Suppose that f is a continuous, non-negative function on the interval [0; 1]. Let M be the supremum of f on the interval. Prove that Suppose that $f$ is a continuous, non-negative function on the interval $[0,1]$. Let $M$ be the supremum of $f$ on the interval. Prove that:
$$\lim_{n\to \infty} \left[\int_0^1f(t)^ndt\right]^{\frac{1}{n}} = M$$
 A: Hint: pick an $\epsilon,$ divide the interval into pieces where $f < M-\epsilon$ and the complement. See what happens.
A: First note that $(\int_{0}^1 f(x)^n dx)^{1/n}\leq (\int_{0}^1 M^n dx)^{1/n} = M \implies \limsup_{n=1}^\infty (\int_{0}^1 f(x)^n dx)^{1/n}\leq M$
Then, by the continuity of $f$, there exists $x_0 \in [0, 1]$ such that $f(x_0) = \sup_{x \in [0, 1]} f(x) = M$.  So there exists a $\delta > 0$ such that 
$|M - f(x)| \leq \epsilon/2$ for all $x \in B_\delta (x_0)$.  That is, $-\epsilon/2 \leq M - f(x) \leq \epsilon/2 \implies f(x) \geq M - \epsilon/2$ for all $x \in B_\delta (x_0)$.
But then $(\int_{0}^1 f(x)^n dx) \geq (\int_{B_\delta(x_0)} f(x)^n dx) \geq (\int_{B_\delta(x_0)} (M-\epsilon/2)^n dx)$.  Hence, taking $n$-th roots of both sides, we get that: $(\int_{0}^1 f(x)^n dx)^{1/n} \geq (M-\epsilon/2)\mu(B_\delta(x_0))^{1/n}$.
Now since we know that $\liminf_{n \to \infty} \mu(B_\delta(x_0))^{1/n} = 1$,
we arrive at the conclusion that $\liminf_{n \to \infty} \int_0^1 f(x) dx \geq M-\epsilon$.  Since $\epsilon > 0 $ is arbitrary we are done.
A: Note that
$$
\left[ \int_0^1 f(t)^ndt \right]^{1/n} \leq [M^n]^{1/n} = M \qquad\qquad\qquad (1)
$$ 
And, if $\epsilon > 0$, $M-\epsilon$ is not an upper bound for $f([0,1])$, and so there is $x \in [0,1]$ such that
$$
f(x) > M - \epsilon
$$
Since $f$ is continuous, $\exists \delta > 0$ such that
$$
|y-x| < \delta \Rightarrow f(y) > M-\epsilon
$$
Hence,
$$
\int_0^1 f(t)^ndt \geq \int_{x-\delta}^{x+\delta}f(t)^ndt \geq (M-\epsilon)^n(2\delta)
$$
Hence,
$$
\left[\int_0^1 f(t)^ndt\right]^{1/n} \geq (M-\epsilon)(2\delta)^{1/n} \to (M-\epsilon) \qquad\qquad (2)
$$
The result follows from $(1)$ and $(2)$
