O inegalitate integrala implica f=0
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O inegalitate integrala implica f=0
Fie \( f:[0,\infty)\to [0,\infty) \) o functie crescatoare, astfel incat \( f(x)\geq\int_0^x f(e^t)dt,\forall x\geq 0 \). Sa se arate ca \( f(x)=0,\forall x\geq 0 \).
An infinite number of mathematicians walk into a bar. The first one orders a beer. The second orders half a beer. The third, a quarter of a beer. The bartender says “You’re all idiots”, and pours two beers.
\( \int_{0}^{x} f(e^t)dt= \int_{0}^{\ln x} f(e^t)dt + \int_{\ln x}^{x} f(e^t) dt \geq \int_{0}^{\ln x} f(e^t)dt + (x-\ln x) f(x) \) (functia \( f \) este crescatoare).
Pentru \( x>1, \) \( \int _{0}^{\ln x} f(e^t) dt \geq 0 \Rightarrow \int_{0}^{x} f(e^t) dt \geq (x-ln x) f(x)
\) si, conform ipotezei, rezulta ca \( f(x)\geq (x-\ln x) f(x), \forall x>1\Leftrightarrow f(x)(1-x+\ln x) \geq 0, \forall x> 1 \). Daca presupunem ca exista \( c\in \mathbb{R} \) cu \( f(c) >0 \Rightarrow f(x)>0 ,\forall x\geq c \Rightarrow 1-x+\ln x \geq 0, \forall x\geq c. \) (fals). Deci \( f=0 \).
Pentru \( x>1, \) \( \int _{0}^{\ln x} f(e^t) dt \geq 0 \Rightarrow \int_{0}^{x} f(e^t) dt \geq (x-ln x) f(x)
\) si, conform ipotezei, rezulta ca \( f(x)\geq (x-\ln x) f(x), \forall x>1\Leftrightarrow f(x)(1-x+\ln x) \geq 0, \forall x> 1 \). Daca presupunem ca exista \( c\in \mathbb{R} \) cu \( f(c) >0 \Rightarrow f(x)>0 ,\forall x\geq c \Rightarrow 1-x+\ln x \geq 0, \forall x\geq c. \) (fals). Deci \( f=0 \).