Is this algebraic proof of the convexity of the exponential correct? Is there any purely algebraic proof?

Question

In order to compute the derivative of the exponential function in an "early transcendentals" approach (where exponentials are defined not via power series or differential equations, but instead arithmetically via repeated multiplication, extended to reals), you first need to prove some fact like convexity of the exponential function or Bernoulli's inequality. In my experience these are analytic facts. Is it possible to give a purely algebraic proof of the convexity of the exponential function?

I found what looks like a purely algebraic proof of the convexity of the exponential function in the textbook Calculus Unlimited by Marsden and Weinstein. But it appears to contain an error or else I am just not following it. Is this proof correct? It's on pages 133 and 134 in chapter 10 here is a link, but I reproduce the proof below.

First, as a warm-up to convexity is a proof of midpoint convexity, in a working backwards style:

$$ \begin{align} b^{\frac{x_1+x_2}{2}}&<\frac{b^{x_1}+b^{x_2}}{2}\\ 2b^{\frac{x_2-x_1}{2}}&<b^{x_2-x_1}+1\\ b^{\frac{x_2-x_1}{2}}-1&<b^{x_2-x_1}-b^{\frac{x_2-x_1}{2}}\\ b^{\frac{x_2-x_1}{2}}-1&<\left(b^{\frac{x_2-x_1}{2}}-1\right)b^{\frac{x_2-x_1}{2}}\\ 1&<b^{\frac{x_2-x_1}{2}}\\ 0 &< \frac{x_2-x_1}{2}\\ x_1&<x_2. \end{align} $$

Since the steps are all equivalences, the inequality $x_1<x_2$ is equivalent to strict midpoint convexity.

The only facts used are basic laws of exponents and that $b^x$ is a strictly increasing function. If we know that $b^x$ is continuous, this is enough to prove full convexity (see for example Midpoint-Convex and Continuous Implies Convex or If $f$ is continuous and $\,f\big(\frac{1}2(x+y)\big) \le \frac{1}{2}\big(\,f(x)+f(y)\big),$ then $f$ is convex). However at this point in the textbook it is not known to be continuous.

So then they repeat the proof for a general convex combination $\lambda x_1+(1-\lambda)x_2$ with $0<\lambda<\frac{1}{2}$:

$$ \begin{align} b^{\lambda x_1+(1-\lambda)x_2}&<\lambda b^{x_1}+(1-\lambda)b^{x_2}\\ \lambda b^{\lambda x_1+(1-\lambda)x_2}-\lambda b^{x_1}&<(1-\lambda)\left(b^{x_2}-\color{red}{b^{\lambda x_1+(1-\lambda)x_2}}\right)\\ \lambda b^{x_1} \left(b^{(1-\lambda) (x_2-x_1)}-1\right)&<(1-\lambda)\color{red}{b^{\lambda x_2+(1-\lambda)x_1}}\left(\color{red}{b^{(1-\lambda)(x_2-x_1)}}-1\right)\\ \lambda b^{x_1} &<(1-\lambda)b^{\lambda x_2+(1-\lambda)x_1}. \end{align} $$

We conclude that since $\lambda<(1-\lambda)$, and also $x_1<\lambda x_1+(1-\lambda)x_2$, and by monotonicity of $b^x$ also $b^{x_1}<b^{\lambda x_1+(1-\lambda)x_2},$ hence the inequality holds, and by equivalence establishes convexity.

The part where I didn't follow is the terms marked in red. It appears as though the authors want to factor out $b^{\lambda x_1+(1-\lambda)x_2}$ from $\left(b^{x_2}-b^{\lambda x_1+(1-\lambda)x_2}\right)$ which should become $b^{\lambda x_1+(1-\lambda)x_2}\left(b^{\lambda(x_2-x_1)}-1\right).$ But instead they factored out $b^{\lambda x_2+(1-\lambda)x_1},$ with the $x_1$ and $x_2$ switched in the convex combination. Factoring that out instead does give the claimed $b^{(1-\lambda)(x_2-x_1)}$ for the first term, but does not leave a $1$ for the second term.

Is this a mistake in the text, or have I misunderstood the computation?

If there is a mistake, can it be repaired? As far as I can tell, in order to conclude that the the corrected inequality $$\lambda b^{x_1} \left(b^{(1-\lambda) (x_2-x_1)}-1\right)<(1-\lambda)b^{\lambda x_1+(1-\lambda)x_2}\left(b^{\lambda(x_2-x_1)}-1\right)$$ holds, we would use $\lambda<1-\lambda,$ $x_1<\lambda x_1+(1-\lambda)x_2,$ and $(1-\lambda)<\lambda$ (so that $b^{(1-\lambda) (x_2-x_1)}<b^{\lambda(x_2-x_1)}.$) But this third inequality is in fact false.

Can we conclude the convexity inequality holds, despite the failure of that last inequality? Is there another proof of the convexity of the exponential in this style, using only algebraic properties and the monotonicity? Or should we toss this proof, and wait instead for a more analytic proof using midpoint convexity?

Answer 1

Suppose the function $b^x$, where $b>1$, is not convex. Then there are $U=b^{x_1}$ and $V=b^{x_2}$, with $x_1<x_2$, such that $$U^\lambda V^{1-\lambda}>\lambda U+(1-\lambda)V.$$ Let $X=\frac{U}{V}$. Then $X<1$ and $$X^\lambda >\lambda X+1-\lambda$$ $$\lambda > \frac{1-X^\lambda}{1-X}. $$

Let $A<B$ be two positive integers such that $$\lambda > \frac{A}{B}>\frac{1-X^\lambda}{1-X}. $$ Then, for $Y=X^{\frac{1}{B}},$

$$ \frac{1-X^{\frac{A}{B}}}{1-X}=\frac{1-Y^A}{1-Y^B}=\frac{1+Y+...Y^{A-1}}{1+Y+...Y^{B-1}} >\frac{A}{B}> \frac{1-X^\lambda}{1-X}. $$

Therefore $X^\lambda>X^\frac{A}{B}$ and so $\frac{A}{B}>\lambda$, a contradiction.

Answer 2

Suppose the function $b^x$, where $b>1$, is not strictly convex. Then there are $U=b^{x_1}$ and $V=b^{x_2}$, with $x_1<x_2$, such that $$U^\lambda V^{1-\lambda}\ge\lambda U+(1-\lambda)V.$$ Let $X=\frac{U}{V}$. Then $X<1$ and $$X^\lambda \ge\lambda X+1-\lambda$$

However, by elementary properties of the $x^n$ function, $X^\lambda \le\lambda X+1-\lambda$ with equality only if $X=1$, $\lambda=0$ or $\lambda=1$. This contradiction proves that $b^x$ is strictly convex.