TITLE: Show the sub level set is convex
QUESTION [5 upvotes]: 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.

REPLY [8 votes]: 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