I take the lebesgue measure on $\mathbb R$ . Take $f:\mathbb R \to \mathbb R$ a continuous function and $\int_{\mathbb R}|f(x)|\mathrm d\mu(x)<\infty$.
Can we say any thing about $\lim_{|x|\to \infty}|f(x)|$ ? or can we say if $f$ is bounded ?
If we can show $f$ is bounded can we generalize this result to $\mathbb R^n$.
Basically I was dealing with a different problem .
It said "If $f:\mathbb C \to \mathbb C$ is a holomorphic function and $\int_{\mathbb R^2}|f(x+iy)|\mathrm dx\mathrm dy <\infty$ then $f\equiv0$".
So my approach was if I can show $f$ is bounded then I am done.
Thanks in advance please help me out and let me know if I am wrong at my approach.
by Cauchy $2\pi i f(z)=\int_{|w-z|=R}\frac{f(w)}{w-z}dw=\int_0^{2\pi}f(z+Re^{it})idt$ so $2\pi|f(z)|\le \int_0^{2\pi}|f(z+Re^{it})|dt$; integrating from say $R=1$ to $R=2$ we get (using that $dxdy=RdRdt \ge dRdt=dtdR, R \ge 1$)
$2\pi|f(z)|\le \int_1^2\int_0^{2\pi}|f(z+Re^{it})|dtdR \le \int_1^2\int_0^{2\pi}|f(z+Re^{it})|RdtdR \le \int \int_{A_z} |f(x+iy)|dxdy \le M $
where $A_z$ is the annulus centered at $z$ between circles of radiuses $1$ and $2$ and $M= \int_{\mathbb R^2}|f(x+iy)|\mathrm dx\mathrm dy <\infty$
so $f$ is constant and hence zero by the finite integral property
As noted in the comments, the same proof works for $f$ harmonic as the integral mean value property is still true as is the fact that a bounded harmonic function in the plane (even bounded one way if the function is real) is constant