- #1
kingwinner
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Homework Statement
Let S C Rd be open and convex.
Let f be C1(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)
Homework Equations
Strictly convex functions
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!