# Is the exponential of the euclidean norm convex?

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

• If $f$ is convex, and $g$ is convex and non-decreasing, does it follow that $g \circ f$ is convex? – Daniel Fischer Nov 12 '16 at 19:55
• 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. – April Nov 12 '16 at 19:57
• 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. – Daniel Fischer Nov 12 '16 at 20:01

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