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Fie \( f:\[0,1]\rightarrow\mathbb{R}^+ \) continua. Demonstrati ca \( \lim_{n\rightarrow\infty}\sqrt[n]{\int_{0}^{1}f^n(x)dx}=max f(x) \), unde max este valoarea maxima a functiei f (evident maximul exista, din teorema lui Weierstrass).

\( f:[0,1]->R_{+} \) continua => f marginita si isi atinge mariginile:
\( min(f)=m\leq f(x) \leq M=max(f) \) oricare ar fi x din [0,1].

\( \sqrt[n]{\int_{0}^{1}f^n(x)dx}\leq\sqrt[n]{M^n}=M \)

Exista \( x_{0} \) a.i. \( f(x_{0})=M \); fie \( \epsilon>0 \) => exista \( \delta>0 \) a.i. \( x\in (x_{0}-\delta, x_{0}+\delta) \) => \( f(x)\in (M-\epsilon, M) \), dar \( x\in [0,1]\Rightarrow \) intersectie si luat pe cazuri esential:
\( \sqrt[n]{\int_{0}^{1}f^n(x)dx}\geq\sqrt[n]{\int_{x_{0}-\delta}^{x_{0}+\delta}f^n(x)dx}\geq\sqrt[n]{\int_{x_{0}-\delta}^{x_{0}+\delta}(M-\epsilon)^ndx}=\sqrt[n]{2\delta}*(M-\epsilon) \)
\( liminf\sqrt[n]{\int_{0}^{1}f^n(x)dx}\geq M-\epsilon\Rightarrow liminf\sqrt[n]{\int_{0}^{1}f^n(x)dx}\geq M \).
Dar \( \sqrt[n]{\int_{0}^{1}f^n(x)dx}\leq \sqrt[n]{M^n}=M \) => \( lim\sqrt[n]{\int_{0}^{1}f^n(x)dx}=M \).