# How to prove that e^x is convex? [duplicate]

For convexity :

$$e^{ta+(1-t)b} \leq t \cdot e^a+(1-t)e^b \Rightarrow e^{(a-b)^t} \leq te^{a-b} + 1 - t$$

Now I'm stuck on how to solve it further.

-

## marked as duplicate by Fly by Night, Rasmus, Sami Ben Romdhane, USER91500, Nick PetersonApr 13 '14 at 14:32

Are you allowed to use derivatives? From Wikipedia: "A twice differentiable function of one variable is convex on an interval if and only if its second derivative is non-negative there".

Since the second derivative of $e^x$ is $e^x$, and $e^x > 0$ for all $x \in \Bbb R$, we have that $e^x$ is convex on any interval.

-
I just want to solve it without using derivative. Can you please help ? – user141256 Apr 13 '14 at 13:22
@user141256 At least for me, the exponential function is defined by the fact that it is its own derivative, namely as the unique(!) solotion of $f'(x)=f(x)$, $f(0)=1$. With that situation, using the derivative would be the most direct approach. There are of course other ways to define the exponential, and for each there may be another notion of "most direct" ... – Hagen von Eitzen Apr 13 '14 at 13:51

The exponential function is infinitely differentiable and hence you can apply the second derivative test which is a lot easier than the method you are employing. Assuming then that you are allowed to use the second derivative:

$$f \prime \prime (x) = e^{x}>0$$

and you see that this is a (strictly) convex function on any interval.

-
I just want to solve it without using derivative. Can you please help ? – user141256 Apr 13 '14 at 13:13

You certainly know that $e^x$ grows faster and faster and that it is always positive. Can you conceive a curve satisfying the first condition without being convex ?

-