Prove that the sequence $\{v_n\}$ converges to the maximum value of $f$ on $[a,b]$

Question

Question :

Assume that $f$ is a continuous not negative function on any point defined on the interval $[a,b]$.
For each natural number $n$ , Assume that $v_n$ is the $n$'th root of $\int_a^b f^n$.
Prove that the sequence $\{v_n\}$ converges to the maximum value of $f$ on $[a,b]$.

Note : The problem is how to connect these things in a formal way. I know that many important things happen on the roots. ( Specially maxima and minima ) But I don't know how to relate them to convergence of a sequence like $\{v_n\}$.

Please, If you have the time, explain your answer a bit more. I'm new to integration. Thanks in advance.

Answer 1

Suppose $f$ is nonnegative on $[a,b]$ and attains a maximum $M$ at $c \in [a,b]$. For any $\epsilon > 0$, there exists $\delta>0$ such that for all $y \in (c - \delta, c + \delta)$, $f(y) > M - \epsilon$. Therefore $[\int_a^b f^n(x) dx]^{\frac{1}{n}} \ge [\int_{c - \delta}^{c + \delta} f^n(x) dx]^{\frac{1}{n}} > (2\delta)^{\frac{1}{n}} (M - \epsilon)$. As $n \to \infty$, we see that the limit is bounded below by $M - \epsilon$. We also obviously have $(b-a)^{\frac{1}{n}} M \ge [\int_a^b f^n(x) dx]^{\frac{1}{n}}$. As $n \to \infty$, the limit is thus bounded above by $M$. Since $\epsilon$ was arbitrary you can use the squeeze theorem to conclude that the limit is $M$. An essentially identical argument works if $c$ is an endpoint of $[a,b]$.

Answer 2

If $f=0$ the result is clear, so assume that $f \neq 0$.

Let $M=\max f$ and $I_\epsilon = \{ x \in [a,b] | f(x) > M-\epsilon \}$. Note that $m I_\epsilon >0$.

Then $\int f^n \ge (M-\epsilon)^n m I_ \epsilon$ and so $\sqrt[n]{ \int f^n } \ge (M-\epsilon) \sqrt[n]{ I_\epsilon }$ and so $\liminf_n \sqrt[n]{ \int f^n } \ge (M-\epsilon)$ from which we have $\liminf_n \sqrt[n]{ \int f^n } \ge M$.

Similarly, we have $m([a,b]) M^n \ge \int f^n$ and so $\sqrt[n]{m([a,b])} M \ge \sqrt[n]{ \int f^n } $. Taking limits gives $M \ge \limsup \sqrt[n]{ \int f^n } $ from which the answer follows.