# Problem 1 (Jensen’s function)

កំនត់អនុគមន៍ $f(x)$ ជាអនុគមន៍ជាប់ចំពោះគ្រប់ $x$ ហើយផ្ទៀងផ្ទាត់ៈ
$\displaystyle f\biggl(\frac{x+y}{2}\biggl)=\frac{f(x)+f(y)}{2}$
សំរាយបញ្ជាក់
របៀបទី 1 post by van khea
យើងពិនិត្យមើលករណី $x=y$ នោះសមភាពកើតមានឡើង។ ដូចនេះយើងពិនិត្យករណី $x\neq y$
សមីការខាងលើសមមូលនឹងៈ
$\displaystyle (f(\frac{x+y}{2})-f(x))-(f(y)-f(\frac{x+y}{2}))=0$
$\displaystyle \Leftrightarrow \frac{y-x}{2}\biggl(\frac{f(\frac{x+y}{2}-f(x)}{\frac{x+y}{2}-x}-\frac{f(y)-f(\frac{x+y}{2})}{y-\frac{x+y}{2}}\biggl)=0$
តាមទ្រឹស្ដីបទ $Lagrange$ នោះយ៉ាងតិចមានចំនួន $\displaystyle c_1\in (x, \frac{x+y}{2}); \& c_2\in (\frac{x+y}{2}, y)$ ដែលធ្វើអោយៈ
$\displaystyle \frac{f(\frac{x+y}{2})-f(x)}{\frac{x+y}{2}-x}=f'(c_1)$
$\displaystyle \frac{f(y)-f(\frac{x+y}{2})}{y-\frac{x+y}{2}}=f'(c_2)$
ដូចនេះយើងបានៈ
$\displaystyle \frac{y-x}{2}(f'(c_1)-f'(c_2))=0$
$\displaystyle \Rightarrow f'(c_1)=f'(c_2)$
តាមទ្រឹស្ដីបទ $Rolle$ នាំអោយមាន $t\in (c_1, c_2)$ ដែលធ្វើអោយ $f'(t)=a; a=const$
$\Rightarrow f(t)=at+b$
ដូចនេះយើងបាន $f(x)=ax+b; a, b=const$
របៀបទីពីរ post by (ceojimmyps)
មាន $\displaystyle f(\frac{x+y}{2})=\frac{f(x)+f(y)}{2}$
យក $y=0$
$\displaystyle f(\frac{x}{2})=\frac{f(x)+f(0)}{2}$
ជំនួស $x$ ដោយ $x+y$
$\displaystyle f(\frac{x+y}{2})=\frac{f(x+y)+f(0)}{2}$
$\displaystyle \Rightarrow \frac{f(x)+f(y)}{2}=\frac{f(x+y)+f(0)}{2}$
$f\left( x+y \right)+f\left( 0 \right)=f\left( x \right)+f\left( y \right)$
$f\left( x+y \right)-f\left( 0 \right)=\left( f\left( x \right)-f\left( 0 \right) \right)+\left( f\left( y \right)-f\left( 0 \right) \right)$
តាង $g\left( x \right)=f\left( x \right)-f\left( 0 \right)$
$g\left( x+y \right)=g\left( x \right)+g\left( y \right)$ ជា​អនុគមន៍ $Cauchy$
នោះ $g\left( x \right)=ax$
ដូចនេះ $f\left( x \right)=ax+b\,,\,\,b=f\left( 0 \right)$

### 3 Responses to Problem 1 (Jensen’s function)

1. ceojimmyps says:

មាន $f\left( \frac{x+y}{2} \right)=\frac{f\left( x \right)+f\left( y \right)}{2}$
យក $y=0$
$f\left( \frac{x}{2} \right)=\frac{f\left( x \right)+f\left( 0 \right)}{2}$
ជំនួស $x$ ដោយ $x+y$
$f\left( \frac{x+y}{2} \right)=\frac{f\left( x+y \right)+f\left( 0 \right)}{2}$
$\frac{f\left( x \right)+f\left( y \right)}{2}=\frac{f\left( x+y \right)+f\left( 0 \right)}{2}$
$f\left( x+y \right)+f\left( 0 \right)=f\left( x \right)+f\left( y \right)$
$f\left( x+y \right)-f\left( 0 \right)=\left( f\left( x \right)-f\left( 0 \right) \right)+\left( f\left( y \right)-f\left( 0 \right) \right)$
តាង $g\left( x \right)=f\left( x \right)-f\left( 0 \right)$
$g\left( x+y \right)=g\left( x \right)+g\left( y \right)$ ជា​អនុគមន៍Cauchy
នោះ $g\left( x \right)=ax$
ដូចនេះ $f\left( x \right)=ax+b\,,\,\,b=f\left( 0 \right)$

2. ceojimmyps says:

សន្មត $f$ មាន​ដេរីវេ
ដេរីវេ​ធៀប$x$ : $\frac{1}{2}f'\left( \frac{x+y}{2} \right)=\frac{1}{2}f'\left( x \right)$
ដេរីវេធៀប$y$ : $\frac{1}{2}f'\left( \frac{x+y}{2} \right)=\frac{1}{2}f'\left( y \right)$
ផ្ទឹម​សមីការ​ទាំងពីរ ទាញ​បាន:
$f'\left( x \right)=f'\left( y \right)$
$\int{f'\left( x \right)dx}=\int{f'\left( y \right)dx}\,,\,\,f'(y)=a$
$f\left( x \right)=ax+b\,\,,\,\,b\in \mathbb{R}$

មិន​បាច់​គួរ​សម​ទេ​បង គ្នា​ឯង​ទេ! (I’m rainy :D)