Q: Suppose that $f$ is a continuous, nonnegative function on the interval $[0,1]$. Let $M$ be the maximum of $f$ on the interval. Prove that:
\begin{align*} \lim\limits_{n \to \infty} \left[ \int_0^1 f(t)^n \, dt \right]^{1/n} &= M \\ \end{align*}
My attempt:
The upper/lower Darboux integrals of $f(t)^n$ are:
\begin{align*} U_{f^n,\mathcal{P}} &= \sum\limits_{j=1}^k M_j \Delta_j \\ U_{f^n} &\le M^n \\ L_{f^n,\mathcal{P}} &= \sum\limits_{j=1}^k m_j \Delta_j \\ L_{f^n} &\ge 0 \\ \end{align*}
\begin{align*} 0 \le L_{f^n} \le \int_0^1 f(t)^n \, dt \le U_{f^n} \le M^n \\ 0 \le \int_0^1 f(t)^n \, dt \le M^n \\ \end{align*}
I suspect that this is the wrong direction. I also believe that it is not necessarily true that $\int_0^1 f(t)^n = M^n$ for finite $n$.