Let $f : [0,1]\to \mathbb R$ be a continuous function. Show that the sequence $$\left(\int_ 0^1 |f(x)|^n dx\right)^{1/n}$$ is convergent.
How can I show that this is a monotone sequence? ( I am thinking to apply monotone convergence theorem but was not able to go any further)
Because $|f|$ is continuous on a closed interval, $|f|$ achieves its maximum $M$ at some point $x_0$. Assume without loss of generality that $f\geq0$ and that $M=1$. Fix $\varepsilon>0$ and let $$E=\{x:\ f(x)>1-\varepsilon\}.$$ By continuity of $f$ we have that $E$ is open, so its measure is not zero. Then \begin{align} \left(\int_0^1f(x)^n\,dx\right)^{1/n} &\geq\left(\int_Ef(x)^n\,dx\right)^{1/n}\geq(1-\varepsilon)\,m(E)^{1/n}. \end{align} Thus \begin{align} 1\geq\limsup_n\left(\int_0^1f(x)^n\,dx\right)^{1/n}\geq\liminf_n\left(\int_0^1f(x)^n\,dx\right)^{1/n} &\geq1-\varepsilon. \end{align} As $\varepsilon$ was arbitrary, the limit exists and is equal to $1$ (that is, to $M$ in the general case).