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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. $f(x) = \sum_{i=1}^n|x_{i + 1} - x_i|$

2 Answers 2

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  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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    @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 $(1/2)f(x)+(1/2)f(y)=(3/2)+(3/2)=3\ne 0=f[(1/2)x+(1/2)y]$.2012-10-08
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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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    @Vahe Musinyan: Yes. Do you agree with my answers. I highly appreciate your comments.2012-10-08