Let $f : (0,\infty) \to \mathbb{R}$ be a convex function. Prove that $e^{f(x)}$ is a convex function on $(0,\infty)$.
My original idea was to try and show that the second derivative is positive, but this will not work since $f(x)$ need not be differentiable. Here's my second attempt:
By definition, since $f$ is convex, we have that $f(\lambda x+(1-\lambda)y)\leq\lambda f(x)+(1-\lambda)f(y)$ for any $\lambda \in (0,1)$ and $x,y \in (0,\infty)$. Then, by applying the exponential to both sides we have that $e^{f(\lambda x+(1-\lambda)y))}\leq e^{\lambda f(x)+(1-\lambda)f(y)}$ . Applying rules of exponents, $e^{f(\lambda x+(1-\lambda)y))} \leq e^{\lambda f(x)}e^{(1-\lambda)f(y)}$. From here, I want to bring the $\lambda$ and $1-\lambda$ terms down in front of the exponential, but I am stuck as to how to do this.
Best Answer
\begin{align} f(\lambda x + (1-\lambda)y) & \le \lambda f(x) + (1-\lambda) f(y) & & \text{because $f$ is convex.} \\[10pt] \text{Therefore } e^{f(\lambda x+(1-\lambda)y))} & \leq e^{\lambda f(x) + (1-\lambda) f(y)} & & \text{because $w\mapsto e^w$ is increasing,} \\[10pt] & = e^{\lambda v + (1-\lambda) w} \\[10pt] & \le \lambda e^v + (1-\lambda)e^w & & \text{because $w\mapsto e^w$ is convex, since} \\ & & & \text{its second derivative is positive,} \\[10pt] & = \lambda e^{f(x)} + (1-\lambda)e^{f(y)}. \end{align}