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题：设$\,f\,$是$\,\mathbb{R}\,$上的凸函数, 则aCN6G"
\(\quad(1)\quad\displaystyle\lim_{x\to\infty}\frac{x}{x}\,\)存在(可为$\infty$).[
\(\quad(2)\quad\)若\(\displaystyle\;\lim_{x\to\pm\infty}\frac{f(x)}{x}=0,\;\)则\(\,f\,\)是常数.W5
证：设\(\,x\cdot h>0,\;\)则\(\,\small\dfrac{f(x+h)-f(0)}{x+h}-\dfrac{f(x)-f(0)}{x}\)'oM>E5
\(\qquad=\small\dfrac{f(x+h)}{x+h}+\dfrac{hf(0)}{x(x+h)}-\dfrac{f(x)}{x}\quad\)(注意\(\frac{x}{x+h}{\small(x+h)+}\frac{h}{x+h}{\scriptsize 0}\small=x\))hc=
\(\qquad=\small\dfrac{1}{x}\big(\dfrac{x}{x+h}f(x+h)+\dfrac{h}{x+h}f(0)-f(x)\big)\)C'
\(\therefore\quad g(x)=\small\dfrac{f(x)-f(0)}{x}\,\)单调不减,\(\;\displaystyle\lim_{x\to\infty}{\small\frac{f(x)}{x}}=\lim_{x\to\infty}g(x)\le\infty\;\)恒存在.6eJ~"
\(\qquad\)若有某\(\,a\ne 0\,\)使\(\,f(a)\ne f(0)\;\text{i.e.}\,g(a)\ne 0,\;\)则qU
\(\qquad g(a)>0\implies\displaystyle\lim_{x\to\infty}{\small\frac{f(x)}{x}}\ge g(a)>0,\)M
\(\qquad g(a)< 0\implies\displaystyle\lim_{x\to-\infty}{\small\frac{f(x)}{x}}\le g(a)< 0.\)(}rrJp
\(\qquad\)综上\((1),\,(2)\,\)得证.$l!,Fj
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2020/12/29 04:30am 
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