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How to prove a Hyperplane and its associated half spaces as convex sets.

I know the convexity condition that if $x,y$ belongs to convex set, then their linear combination should also lie in the set. (Linear combination such that their coefficients are positive and sum to $1$). In this case, it is given $x\in \mathbb{R}^n$. I 'm confused in taking coefficients. For the case of $\mathbb{R}^2$, we can take any $(t, 1-t)$ such that $t<1$. I just couldn't start the proof in this case.

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if $x, y$ in $H$, then $$\sum_i a_i x_i = \sum_i a_i y_i = b$$

Now assume $z= tx+ (1-t)y$. Then \begin{eqnarray} \sum_i a_iz_i &=& \sum_i (ta_i x_i + (1-t)a_iy_i )\\ &=& t \sum_i a_i x_i + (1-t) \sum_i a_i y_i \\ &=& tb + (1-t)b = b \end{eqnarray}

A similar calculation applies to the halfspaces, which are defined by $$ \{x:\,\sum_i a_i x_i \ge b\}$$ resp $$ \{x:\, \sum_i a_i x_i \le b\}$$

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  • $\begingroup$ (i used $a$ instead of $\alpha$ and $b$ instead of $\beta$) $\endgroup$ – Thomas Feb 10 '18 at 12:29

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