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For $\alpha, \beta$ define $f_{\alpha, \beta}$: $\mathbb{R} \rightarrow \mathbb{R}$ by

  • $f_{\alpha, \beta} (x) = \alpha x + \beta$ if $x < -2$
  • $f_{\alpha, \beta} (x) = 2x^2 + x + 3$ if $x \geq -2$

Give all $\alpha \in \mathbb{R}$, and $\beta \in \mathbb{R}$ for which $f_{\alpha, \beta}$ is convex.

Give all $\alpha \in \mathbb{R}$, and $\beta \in \mathbb{R}$ for which $f_{\alpha, \beta}$ is quasiconvex

Answers:

For all $\alpha, \beta \in \mathbb{R}$ sich that $\beta - 2\alpha = 9$ and $\alpha \leq -7$, $f_{\alpha, \beta}$ is convex

For all $\alpha, \beta \in \mathbb{R}$ such that $\beta - 2\alpha \geq 9$ and $\alpha \leq 0$, $f_{\alpha, \beta}$ is quasiconvex

These are the answers, I know the function f is convex iff the set Epi subset $\mathbb{R}^n \times \mathbb{R}$ is convex

However, the fact that $\alpha \leq -7$ confuses me, since I do not know how to get there. I am missing a simple step. could please someone tell me how to come up with the conclusion why it the answers are like this..

Second example: For $\alpha, \beta$ define $f_{\alpha, \beta}$: $\mathbb{R} \rightarrow \mathbb{R}$ by

  • $f_{\alpha, \beta} (x) = \alpha x + \beta$ if $x < 0$
  • $f_{\alpha, \beta} (x) = 4x^3 + 2x + 5$ if $x \geq 0$

Give all $\alpha \in \mathbb{R}$, and $\beta \in \mathbb{R}$ for which $f_{\alpha, \beta}$ is convex.

  1. $f$ is continuous
  2. $f'$ is increasing

These requirements amount to:

  1. $(\alpha x+\beta)_{x=0} = (4x^3+2x+5)_{x=0} = 5$ Hence, $\beta = 5$
  2. $\alpha \leq ((4x^3+2x+3)')_{x=0} = (8x^2+2)_{x=0} = 2$. Hence, $\alpha \leq 2$

Give all $\alpha \in \mathbb{R}$, and $\beta \in \mathbb{R}$ for which $f_{\alpha, \beta}$ is quasiconvex

For quasiconvexity, considering that $f$ has a (global?) minimum at $x=0$

, you need that

  1. $f$ is decreasing on $(-\infty,0)$
  2. $f$ is increasing on $(0, \infty)$

Property 1 requires $\alpha\le 0$ and $(\alpha x+\beta)_{x=0} \ge (4x^3+2x+5)_{x=0} = 5$. And also $\alpha > 0$ with $\beta \leq 5$ (since, $\alpha \le 2$ is still convex at $\beta = 5$.

Property 2 is automatically fulfilled by the given $x \ge 0$.

Gauss
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1 Answers1

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You have $f$ that is piecewise defined, with differentiable pieces. For convexity of such a function, you need to make sure that:

  1. $f$ is continuous
  2. $f'$ is increasing

These requirements amount to:

  1. $(\alpha x+\beta)_{x=-2} = (2x^2+x+3)_{x=-2} = 9$
  2. $\alpha \le ((2x^2+x+3)')_{x=-2} = (4x+1)_{x=-2} = -7$.

For quasiconvexity, considering that $f$ has a local minimum at $x=-1/4$ (which would have to be global under quasiconvexity), you need that

  1. $f$ is decreasing on $(-\infty,-1/4)$
  2. $f$ is increasing on $(-1/4, \infty)$

Property 1 requires $\alpha\le 0$ and $(\alpha x+\beta)_{x=-2} \ge (2x^2+x+3)_{x=-2} = 9$. Property 2 is automatically fulfilled by the given quadratic polynomial.

Let's turn to your second example:

  • $f_{\alpha, \beta} (x) = \alpha x + \beta$ if $x < 0$
  • $f_{\alpha, \beta} (x) = 4x^3 + 2x + 5$ if $x \geq 0$

Your analysis of convexity is correct. The situation with quasiconvexity is different now, because we can't say for sure that $f$ has a minimum at some point. (Previously, the quadratic polynomial had a local minimum, which under the quasiconvexity assumption has to be global, with no other local minima.) Let's consider two cases:

  1. $\alpha>0$. Then $f$ has to be increasing on $\mathbb{R}$, so the inequality $(\alpha x+\beta)_{x=0}\le (4x^3 + 2x + 5)_{x=0} $ has to hold.
  2. $\alpha\le 0$. Then $f$ decreases for $x<0$ and increases for $x>0$. Such a function is quasiconvex provided that $f(0)\le \max (f(0+), f(0-))$. For the given function, this is always the case, since $f(0)=f(0+)$. Thus, no other conditions on $\alpha,\beta$ are needed.