# Property of convex functions

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?

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 $$f(x)-f(y)\geq \nabla f(y)(x-y).$$
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 $x_0$...
• 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 $f(tx+(1-t)y)\leq tf(x)+(1-t)f(y)$. Commented Mar 2, 2013 at 16:06