I am having a bit of trouble solving the following convexity problem:
Let $f:X\rightarrow (-\infty,+\infty)$ be convex and let $\alpha\in\mathbb{R}$. Show that the sub level set $c= \lbrace x\in X:f(x)\leq\alpha\rbrace$ is convex.
Given that $f(x)$ is convex we know
$$f((1-\lambda)x_1+\lambda x_2)\leq (1-\lambda)f(x_1)+\lambda f(x_2)$$
for $x_1,x_2\in X, 0\leq\lambda\leq1$.
However, I am having trouble using this to show what the question is asking to be shown. Any hints or suggestions for this question is much appreciated.
Thank you.
Best Answer
You need to show that $f(x) \leq \alpha$, where $x$ is chosen as convex combination of the points $x_1$ and $x_2$, i.e. $x=(1-\lambda)x_1 + \lambda x_2$. Now, \begin{align*} f(x)& =f((1-\lambda)x_1 + \lambda x_2) \\ &\leq (1-\lambda)f(x_1) + \lambda f(x_2) \;\; \text{(using convexity of $f(\cdot)$)}\\ & \leq(1-\lambda)\alpha + \lambda \alpha \;\;\text{(using epigraph definition)}\\ & = \alpha \end{align*} Thus, epi $f$ is convex. q.e.d