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I am having a bit of trouble solving the following convexity problem:

Let $f : X \to (-\infty,+\infty)$ be convex and let $\alpha \in \mathbb{R}$. Show that the sublevel 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.

Rodrigo de Azevedo
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Jeremy
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1 Answers1

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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

CKM
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