Direct proof for convexity of $e^x$

Question

Is there any direct proof without using second derivative for convexity of $e^x$?

Answer 1

Just expanding Daniel Fischer's comment, given that $f(x)=e^x$ is a positive and continuous function for which $f(x+y)=f(x)\,f(y)$, we have:

$$f\left(\frac{x+y}{2}\right) = \sqrt{f(x)\cdot f(y)}\color{red}{\leq} \frac{f(x)+f(y)}{2}\tag{1}$$ where $\color{red}{\leq}$ follows from the AM-GM inequality. But $(1)$ just gives the midpoint-convexity of $f(x)$, that together with continuity gives full convexity.

Answer 2

Hint Here is a first step. Direct definition of convexity between points $x,y$ is $$f(hx + (1-h)y) \le hf(x) + (1-h) f(y)$$

Let $y=0$ and use Taylor expansion $e^x = \sum_{k=0}^\infty x^n/n!$ to note that $$ e^{hx} = 1 + (hx) + h^2 x^2 + h^3 x^3 + \ldots $$ and $$ (1-h) + he^x = 1+h + (h + hx + hx^2 + hx^3 + \ldots) = 1 + hx + hx^2 + hx^3 + \ldots $$ so now the desired inequality follows assuming $h > 0, x > 0$.

We can generalize this argument for other $x,y$...

Answer 3

Let's check condition $$ f\Big(\frac{x+y}{2}\Big) \le \frac12 (f(x) + f(y)) $$ for $f(x)=e^x$. So, $$ e^{(x+y)/2} = e^{x/2} e^{y/2} \le \frac12 (e^x + e^y). $$ Denote $a=e^{x/2}$, $b=e^{y/2}$; then, $$ ab \le \frac12 (a^2 + b^2) \Longrightarrow 2ab \le a^2 + b^2 \Longrightarrow (a-b)^2\ge 0 $$ And... it's true!

Answer 4

We can use the definition of convexity itself.

A space S is convex if for any $u,v \in S$ $$\lambda u + (1- \lambda)v \in S \ \forall \ \lambda \in [0,1] $$

Intuitively this means if two points are in the space, then every point between them is in the space. (I can explain further if requested in comments). From here note that the space we want to consider is the set of points $(x,y)$ such that

$$ y \ge e^x $$

So let us consider any pair of points $ (x_1, y_1) $, $(x_2, y_2)$ satisfying $$ y_1 \ge e^{x_1} $$ and $$ y_2 \ge e^{x_2} $$

From we wish to establish that

$$ \lambda y_1 + (1- \lambda) y_2 \ge e^{\lambda x_1 + (1- \lambda) x_2 } $$

Where again $0 \le \lambda \le 1 $. Recall that by definition $$ \lambda y_1 + (1- \lambda) y_2 \ge \lambda e^{x_1} + (1- \lambda) e^{x_2} $$

We then wish to show that

$$\lambda e^{x_1} + (1- \lambda) e^{x_2} \ge e^{\lambda x_1 + (1- \lambda) x_2 } $$

Without loss of generality we assume $x_2 > x_1$ and divide through by $e^{x_1}$ to find

$$ \lambda + (1- \lambda)e^{x_2-x_1} \ge e^{(\lambda - 1)x_1 + (1 - \lambda )x_2 } $$

Focus on the right side, we can rewrite that as,

$$ \lambda + (1- \lambda)e^{x_2-x_1} \ge e^{ (1 - \lambda )(x_2- x_1) } $$

Let us denote $$ x_2 - x_1 = T$$ whereas $T \ge 0 $

$$ \lambda + (1- \lambda)e^{T} \ge e^{ (1 - \lambda )(T) } $$

I wish I knew how to finish this. But I will leave it as an exercise to the reader ;)

addendum

We can consider the taylor series of $e^x$ noting that

$$ \lambda + (1- \lambda)e^{t} = \lambda + (1 - \lambda)(1 + t + \frac{1}{2}t^2 ... ) = $$

$$ 1 + (1 - \lambda)t + (1 - \lambda) \frac{1}{2}t^2 ... $$

And

$$ e^{(1 - \lambda)t} = 1 + (1 - \lambda)t + \frac{1}{2}(1- \lambda)^2t^2 ... $$

Note that since $$ 0 \le 1 - \lambda \le 1$$

$$(1- \lambda)^n \le (1 - \lambda) \forall n \ge 0 $$

Therefore

$$ e^{(1 - \lambda)t} = 1 + (1 - \lambda)t + \frac{1}{2}(1- \lambda)^2t^2 ... \le 1 + (1 - \lambda)t + (1 - \lambda) \frac{1}{2}t^2 ...$$

Showing the desired result. Although, the use of taylor series, breaks your requirement of avoiding derivatives all together.