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From: Kaba on 26 Nov 2009 03:30
keith wrote:
> > Thus f is convex [].
>
> I am no mathematician, but is convexity not determined by an inequality?

For some reason my first post didn't appear on the server I use so here
goes again to be sure.

There are inequalities at the start of some rows. However, it seems
newsreaders interpret them as quotes. Reformatting:

Proof:

Let
f_j : R^n -> R: f_j(x) = ||x - p_j||^2
f = max {f_j}

If you accept that each f_j is convex, then:

(1 - t)f(x_1) + tf(x_2) >= (definition of f)
(1 - t)f_j(x_1) + tf_j(x_2) >= (convexity of f_j)
f_j((1 - t)x_1 + tx_2)

Because the previous holds for arbitrary j:

(1 - t)f(x_1) + tf(x_2) >=
max {f_j((1 - t)x_1 + tx_2)} =
f((1 - t)x_1 + tx_2)

Thus f is convex [].

> Furthermore, how can you be certain that the same f_j applies for both
> x_1 and x_2?

Perhaps you mean this step:

(1 - t)f(x_1) + tf(x_2) >= (definition of f)
(1 - t)f_j(x_1) + tf_j(x_2)

Everything's positive, and f(x_i) >= f_j(x_i) by definition.

--
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