Question: Prove that there does not exist a non-negative continuous function $f:[0,1] \to \mathbb{R}$ such that $\int_0^1f^n(x)dx \to 2 $ as $n \to \infty$.
Attempt:
Case 1: $\sup f(x) \leq 1 \implies \sup f^n(x) \leq 1$.
$\int_0^1f^n(x)dx\leq 1 \not\to 2$ as $n\to \infty$.
Case 2: $1<\sup f(x) \implies 1< \sup f^n(x)$
By the property of continuous functions on closed intervals, $\exists \ c\in[0,1]$ , such that $f(c)=\sup f $ and $\exists \ \delta>0$ such that $\forall x\in[c-\delta,c+\delta]\cap [0,1]$, $1<k \leq f(x)$ for some $k$. Now, if $c$ is interior to $[0,1]$, $\delta$ can be taken such that $[c-\delta,c+\delta]\subset[0,1]$. For the boundary points, we take $[c,c+\delta]$ or $[c-\delta,c]$.
So, $ \int_{c-\delta}^{c+\delta}k^ndx=2\delta k^n \leq \int_0^1f^n(x)dx$.
Fixing $\delta$, we can take $n\to \infty$ which results in $2\delta k^n\to \infty $.
So, in either cases, $\int_0^1f^n(x)dx \not\to 2$ as $n \to \infty$.
Is this correct? Kindly verify.