Show that every polytope is bounded

Question

The definition of polytope is the convex hull of a finite set. Thus: $$ \parallel\sum_j\lambda _j x_j\parallel\le\sum_j\lambda_j\parallel x_j\parallel\le\sum_j\lambda_j\max_j \parallel x_j\parallel=M\sum_j\lambda_j=M $$ where $M=\max_j \parallel x_j\parallel$

And how can I conclude from above that every polytope is bounded?

What is $x^*$ here? Since $0\le\lambda_i\le1$, you might also simply bound by $\left|\sum_j\lambda_jx_j\right|\le \sum_j|\lambda_j|,|x_j|\le \sum_j |\lambda_j|M=M\sum_j \lambda_j=M$ where $M=\max{,|x_j|\colon 1\le j\le n,}$. — Hagen von Eitzen, Nov 27 '13 at 13:38
I was using x* to indicate maximum value of x but now I modified it to $\max_j x_j$ hope this is clear now — dresden, Nov 27 '13 at 13:46

score 1 · Accepted Answer · answered Nov 27 '13 at 13:32

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It looks correct. And it seems to me that you proved your assertion. You assumed $M = \max_{1 \le i \le m} \|x_i\|$, of course.

answered Nov 27 '13 at 13:32

Stephen Montgomery-Smith

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Is it because the maximum of x lies inside convex hull and hence it is bounded (by the maximum value) ? – dresden Nov 27 '13 at 13:33
Not really. It is because the maximum of a finite number of real numbers is itself a real number, and hence bounded. – Stephen Montgomery-Smith Nov 27 '13 at 13:35
And where you had the superscript *, that part makes no sense. Just leave it out of the chain of inequalities. – Stephen Montgomery-Smith Nov 27 '13 at 13:36

Show that every polytope is bounded

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