I know that the euclidean norm is a convex function, and that exponential functions are also convex. I want to know whether or not a mix of the two would also be a convex function. i.e.

$$f(x) = e^{\sqrt{\sum_{i=1}^{n}{x_i^2}}}$$ where $$x = (x_1,x_2,...,x_n)$$

So my approach so far has been to prove that

$$f(\lambda x + (1-\lambda y)) \leq \lambda f(x) + (1-\lambda) f(y)$$ so $$e^{\lVert \lambda x + (1-\lambda)y\rVert_2}$$

but i'm not sure how to proceed from here to get: $$e^{\lVert \lambda x + (1-\lambda)y\rVert_2} \leq \lambda e^{\lVert x \rVert_2} + (1-\lambda)e^{\lVert y \rVert_2} $$

I've thought about using the triangle inequality but I think that just gets me to: $$e^{\lVert \lambda x + (1-\lambda)y\rVert_2} \leq e^{\lambda \lVert x \rVert_2 + (1-\lambda)\lVert y\rVert_2} = e^{\lambda \lVert x \rVert_2} + e^{(1-\lambda)\lVert y\rVert_2} $$

  • 2
    $\begingroup$ If $f$ is convex, and $g$ is convex and non-decreasing, does it follow that $g \circ f$ is convex? $\endgroup$ – Daniel Fischer Nov 12 '16 at 19:55
  • $\begingroup$ right, but for this particular problem i have to prove it the way that I started, they're not going to accept your way, which is simpler. $\endgroup$ – April Nov 12 '16 at 19:57
  • $\begingroup$ Well, it's just replacing the abstract functions with concrete ones. Note that in the last line you have an error, $e^{a+b} = e^a \cdot e^b$, not $e^a + e^b$. But with $e^{\lambda \lVert x\rVert_2 + (1-\lambda)\lVert y\rVert_2}$, you are in a situation where using the convexity of the exponential is very very tempting. $\endgroup$ – Daniel Fischer Nov 12 '16 at 20:01

$$e^{\lVert \lambda x + (1-\lambda)y\rVert_2} \leq e^{\lambda \lVert x \rVert_2 + (1-\lambda)\lVert y\rVert_2} \leq \lambda e^{ \lVert x \rVert_2} + (1-\lambda)e^{\lVert y\rVert_2} $$

Where the first inequality is due to convexity of norm and exponential is an increasing function. The second inequality is due to convexity of exponential function.

Remark: $\exp$ being convex means $\exp(\lambda \alpha +(1-\lambda) \beta) \leq \lambda \exp(\alpha) + (1-\lambda) \exp(\beta)$. $\alpha = \left\| x\right\|, \beta = \left\| y\right\|$ for your question.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.