Maths and Economics

Question

The function $f:\mathbb{R} \to \mathbb{R}$ defined by $$ f(x)=\begin{cases} 1\,,\qquad &\text{if $|x| \le 1$}\\ 2x\,, &\text{if $|x|>1$}. \end{cases} $$ Is this function convex and contionous on a) [-1, infinity) b) [-1,1] c) [-infinity,1) d)none what I think is this function is convex

Answer 1

As said in the comment, convexity implies Lipschitzianity which in turn implies continuity. Then you have $$\text{convexity} \implies \text{continuity}$$ and $$\text{not continuous} \implies \text{not convex}$$ Clearly it can happen that your given function is continuous but not convex.

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Is your function continuous? Given its definition, can it have discontinuity points? If so, where?

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Or is it convex? For any $x, y$ and $\lambda \in \mathbb{R}$ does

$$f(\lambda x +(1-\lambda)y) \le \lambda f(x)+(1-\lambda)f(y)$$

hold?

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As a starter, can you answer at least one of those two questions?