}\int_{B_{r}(x,y)}|u(s,t)|^{2}dsdt\right)^{1/2}\leq\dfrac{\|u\|_{L^{2}}}{r},\qquad\forall r>0$$
where the implied constants are independent of $r>0$ and $(x,y)$. Letting $r\rightarrow\infty$, we obtain the desired conclusion.
