let the Poisson kernel $P_y(x)$ be defined by $\frac{1}{\pi}\frac{y}{x^2+y^2}$ on the upper half plane $y>0$.
I was actually trying to prove that the Poisson integral of $f\in L^p(-\infty, +\infty)$ is harmonic, that is, $(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2})f(x,y)=0$, where $f(x,y)=(f*P_y)(x)$ is the Poisson integral of $f$.
in doing this, I divided the integral in the convolution into two parts: $I=(-N,N)$ and $I'=(-\infty,-N)\cup (N,+\infty)$.
for $I$, I could put the Laplace operator inside the integral since the domain is compact, by bounded convergence theorem, and got $0$ since the Poisson kernel is harmonic.
but for $I'$, I couldn't do it because the domain is not compact, so I used Holder's inequality:
$\int_{I'}f(t)P_y(x-t)dt\leq ||f||_p(\int_{I'}P_y(x-t)^{\frac{p}{p-1}}dt)^{1-\frac{1}{p}}$
since $f\in L^p$, $||f||_p$ is finite. but I'm not sure that so is $\int_{\mathbb{R}}P_y(x-t)^{\frac{p}{p-1}}dt$. if so, then I can choose large $N$ s.t. $(\int_{I'}P_y(x-t)^{\frac{p}{p-1}}dt)^{1-\frac{1}{p}}<\epsilon$ for given $\epsilon>0$ and conclude.
but it's kind of complicated to compute directly, so I don't know what to do from here.
For each fixed $y>0$, the function $P_y$ is bounded on $\mathbb{R}$ and also decays like $1/x^2$ at infinity. This implies $\int_{\mathbb{R}} P_y(x)^q\,dx$ converges for all $q>1/2$. In particular, Hölder's inequality can be used to show $f*P_y$ is well defined for all $f\in L^p$, where $1\le p\le \infty$.
Concerning the rest of your proof: if you find the justification of moving the Laplacian insider the integral tedious, consider using the Mean Value Property instead. (A continuous function in a domain $\Omega$ is harmonic if and only if its value at any point $p$ is equal to its average over the boundary of any closed disk centered at $p$ and contained in $\Omega$.) The proof of Mean Value Property for the Poisson integral amounts to applying the Fubini theorem, since the Poisson kernel itself satisfies it.