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Fie \( f:[0,1]\to\mathbb{R} \) o functie continua. Sa se arate ca

\( \lim_{n\to\infty}\frac{1}{n}\int_0^1f(x)[nx]dx=\int_0^1xf(x)dx \).

Virgil Nicula, Olimpiada Judeteana Bucuresti 1991

Fara a pierde generalitatea presupunem ca \( f \geq 0 \). Avem ca \( nx-1 \leq [nx] \leq nx ,\forall x \in [0,1], \forall n \in N \Rightarrow (nx-1)f(x) \leq f(x)[nx] \leq f(x)nx\Rightarrow \)
\( \Rightarrow \frac{(nx-1)f(x)}{n} \leq \frac{f(x)[nx]}{n} \leq xf(x) \). Integram, trecem la limita si conform criteriului clestelui obtinem concluzia, adica \( \lim_{n\to\infty}\frac{1}{n}\int_0^1f(x)[nx]dx=\int_0^1xf(x)dx \).

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Am luat acest caz ca sa nu schimb semnul inegalitatii...dar in fine sa zicem ca am comis o greseala tinand cont ca functia poate lua si valori negative... Probabil ca trebuia o suma Riemann, o teorema de medie... Ma mai gandesc! Multumesc pt sesizare!

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Virgil Nicula wrote:O problema similara (own). Daca \( f,\ g\ :[0,1]\to\mathbb{R} \), \( g \) este continua si \( f \) este cu variatie
marginita (Lipschitz), atunci \( \lim_{n\to\infty}\ \int_0^1f(x)\cdot g\left(\frac 1n\cdot [nx]\right)\ \mathrm {dx}=\int_0^1f(x)\cdot g(x)\ \mathrm {dx} \).

Demonstratie. \( \int_0^1f(x)g\left(\frac 1n[nx]\right)\mathrm {dx}= \) \( \sum_{k=0}^{n-1}g\left(\frac kn\right)\int_{\frac kn}^{\frac {k+1}{n}}f(x)\mathrm {dx}= \) \( \frac 1n\sum_{k=0}^{n-1}g\left(\frac kn\right)f(c_k) \) , unde \( 0\le\frac kn\le c_k\le\frac {k+1}{n}\le 1 \) , \( k\in\overline {0,n-1} \). Asadar, \( \left|\int_0^1f(x)g\left(\frac 1n[nx]\right)\mathrm {dx}-\frac 1n\sum_{k=0}^{n-1}f\left(\frac kn\right)g\left(\frac kn\right)\right|= \) \( \left|\frac 1n\sum_{k=0}^{n-1}g\left(\frac kn\right)\left[f\left(c_k\right)-f\left(\frac kn\right)\right]\right|\le \) \( \frac An\sum_{k=0}^{n-1}\left|f\left(c_k\right)-f\left(\frac kn\right)\right|\le \) \( \frac {AB}{n}\sum_{k=0}^{n-1}\left|c_k-\frac kn\right|\le \) \( \frac {AB}{n}\sum_{k=0}^{n-1}\frac 1n=\frac {AB}{n} \) , unde \( A=\max_{0\le x\le 1}g(x) \)

si exista \( B \) ca pentru orice \( x \) , \( y \) din \( [0,1] \) avem \( |f(x)-f(y)|\le B|x-y| \). In concluzie,

\( \lim_{n\to\infty}\left|\int_0^1f(x)g\left(\frac 1n[nx]\right)\mathrm {dx}-\frac 1n\sum_{k=0}^{n-1}f\left(\frac kn\right)g\left(\frac kn\right)\right|=0 \), ceea ce inseamna \( \lim_{n\to\infty}\int_0^1f(x)g\left(\frac 1n[nx]\right)\mathrm {dx}=\lim_{n\to\infty} \frac 1n\sum_{k=0}^{n-1}f\left(\frac kn\right)g\left(\frac kn\right) \), adica \( \lim_{n\to\infty}\ \int_0^1f(x)\cdot g\left(\frac 1n\cdot [nx]\right)\mathrm {dx}=\int_0^1f(x)\cdot g(x)\mathrm {dx} \).

Virgil Nicula wrote: Fie \( f:[0,1]\to\mathbb{R} \) o functie continua. Sa se arate ca \( \lim_{n\to\infty}\frac{1}{n}\int_0^1f(x)[nx]dx=\int_0^1xf(x)dx \).

Demonstratie. Vom considera "translatia" nenegativa \( f-m \) a functiei \( f \), unde \( m=\min_{0\le x\le 1}f(x) \)

(care exista deoarece functia \( f \) este continua) sau oricare alt minorant al functiei \( f \) .

Asadar, din inmultirea lantului \( x-\frac 1n<\frac {[nx]}{n}\le x \) cu functia nenegativa \( f-m \) obtinem :

\( \left[f(x)-m\right]\cdot\left(x-\frac 1n\right)<\left[f(x)-m\right]\cdot \frac {[nx]}{n}\le x\cdot\left[f(x)-m\right] \) \( \Longleftrightarrow \)
\( \left\|\ m\cdot\frac {[nx]}{n}-\frac 1n\cdot f(x)-mx+\frac mn<f(x)\cdot\frac {[nx]}{n}-xf(x)\le m\cdot\frac {[nx]}{n}-mx\ \right\|\ \ \ \int_0^1\ \) \( \Longleftrightarrow \)
\( m\cdot\int_0^1\frac {[nx]}{n}\mathrm {dx}-\frac 1n\cdot\int_0^1f(x)\mathrm {dx}-\frac m2+\frac mn<\int_0^1f(x)\frac {[nx]}{n}\mathrm {dx}-\int_0^1xf(x)\mathrm {dx}\le m\cdot\int_0^1\frac {[nx]}{n}\mathrm {dx}-\frac m2 \) .
Pentru \( n\rightarrow\infty \) in lantul precedent obtinem : \( \lim_{n\to\infty}\ \left[\ \int_0^1f(x)\frac {[nx]}{n}\mathrm {dx}-\int_0^1xf(x)\mathrm {dx}\ \right]=0 \) deoarece
\( \lim_{n\to\infty}\ \int_0^1\frac {[nx]}{n}\mathrm {dx}=\frac 12 \) (se arata usor !). In concluzie, \( \lim_{n\to\infty}\frac{1}{n}\int_0^1f(x)[nx]dx=\int_0^1xf(x)dx \) .