It is known that If $f:U→ R$ is a real-valued convex function defined on a convex open set in the Euclidean space $R^n$, a vector v in that space is called a subgradient at a point $x_0$ in $U$ if for any $x$ in U one has $f(x)-f(x_0)\geq v\cdot(x-x_0)$

What if for function $f$, at any $x_0$ I can find $v$, such that $f(x)-f(x_0)\geq v\cdot(x-x_0)$ for any $x$, does this show that $f$ is convex?