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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.

Rodrigo de Azevedo
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mathqueen459
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    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 https://math.stackexchange.com/questions/287716/the-composition-of-two-convex-functions-is-convex – 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

2 Answers2

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\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}

Michael Hardy
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"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.

A.Γ.
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  • And how does one show that the logarithm is concave without exploiting its smoothness? – Mark Viola Apr 20 '17 at 20:17
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    @Dr.MV If one does not want to use calculus, one may try AM-GM inequality as it explained here, but it is easier with calculus, of course. – A.Γ. Apr 20 '17 at 20:28
  • Yes, calculus makes it much easier. It just seemed that the OP might have wanted to see somewhere a more elementary, pre-calculus approach. – Mark Viola Apr 20 '17 at 20:33