5
$\begingroup$

I need a help with proving convexity of $e^x$ function. $f(x)$ is a convex function, If it satisfies following condition.

$$f(c\cdot x_1+(1-c)\cdot x_2) ≤ c\cdot f(x_1)+(1-c)\cdot f(x_2), \quad 0 ≤ c ≤ 1$$

$\endgroup$

closed as unclear what you're asking by Pedro Tamaroff, user63181, Yiorgos S. Smyrlis, user127.0.0.1, Andrew D. Hwang Mar 7 '14 at 0:01

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 7
    $\begingroup$ The second derivative of $e^x$ is $e^x$ which is greater than $0$ for all $x$ $\endgroup$ – Rustyn Mar 6 '14 at 21:27
  • 3
    $\begingroup$ Or just the first derivative which is increasing. $\endgroup$ – user119228 Mar 6 '14 at 21:29
  • $\begingroup$ Can this be proved just by using definitions and not using second derivatives $\endgroup$ – user345777 Sep 25 '17 at 13:42
14
$\begingroup$

A continuously differentiable function $\mathrm{f}$ is convex on an interval $I$ if, and only if $$\mathrm{f}(x) \ge \mathrm{f}(y)+\mathrm{f}'(y)\cdot(x-y)$$ for all $x,y \in I$. In your case $\mathrm{f}(x) = \mathrm{e}^x$ and $I = \mathbb{R}$. Hence, we need to show that $$\mathrm{e}^x \ge \mathrm{e}^y + \mathrm{e}^y \cdot(x-y)$$ for all $x,y\in \mathbb{R}$. This inequality can be rearranged to give $$\mathrm{e}^{x-y} \ge 1+x-y$$ This is equivalent to $\mathrm{e}^z \ge 1+z$ for all $z \in \mathbb{R}$. Since $\mathrm{e}^z > 0$ for all $z \in \mathbb{R}$ it is clear that $\mathrm{e}^z \ge 1+z$ for all $z \le -1$. It remains to prove that $\mathrm{e}^z \ge 1+z$ for all $z \ge -1$.

The Taylor Series of $\mathrm{e}^z$ is $$\mathrm{e}^z = 1+z + \frac{z^2}{2!} + \frac{z^3}{3!} + \cdots$$

If $z > 0$ then all of the terms $\frac{z^k}{k!}$ are positive and hence $$ \mathrm{e}^z = 1+z + \frac{z^2}{2!} + \frac{z^3}{3!} + \cdots \ge 1 + z$$

If $-1 \le z \le 0$ then $z^2+z^3 \ge 0$, $z^4 + z^5 \ge 0$ and, in fact, $z^{2k}+z^{2k+1} \ge 0$ for all positive integers $k$. It follows that $\frac{z^2}{2!}+\frac{z^3}{3!} \ge 0$, $\frac{z^4}{4!}+\frac{z^5}{5!} \ge 0$ and, in fact, $$\frac{z^{2k}}{(2k)!} + \frac{z^{2k+1}}{(2k+1)!} \ge 0$$ Since the Taylor Series for $\mathrm{e}^z$ is absolutely convergent for all $-1 \le z \le 0$ (the exponential function converges on the entire complex plane, and power series converge absolutely within their radius of convergence) it follows that $$ \mathrm{e}^z = 1+z + \frac{z^2}{2!} + \frac{z^3}{3!} + \cdots \ge 1+z$$ for all $-1 \le z \le 0$.

$\endgroup$
  • $\begingroup$ Well done, although I would be a little more careful in the region $[-1,0]$. $\endgroup$ – Emily Mar 6 '14 at 21:40
  • $\begingroup$ @Arkamis You're right. I'm just trying to edit that now. $\endgroup$ – Fly by Night Mar 6 '14 at 21:42
  • $\begingroup$ @Arkamis I think I've fixed it. Do you agree? $\endgroup$ – Fly by Night Mar 6 '14 at 22:17
  • 1
    $\begingroup$ That's great and actually does not rely on derivative arguments: A function $f$ is convex if for every $x$ there exists $c$ such that for all $y$ it holds $f(y) \geq f(x) + c(y-x)$. You showed that $c=e^x$ works. This does not only show that $f$ is convex but also that $e^x$ is a subgradient of $\exp$ at $x$.) $\endgroup$ – Dirk Jul 20 '15 at 8:08
  • 1
    $\begingroup$ @Dirk Thank you for your kind words :-) $\endgroup$ – Fly by Night Jul 22 '15 at 22:40
7
$\begingroup$

The second derivative describes where a function is concave or convex. $(e^x)'' > 0$ for all $x$ so $e^x$ is convex everywhere.

$\endgroup$
3
$\begingroup$

A function of real variable is convex on an interval if it has nonnegative second derivative on this interval. This is a simplification of the Hessian condition for convexity to the case $\mathbb{R}\rightarrow \mathbb{R}$. The second derivative of $e^{x}$ is $e^{x}$, and this is of course nonnegative on the entire real line.

$\endgroup$

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