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I am trying to show that if a function $f$ defined in $\mathbb R^n$ is differentiable and convex then $f(y)-f(x)\ge \nabla f(x)(y-x).$ for each $x,y\in\mathbb R^n$

Using differentiability of $f$ I have got $f(y) = f(x+(y-x)) = f(x)+\nabla f(x)(y-x) + o(y-x)$. How to continue?

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It should be $\nabla f(x)$. It remains one. – 1015 Mar 2 '13 at 15:10
up vote 5 down vote accepted

Let $f:\mathbb{R}\longrightarrow \mathbb{R}$ differentiable and convex first. For every $x> y$ and every $t\neq 0$, convexity yields $$ f(tx+(1-t)y)\leq tf(x)+(1-t)f(y)\quad\Leftrightarrow\quad f(y+t(x-y))-f(y)\leq t(f(x)-f(y). $$ So $$ \frac{ f(y+t((x-y))-f(y)}{t(x-y)}\leq \frac{f(x)-f(y)}{x-y} $$ for all $t\neq 0$. Letting $t$ tend to $0$, this entails $$ f'(y)\leq \frac{f(x)-f(y)}{x-y}\quad\Leftrightarrow\quad f(x)-f(y)\geq f'(y)(x-y). $$ for all $x>y$. In the case $x<y$, one follows the same steps reversing the inequality twice.

Now in the general case, fix $x\neq y$ and consider the function $g:\mathbb{R}\longrightarrow\mathbb{R}$ $$ g:t\longmapsto f(y+t(x-y)). $$ Then $g$ is convex (check) and differentiable so in particular $$ g(1)-g(0)\geq g'(0)(1-0)=g'(0). $$ Now by the chain rule $$ g'(t)=\nabla f(y+t(x-y))(x-y)\quad\Rightarrow \quad g'(0)=\nabla f(y)(x-y). $$ And $g(1)=f(x)$, $g(0)=f(y)$, so $$ f(x)-f(y)\geq \nabla f(y)(x-y). $$

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The graph of a convex function is above any tangent plane, and $$ L(y) = f(x_0) + \nabla f(x_0)(y-x_0) $$ is the tangent plane in the point $x_0$...

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I think that the question is precisely about that: prove that the graph of a convex function is above any tangent plane. Given that the most common definition of convex (real-valued) is $f(tx+(1-t)y)\leq tf(x)+(1-t)f(y)$. – 1015 Mar 2 '13 at 16:06

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