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This question is arised when I study the content in the following question entitled by

" lower bound of a special type of convex functions " in here.

Let $f: {\bf R}^n \rightarrow {\bf R}$ be a function in $C^1$ with

$f(sx+(1-s)y)< s f(x) + (1-s) f(y)$, $s\in (0,1)$.

Then $f(y) > f(x) + \nabla f(x) \cdot (y-x)$.

This inequality is followed from the intuition. But I want to know the rigorous proof. Thank you in advance.

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up vote 1 down vote accepted

For $1>t>0$, $$f(x)+ t \nabla f(x) \cdot (y-x) + o(t) = f(x+t(y-x)) <(1-t)f(x)+tf(y) $$ Whence, $$t \nabla f(x) \cdot (y-x) + o(t)<t(f(y)-f(x))$$ Then, divide by $t$. For $t \to 0$, you get your inequality: $$f(y)>f(x)+ \nabla f(x) \cdot (y-x)$$

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Read section 3.1.3 in this book. It is a very standard fact in convex analysis.

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Thanks for sharing the book. – elaRosca Nov 12 '13 at 11:35

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