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August 8th, 2017, 03:42 AM   #1
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Convex function

Quote:
 A function $f : \mathbb{R}^n \to \mathbb{R}$ is called convex if for every pair of vectors $\mathbf{x}, \mathbf{y} \in \mathbb{R}^n$ and every $\lambda \in [0, 1]$, we have, $$f(\lambda \mathbf{x} + (1 - \lambda)\mathbf{y}) \le \lambda f(\mathbf{x}) + (1 - \lambda)f(\mathbf{y})$$
I don't understand the role of $\lambda$ here. I can see that addition is preserved but that's about it. Perhaps I'm being daft. Any insight? Thanks .

August 9th, 2017, 10:02 PM   #2
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Quote:
 Originally Posted by Joppy Perhaps I'm being daft.
Yes very daft Joppy.

The argument of $f$ merely denotes all points between $\mathbf{x}$ and $\mathbf{y}$. Hence if we consider a function whose argument is all of these points, and is less than the points themselves, the function is convex.

 August 16th, 2017, 05:52 AM #3 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 When $\displaystyle \lambda$ is 0, $\displaystyle \lambda x+ (1- \lambda)y= (0)x+ (1)y$ is just y. When $\displaystyle \lambda$ is 1, $\displaystyle \lambda x+ (1- \lambda)y= (1)x+ (0)y$ is just x. Since that is a linear formula, as $\displaystyle \lambda$ goes from 0 to 1, the point moves on the straight line from y to x. Last edited by greg1313; August 16th, 2017 at 07:02 AM.
August 16th, 2017, 03:40 PM   #4
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Quote:
 Originally Posted by Country Boy When $\displaystyle \lambda$ is 0, $\displaystyle \lambda x+ (1- \lambda)y= (0)x+ (1)y$ is just y. When $\displaystyle \lambda$ is 1, $\displaystyle \lambda x+ (1- \lambda)y= (1)x+ (0)y$ is just x. Since that is a linear formula, as $\displaystyle \lambda$ goes from 0 to 1, the point moves on the straight line from y to x.
Thanks . Don't know why it took me so long to see it.

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