How can we show that if the function $y=f(x)$ is convex in $0\leq x<\infty$, then the points $a_{n}=f(n)$ form a convex sequence.
To show if function $f(x), 0\leq x<\infty$ is convex then $a_{n}=f(n)$ is convex sequence.
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real-analysis
convex-analysis
1 Answers
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Is it true that $$f(n+1)\leq\frac{1}{2}\left(f(n)+f(n+2)\right)\,\,?$$Hint: take $\,\,x_1:=n\,\,,\,x_2:=n+2\,$ and apply convexity of $f$:$$f\left(\frac{x_1+x_2}{2}\right)\leq\frac{1}{2}\left(f(x_1)+f(x_2)\right)$$
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0Yes it is true. – 2012-06-05
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0There...instead of $\,f(n)\,$ and etc. put $\,a_n\,$ and etc. and you get the definition of convex sequence...:) – 2012-06-05
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0A function f is said to be convex on $(a,b)$ if $$ f(tx_{1}+(1-t)x_{2})\leq tf(x_{1})+(1-t)f(y_{1})$$, where $0
and $x,y\in (a,b)$. As you say if we take $x_{1}=n, x_{2}=n+2$ and take $t=\frac{1}{2}$, we get, $$ f(\frac{1}{2}n+(1-\frac{1}{2})(n+2))\leq \frac{1}{2}f(n)+(1-\frac{1}{2}f(n+2)$$ $$\therefore f(n+1) \leq \frac{1}{2}(f(n)+f(n+2))$$. So i think i show what you suggest. So from this what we say? – 2012-06-05 -
0Do what I wrote in my last comment...and read carefully and understand the definition of *convex sequence* – 2012-06-05
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0A sequence $(a_{n})$ is said to be convex sequence if $(\Delta a_{n})$ is decreasing or $\Delta^{2}a_{n}\geq 0, \forall n\geq 0$. It means $$\Delta^{2} a_{n}\geq 0$$ $$\Rightarrow a_{n+1}\leq \frac{1}{2}(a_{n}+a_{n+2})$$ Am i right? – 2012-06-05
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0Yup, exactly that. – 2012-06-05
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0Thanku so much! And it means to write the proof of given result i have to show first$$ f(n+1)\leq \frac{1}{2}(f(n)+f(n+2)).$$ – 2012-06-05