Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
add comment

1 Answer 1

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.