= Supporting inequality for a differentiable convex function
{title2=$f(y)\ge f(x)+\nabla f(x)^{\mathsf T}(y-x)$}
For a differentiable real-valued <function> on an open <convex set>, <convexity> is equivalent to the displayed inequality for all $x,y$ in its domain. To prove the forward direction, convexity gives $[f(x+t(y-x))-f(x)]/t\le f(y)-f(x)$ for $0<t\le1$. Take the <limit> as $t\downarrow0$ using first-order <Taylor expansion>.
Conversely, put $z=(1-t)x+ty$ and apply the supporting inequality at $z$ to both $x$ and $y$. Multiply the two inequalities by $1-t$ and $t$ and add them. The gradient terms cancel because $(1-t)(x-z)+t(y-z)=0$, leaving $f(z)\le(1-t)f(x)+tf(y)$. This is the defining <convexity> inequality. The endpoint values $t=0,1$ hold as equalities.
Back to article page
