# Show that $e^{f(x)}$ is convex.

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.

• More generally, if $g$ is a convex non-decreasing function, and $f$ is a convex function on a convex domain, then $g(f)$ is convex on the domain of $f$. This can be adapted to your specific problem, see math.stackexchange.com/questions/287716/… – David Apr 20 '17 at 19:49
• Is there a way for me to do this without showing the more general fact that the composition of two convex functions is convex? I would like to avoid proving this and proving the fact that $e^x$ is convex. It seems like it would be less work to continue with the proof I started in the question. – mathqueen459 Apr 20 '17 at 19:56
• @britgirl5: you have to exploit the convexity of $e^x$ at some point, so I do not see why you should avoid the one-line proof already provided. – Jack D'Aurizio Apr 20 '17 at 19:57
• If $f$ is non-decreasing and convex, and if $g$ is convex, then $gof$ is convex. See the second answer here for a proof. – Mark Viola Apr 20 '17 at 20:12

\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}
"From here, I want to bring the $\lambda$ and $1−λ$ terms down in front of the exponential, but I am stuck as to how to do this."
If you wish to continue from where you stuck: denoting $e^{f(x)}=A$, $e^{f(y)}=B$, you have to prove that $$A^\lambda B^{1-\lambda}\le\lambda A+(1-\lambda)B.$$ Taking the logarithm on the both sides, it is equivalent to $$\lambda\ln A+(1-\lambda)\ln B\le\ln(\lambda A+(1-\lambda)B)$$ which is the same as the definition of the logarithm being a concave function.
For the proof, we need only the elementary inequality $e^z\ge 1+z$ for all real $z$. Then, we have $$e^x=e^{x-y}e^y\ge e^y\,[1+(x-y)]=x\,e^y+(1-y)\,e^y,$$ and there's equality for $y=x$. Thus, $$e^x=\sup_{y\in\mathbb{R}}\,\left[x\,e^y+(1-y)\,e^y\right].$$ Substituting $f(x)$ instead of $x$, we have $$e^{f(x)}=\sup_{y\in\mathbb{R}}\,\left[f(x)\,e^y+(1-y)\,e^y\right],$$ and the RHS is a supremum of convex functions, since the coefficient $e^y$ of $f(x)$ is always positive.