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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?

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up vote 2 down vote accepted

Yes. Let $x, y \in U$. Let $z = \lambda x + (1 - \lambda) y$, for $\lambda \in [0,1]$. Let $v_z$ be a subgradient for $z$.

Then $$f(x) \geq f(z) + v_z \cdot (x - z) = f(z) + v_z \cdot \left(x - ( \lambda x + (1 - \lambda) y)\right) $$ $$= f(z) + (1 - \lambda) v_z \cdot (x - y).$$ Similarly, $$f(y) \geq f(z) - \lambda v_z \cdot (x - y).$$

Multiplying the first inequality by $\lambda$ and the second by $1 - \lambda$ and adding the two, we obtain

$$\lambda f(x) + (1 - \lambda)f(y) \geq f(z),$$ proving that $f$ is convex.

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