(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Let S C R^{d}be open and convex.

Let f be C^{1}(S).

Prove that if f is strictly convex, then f(y) > f(x) + grad f(x) o (y-x) for all x,y in S such that x≠y.

(note: "o" means dot product)

2. Relevant equations

Strictly convex functions

3. The attempt at a solution

Suppose f is stirctly convex.

Then for all x,y in S such that x≠y, 0<s<1,

f[(1-s)x+sy] < (1-s)f(x) + s f(y)

=> f[x+s(y-x)] < f(x) + s[f(y)-f(x)]

=> [f(x+s(y-x)) - f(x)] /s < f(y)-f(x)

=> lim [f(x+s(y-x)) - f(x)] /s≤lim [f(y)-f(x)]

s->0------------------------s->0

(recall from calculus, we know that a strict inequality, IN THE LIMIT, becomes a loose inequality. And that's why I must write ≤ when I take the limit)

=> grad f(x) o (y-x) ≤ f(y)-f(x)

=> f(y)≥f(x) + grad f(x) o (y-x) for all x,y in S such that x≠y.

BUT what I actually want to prove is >, not ≥. How can I rigorously justify that it should be >? I'm really running out of ideas regarding proving this subtle point.

Any help will be much appreciated!

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# Strictly Convex Functions

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