May anyone help me on this problem? If f is differentiable and convex ,$\forall$ x,x+v $\in$ $\mathbb{R}^{m}$ $\iff$ $ f(x+v) \geq f(x)+\nabla f(x) \cdot v$?
F differential and convex
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real-analysis
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0Please, share your thoughts on this problem – 2017-02-02
1 Answers
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Consider $tx+(1-t)y$, $t \in [0,1]$. Take a look at $$ g(t)=f(tx+(1-t)y). $$ We know get that $$ g(t)=f(tx+(1-t)y)=g(0)+ g'(0)t+o(t)=f(y)+t\nabla f(x)(x-y)+o(t) $$ Using convexity, we now get for $t \neq 0$ (why can we do this?): $$ f(y)+t\nabla f(y)(x-y)+o(t) \leq tf(x)+(1-t)f(y) \\ -t f(y)+t\nabla f(y)(x-y)+o(t) \leq tf(x) \\ f(y)+\nabla f(y)(x-y)+\frac{o(t)}{t} \leq f(x) \\ \nabla f(y)(x-y)+\frac{o(t)}{t} \leq f(x)-f(y) \\ $$ Now letting $t \to 0$ gives us the result. Try now to re-write it for $x$ and $x+v$.