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, we obtain

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

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.