Točno
17. siječnja 2014. 14:52 (10 godine, 6 mjeseci)
Upozorenje: Ovaj zadatak još niste riješili!
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Kliknite ovdje kako biste prikazali rješenje.
![(x,y)](/media/m/c/9/1/c91aec4078b932368ded863349deaec5.png)
![(x,y)](/media/m/c/9/1/c91aec4078b932368ded863349deaec5.png)
![f(\frac{x+y}{2}) \le af(x) + (1-a)f(y)](/media/m/e/3/5/e3552f922a441df30351f130b83f5e31.png)
![(y,x) \implies f(\frac{x+y}{2}) \le af(y) + (1-a)f(x)](/media/m/c/7/4/c741c63ed150b1fe06cc33aad19517c5.png)
Zbrajajući te 2 jednadžbe dobivamo
![f(\frac{x+y}{2}) \le \frac{f(x)+f(y)}{2}](/media/m/2/1/a/21ad8ed62e33d60c2b59c430d554480b.png)
Očito je da je
![f(0)=0](/media/m/2/9/7/297be37ff10e3b2aa5ab3d32c2202c49.png)
![a \in \mathbb{R}_{+}](/media/m/0/9/0/0905b8a27dee1ed13dd459fcb4d20122.png)
![f(a)=a](/media/m/a/1/c/a1cd8712a291c8c1838ac0ad0e45a4ab.png)
![0 \le b \le a](/media/m/f/d/8/fd8c0f3a65fb035d224347c5266fd480.png)
![(a-b, a+b) \implies (a-b) + (a+b) \le f(a-b) + f(a+b) \le (a-b) + (a+b) \implies f(x)=x \forall x \in [0, 2a]](/media/m/1/7/c/17c87269d25bc19bf728e4f1e5c2c4ff.png)
Lako se indukcijom pokaže (isti argument,
![a:=2a](/media/m/3/a/9/3a9be7adc13b1977f0b41e4cf3a79cf6.png)
![f(x)=x \forall x \in \mathbb{R}_{+}](/media/m/1/0/3/10317d64c57af5f823a317abf2fcda4b.png)
Međutim, ta funkcija nije rješenje za
![0<a<\frac{1}{2}](/media/m/a/7/1/a71fb987bf9dec902d08ee70cb25e8fd.png)
![(2, 1)](/media/m/e/b/6/eb68db6f5b8a0b82c4a1b953d09f44d4.png)
Sada znamo da ne postoji
![a \in \mathbb{R}_{+}](/media/m/0/9/0/0905b8a27dee1ed13dd459fcb4d20122.png)
![f(a)=a](/media/m/a/1/c/a1cd8712a291c8c1838ac0ad0e45a4ab.png)
![f(x)<x \forall x \in \mathbb{R}_{+}](/media/m/5/1/b/51bb666a792f2c86483c84377278b4bb.png)
![(2x,0) \implies f(x) \le af(2x) + (1-a)f(0) < 2ax](/media/m/9/6/6/9661c77d809eaace63d3fa2fca2a0a9e.png)
![f(x) < 2ax \forall x \in \mathbb{R}_{+}](/media/m/2/6/9/2693d9ef450fd6c25274a1220d304f1c.png)
![(2x, 0) \implies f(x) \le af(2x) + (1-a)f(0) < 4a^2x](/media/m/a/1/3/a13ad84373b821dffff8d6c3f776ad78.png)
![f(x) < 4a^2x \forall x \in \mathbb{R}_{+}](/media/m/a/b/1/ab1d92d1cb6bdc79c617f686212debf1.png)
Indukcijom se lako pokaže
![f(x)<2^na^nx](/media/m/f/c/e/fce44362b05def7340d88ca5e9f78b61.png)
![n](/media/m/a/e/5/ae594d7d1e46f4b979494cf8a815232b.png)
![2^na^n](/media/m/7/a/b/7ab33b7b5168c7db22f9fa7fb1493f73.png)
![\boxed{ f(x)=0 \forall x \in \mathbb{R}_{\ge 0}}](/media/m/9/a/f/9af895b5572d950bd120b3b8d8534841.png)
Ocjene: (2)
Komentari:
markota, 18. siječnja 2014. 23:46