Given $1\le p\le\infty$ and $f \in L^{p}([0.\infty))$, prove $\lim_{n\to\infty} \int_{0}^{\infty}f(x)e^{-nx}dx=0$.
If $p=1$, I can use Dominated Convergence theorem directly because $f(x)e^{-nx} \rightarrow 0 \quad a.e.$ and $|f(x)e^{-nx}| \le |f(x)| \in L^{1}$.
But how about $1\lt p$ ?
Let $q$ be defined by $\frac{1}{p}+\frac{1}{q}=1$. $e^{-nqx}$ is integrable for all $n\ge 1$, i.e. $e^{-nx}$ belongs to the conjugate space $(L_q)$ of $L_p$. Therefore $\int_0^\infty f(x)e^{-nx}dx$ exists.
To get the $0$ limit, $|\int_0^\infty f(x)e^{-nx}dx|\le (\int_0^\infty|f(x)|^pdx)^\frac{1}{p}(\int_0^\infty e^{-qnx}dx)^\frac{1}{q}\to 0$ since the $q$ integral term =$(\frac{1}{nq})^\frac{1}{q}\to 0$, as $n\to \infty$.