Why do people care about $L^p$-convergence $f_n \rightarrow f$? Are there any interesting application of $L^p$-convergence?
For example, if $p=\infty$, then the limit $f$ of the sequence $f_n$ of continuous function is again continuous. are there any other interesting, useful applications?
It's difficult to know what you find interesting without further information. But here's an application from partial differential equations, anyway.
Many undergraduates learn how to solve the initial value problem $$ \begin{cases} u_t + u_{xxx} = 0 \qquad x,t\in\mathbb{R}\\ u(x,0) = u_0(x). \end{cases} $$ Using the Fourier transform, the solution is written $$ u(x,t) = U(t)u_0(x) = [e^{it\xi^3}\hat{u}_0]^\vee(x). $$ A basic calculus class will usually provide a description of the solution procedure, side stepping convergence issues. What are these issues?
For concreteness, suppose $u_0 \in L^2(\mathbb{R})$. Since $|e^{it\xi^3}|=1$, two applications of Parseval's theorem shows $$ \|U(t)u_0\|_2 = \|e^{it\xi^3}\hat{u}_0\|_2 = \|\hat{u}_0\|_2 = \|u_0\|_2. $$ That is, if the initial data lies in $L^2$, then at each time $t$, the solution (considered as a function in $x$) also lies in $L^2$. The operator $U(t)$ simply pushes the data around the closed ball in $L^2$ of radius $\|u_0\|_2$; the $L^2$ norm of the solution is conserved.
That's nice, but what happens as $t\rightarrow0$ with the solution? Do you actually recover the initial data? One would hope! What do you mean by recover? Ah, a convergence issue!
In the $L^2$-sense, the above calculation will show that $U(t+h)u_0 \rightarrow U(t)u_0$ as $h\rightarrow0$. In particular, this will hold for $t=0$ and we simply note that $U(0)u_0=u_0$. Here are some details: \begin{align*} \|U(t+h)u_0 - U(t)u_0\|_2 &= \|e^{i(t+h)\xi^3}\hat{u}_0 - e^{it\xi^3}\hat{u}_0\|_2 \\ &= \|e^{it\xi^3}[e^{ih\xi^3}-1]\hat{u}_0\|_2 \\ &= \|[e^{ih\xi^3}-1]\hat{u}_0\|_2. \end{align*} Now apply the Lebesgue Dominated Convergence Theorem to the final line to reach the conclusion. This result is abbreviated by saying $$ u(t) \in C(\mathbb{R} : L^2(\mathbb{R})), $$ which means the solution to the IVP describes a continuous curve $t \mapsto u(t)$ in $L^2$, where we have suppressed the spatial variable.
Two comments: