Is the following function convex? \begin{align} f(\alpha, \beta) = \exp\left( -j \cdot \alpha \cdot d^{\beta} \right), \end{align} where $j = \sqrt{-1}$, $\alpha \geq 0$, $\beta \geq 0$, and $d \in \mathbb{R}$.

If yes, how to show it?

If the function was a single variable dependent and twice differentiable, then one can show that the $f^{\prime\prime} \geq 0$ or show that $f(\lambda x_1 + (1-\lambda)x_2) \leq \lambda f(x_1) + (1-\lambda) f(x_2)$ is true.

But here a function of two variables $\alpha$ and $\beta$ is difficult for me. Can you please help me? Thank you so much in advance.

  • $\begingroup$ How is convexity defined for complex valued functions? Also, please do not abuse the square root function. $\endgroup$ – LinAlg Nov 4 '18 at 14:05
  • $\begingroup$ Well, since $j=\sqrt{-1}$, we have that $f(\alpha,\beta)=\exp(-j\alpha d^\beta)=\cos(-\alpha d^\beta)+ i\sin(-\alpha d^\beta)$. So $f$ is a function from $\mathbb R^2$ to $\mathbb C$. I don't know the definition of convex function when they have image in $\mathbb C$. $\endgroup$ – André Porto Nov 4 '18 at 14:09

I think you can't define a convex function unless it takes real values. But there is a way to get around that by trying to prove that :

g(α,β):= |exp(−j⋅α⋅d^β)|is convex with the following definition :

g(λx1+(1−λ)x2)≤λg(x1)+(1−λ)g(x2) with x1 = (α1,β1) and x2 having the appropriate coordinate.

You can generalize the same line of thought if your function goes into any vector space, you introduce a norm.

First post here, make sure to tell me if i didn't awnser properly.

  • $\begingroup$ thanks for your contribution, may may want to look into using mathjax $\endgroup$ – LinAlg Nov 4 '18 at 14:53
  • $\begingroup$ Thank you for your efforts. I was working on a problem, but I think my formulation is not making sense (that's why I am not able to even make it run on CVX for instance). Nevertheless, I will accept your answer for your efforts. Thank you. $\endgroup$ – user550103 Nov 5 '18 at 10:26

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