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so i am having hard time understanding the idea of coming up with random n-vectors to disprove the superposition equality $f(\alpha x + \beta y) = \alpha f(x) + \beta f(y)$.

or to prove that a function is linear or affine???

can someone explain to me for the following function examples?

  1. $f(x) = \min x_i$
  2. $f(x) = \sum_{i=1}^n|x_i|$
  3. $f(x) = \sum_{i=1}^n|x_{i + 1} - x_i|$
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up vote 1 down vote accepted
  1. $\displaystyle f(x)=\min_{i=\overline{1,n}}x_i$ is neither linear nor affine. Indeed, let $x=(-1,1, 0, \ldots, 0), y=(1,-1, 0, \ldots, 0)\in \mathbb{R}^n$ we have $$ f(x)+f(y)=-1-1=-2\ne 0 = f(x+y), $$ $$ \frac{1}{2}f(x)+\frac{1}{2}f(y)=-\frac{1}{2}-\frac{1}{2}=-1\ne 0 = f[(1/2)x+(1/2)y]. $$

  2. $\displaystyle f(x)=\sum_{i=1}^{n}|x_i|$ is neither linear nor affine. Indeed, let $x=(-1,1, 0, \ldots, 0), y=(1,-1, 0, \ldots, 0)\in \mathbb{R}^n$ we have $$ f(x)+f(y)=2+2=4\ne 0 = f(x+y), $$ $$ \frac{1}{2}f(x)+\frac{1}{2}f(y)=1+1=2\ne 0 = f[(1/2)x+(1/2)y]. $$

  3. $\displaystyle f(x)=\sum_{i=1}^{n}|x_{i+1}-x_i|$ can not be defined since we do not have $x_{n+1}$ in the expression of $x$.
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the last one is from 1 to n-1 – ASROMA Oct 8 '12 at 8:10
and WHAT ABOUT AFFINE? for your example 1 if i take alfa=betta=1/2, the formula hold – ASROMA Oct 8 '12 at 8:16
@Vahe Musinyan: $\displaystyle f(x)=\sum_{i=1}^{n-1}|x_{i+1}-x_i|$ is not a linear function. Indeed, we can choose $$x=(-1, 1, 0, \ldots, 0) y=(1, -1, 0, \ldots, 0)\in \mathbb{R}^n$$. Then we have $$f(x)+f(y)=3+3=6\ne 0=f(x+y)$$. – blindman Oct 8 '12 at 8:20
Are you forgetting about ALFA and BETTA, this is not linear, but can you show whether they are AFFINE or not? – ASROMA Oct 8 '12 at 8:27
@Vahe Musinyan: $\displaystyle f(x)=\sum_{i=1}^{n-1}|x_{i+1}-x_i|$ is not an affine function. We choose $x=(-1, 1, 0, \ldots, 0) y=(1, -1, 0, \ldots, 0)\in \mathbb{R}^n$ and we have $(1/2)f(x)+(1/2)f(y)=(3/2)+(3/2)=3\ne 0=f[(1/2)x+(1/2)y]$. – blindman Oct 8 '12 at 8:32

To prove that $f:\mathbb{R}^n\rightarrow \mathbb{R}$ is not a linear function we only need to choose two points $x,y\in\mathbb{R}^n$ such that $$ f(x+y)\ne f(x)+f(y) $$ or $$ f(\lambda x)\ne \lambda f(x) $$ for some $\lambda$.

To prove that $f:\mathbb{R}^n\rightarrow \mathbb{R}$ is not an affine function we only need to choose two points $x, y\in\mathbb{R}^n$ and $\alpha\in \mathbb{R}$ such that $$ f[\alpha x+ (1-\alpha)y]\ne \alpha f(x)+ (1-\alpha)f(y). $$

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SO non of these exercises are AFFINE? – ASROMA Oct 8 '12 at 8:42
@Vahe Musinyan: Yes. Do you agree with my answers. I highly appreciate your comments. – blindman Oct 8 '12 at 8:44

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