Suppose that $f, g: \mathbb R^n \to \mathbb R $ are $C^1$ convex functions. Show that $C = ${$\mathbf x \mid g(\mathbf x) \leq 0$} is a convex subset of $\mathbb R^n$. Show that if $\nabla f(\mathbf x^*) + \lambda \nabla g(\mathbf x^*)=0 $ with $\lambda \geq 0$, $g(\mathbf x ^*) =0$ then $f(\mathbf x) \geq f(\mathbf x^*)$ for all $\mathbf x \in C$.
To be a convex function it must satisfy the following inequality: $f[\lambda x_1 +(1-\lambda)x_2] \leq \lambda f(x_1) + (1-\lambda)f(x_2)$ and in my case $x_1$ will be $x$ and $x_2$ will be $x^*$
I have the two equations for f and g because they are convex functions: $$f[\lambda x +(1-\lambda)x^*] \leq \lambda f(x) + (1-\lambda)f(x^*)\text{ and }g[\lambda x +(1-\lambda)x^*] \leq \lambda g(x) + (1-\lambda)g(x^*)$$
I'm not sure how to use these two equations to show what I want to show. Any advice would be great!
The condition $\nabla f(x^\star) + \lambda \nabla g(x^\star) = 0$ tells us that $x^\star$ is a minimizer of $f(x) + \lambda g(x)$.
If $x \in C$, then \begin{align*} f(x) & \geq f(x) + \lambda g(x) \\ & \geq f(x^\star) + \lambda g(x^\star) \\ &= f(x^\star). \end{align*} (In the first step, I used the fact that $\lambda g(x) \leq 0$, which follows from $\lambda \geq 0$ and $g(x) \leq 0$.)
To see that $C$ is convex, assume that $x_1, x_2 \in C$ and $0 < \theta < 1$. Then $g(\theta x_1 + (1-\theta) x_2) \leq \theta g(x_1) + (1-\theta) g(x_2) \leq 0$, so $\theta x_1 + (1-\theta) x_2 \in C$.