# If $0 \le f(x) \le M$, prove $\lim\limits_{n \to \infty} \left[ \int_0^1 f(t)^n \, dt \right]^{1/n} = M$ [duplicate]

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 $$n$$.

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• Obviously $\int f^n \le M^n$. Take $x$ with $f(x) = M$. Take $\epsilon > 0$ and an interval $U$ of $x$ s.t. $f(y) > M-\epsilon$ for $y \in U$. Then $\int f^n \ge \int_U (M-\epsilon)^n = (M-\epsilon)^nl(U)$ and so $\liminf_n (\int f^n)^{1/n} \ge M-\epsilon$. This holds for all $\epsilon > 0$. – mathworker21 Nov 25 '18 at 6:40