# Proof (without use of differential calculus) that $e^{\sqrt{x}}$ is convex on $[1,+\infty)$.

Prove (without use of differentiation) that $$f(x)=e^{\sqrt{x}}$$ is convex on $$[1,+\infty)$$.

Attempt. Function $$x\mapsto e^x$$ is convex and increasing, but $$x\mapsto \sqrt{x}$$ is concave, so we cannot use the proposition of composition: $$(convex~\&~increasing)\circ convex=convex.$$ Definition would require to prove for all $$x,~y\geqslant 1$$ and $$\lambda \in [0,1]$$: $$e^{\sqrt{\lambda x+(1-\lambda) y}}\leqslant \lambda e^{\sqrt{ x}}+(1-\lambda)e^{\sqrt{y}}$$ but squaring doesn't work here.

If we used continuity in order to prove mid-convexity, the problem would go like: $$e^{\sqrt{\frac{x+y}{2}}}\leqslant \frac{e^{\sqrt{x}}+e^{\sqrt{y}}}{2},$$ equivalently: $$e^{\sqrt{\frac{x+y}{2}}}-e^{\sqrt{x}} \leqslant e^{\sqrt{y}}-e^{\sqrt{\frac{x+y}{2}}}$$ (but without MVT what would we do with these differences?)

Thanks in advance for the help.

• Thank you for your initial comment. Very nice method of turning the problem to logarithms. I am looking for an elementary proof, if such a proof exits of course(!) (which would not make use of differention at all). In other words, more approaches are more than welcomed. Thanks again! Commented Feb 18, 2019 at 13:30
• (Deleted most of earlier comment; added a comment to my answer instead.) I'm still curious as to the source of the problem: it can hardly have been set unless there is a nice trick for doing it. Commented Feb 18, 2019 at 22:02

To show convexity it is enough to prove the property of a supporting hyperplane at each point, i.e. $$e^{\sqrt{x+t}}\geq e^{\sqrt x}+t\cdot \tfrac{1}{2\sqrt x}e^{\sqrt x}\tag{*}$$ whenever $$x\geq 1$$ and $$t+x\geq 1.$$

I will just recall here how your definition of convexity follows from (*), to show that it does not use calculus. Given $$y,z\geq 1$$ and $$0\leq \lambda\leq 1,$$ set $$x=\lambda y+(1-\lambda)z.$$ Then (*) gives

\begin{align*} e^{\sqrt y}&\geq e^{\sqrt{x}} + (y-x)\cdot \frac{1}{2\sqrt x}e^{\sqrt x}\\ e^{\sqrt z}&\geq e^{\sqrt{x}} + (z-x)\cdot \frac{1}{2\sqrt x}e^{\sqrt x}\\ &\implies\\ \lambda e^{\sqrt y}+(1-\lambda)e^{\sqrt z}&\geq e^{\sqrt{x}} + (\lambda(y-x)+(1-\lambda)(z-x))\cdot \frac{1}{2\sqrt x}e^{\sqrt x}\\ &= e^{\sqrt{x}}. \end{align*}

It remains to show (*). For $$t\leq 0$$ we can use $$\sqrt{x+t}-\sqrt{x}=\frac{t}{\sqrt{x+t}+\sqrt{x}}\geq \frac{t}{2\sqrt x}$$ to get $$e^{\sqrt{x+t}-\sqrt x}\geq e^{t/2\sqrt x}\geq 1+t/2\sqrt x,$$ which is (*).

For $$t\geq 0$$ we can find an $$h\geq 0$$ such that $$1+\tfrac{t}{2\sqrt x}=e^h.$$ Since $$h\geq 0$$ we have $$e^h\geq 1+h+h^2/2,$$ so

\begin{align*} x+t&=x+2\sqrt{x}(e^h-1)\\ &\geq x+2h\sqrt x+h^2\sqrt x\\ &\geq x+2h\sqrt x+h^2\\ &=(h+\sqrt x)^2. \end{align*} (This is where we need $$x\geq 1.$$) Taking square roots and exponentiating gives $$e^{\sqrt{x+t}}\geq e^{h+\sqrt x},$$ which is (*).

• @CalumGilhooley: yes, my edit wasn't a response to your comment. The edit shows how my answer relates to the definition of convexity in the question, which wasn't clear before. The choice of slope is by magic - the point is that the proof can be verified without calculus (modulo some facts about $\exp$).
– Dap
Commented Feb 20, 2019 at 19:32
• Damn, I wish I could do magic! :) Commented Feb 20, 2019 at 19:33
• @CalumGilhooley Actually you can: The magic slope choice is disenchanted by the fact that $e^{\sqrt x}\big/2\sqrt x$ is the slope of $e^{\sqrt x}$ at $x$. Hence the 'supporting hyperplane' is the tangent to $e^{\sqrt x}$ at $x$ (Beware of calculus lurking behind the corner.), and the graph of a (differentiable) convex function never falls below its tangents. Commented Mar 22, 2019 at 10:46
• @Hanno That was my point (in a comment now deleted, because it had become largely irrelevant): how do you guess the slope without using differentiation (or magic, which is presumably also disallowed!)? Commented Mar 22, 2019 at 15:44

I'm going to assume we are permitted to use differentiation to prove an intermediate result.

(This seems reasonable, as the exponential function is normally treated using differential calculus; but if it is not acceptable, then perhaps the intermediate result can be derived in some other way.)

Stop Press In the addendum below, the inequality is proved without using differentiation.

Lemma If $$\varphi(t) = [\log(1 + t)]^2 + 2\log(1 + t)$$, then $$\varphi(2t) < 2\varphi(t)$$ for all $$t > 0$$.

Proof Sneakily differentiating: \begin{align*} \tfrac{1}{2}\varphi'(t) & = \frac{1 + \log(1 + t)}{1 + t}, \\ \tfrac{1}{2}\varphi''(t) & = -\frac{\log(1 + t)}{(1 + t)^2} < 0, \end{align*} therefore $$\varphi'(t)$$ is strictly decreasing for all $$t \geqslant 0$$. In particular, $$\varphi'(t) > \varphi'(2t)$$ for all $$t > 0$$.

Putting $$\psi(t) = 2\varphi(t) - \varphi(2t)$$, we have $$\tfrac{1}{2}\psi'(t) = \varphi'(t) - \varphi'(2t) > 0$$ for all $$t > 0$$, i.e. $$\psi$$ is strictly increasing. Since $$\psi(0) = 0$$, we have $$\psi(t) > 0$$ for all $$t > 0$$. $$\square$$

Because $$f(x) = e^{\sqrt{x}}$$ is a strictly increasing function from $$[1, \infty)$$ to $$[e, \infty)$$, its inverse $$g$$ is the strictly increasing function from $$[e, \infty)$$ to $$[1, \infty)$$ given by $$g(y) = (\log y)^2$$. The strict convexity of $$f$$ is equivalent to the strict concavity of $$g$$, so we prove the latter. Since $$g$$ is continuous, it is sufficient to prove strict midpoint concavity. That is, it is enough to prove: $$g[(1 + t)y] - g(y) > g[(1 + 2t)y] - g[(1 + t)y] \quad (y \geqslant e, \ t > 0).$$ This simplifies to: $$\log(1 + t)[\log(1 + t) + 2\log y] > [\log(1 + 2t) - \log(1 + t)][\log(1 + 2t) + \log(1 + t) + 2\log y].$$ Suppose for the moment that this holds in the particular case $$y = e$$, $$\log y = 1$$. Then for arbitrary $$y \geqslant e$$ we have: $$2\log(1 + t) = \log\left[(1 + t)^2\right] > \log(1 + 2t),$$ whence: $$\log(1 + t)(2\log y - 2) \geqslant [\log(1 + 2t) - \log(1 + t)](2\log y - 2),$$ and the required inequality therefore holds for all $$y \geqslant e$$. We have thus reduced the desired inequality to one not involving $$y$$: $$\log(1 + t)[\log(1 + t) + 2] > [\log(1 + 2t) - \log(1 + t)][\log(1 + 2t) + \log(1 + t) + 2],$$ which simplifies to: $$2[\log(1 + t)]^2 + 4\log(1 + t) > [\log(1 + 2t)]^2 + 2\log(1 + 2t).$$ The above Lemma now completes the proof. $$\square$$

Changing notation: let $$j \colon \mathbb{R}_{\geqslant0} \to \mathbb{R}_{\geqslant0}$$ be the strictly increasing continuous function given by $$j(y) = (\log(1 + y) + 1)^2 - 1 \quad (y \geqslant 0).$$ This function $$j$$, which is denoted by $$\varphi$$ above, is the inverse of the strictly increasing continuous function $$h \colon \mathbb{R}_{\geqslant0} \to \mathbb{R}_{\geqslant0}$$ given by $$h(x) = \frac{f(1 + x)}{e} - 1 = e^{\sqrt{1 + x} - 1} - 1 \quad (x \geqslant 0).$$ By the lemma, $$f$$ is strictly convex if and only if: $$$$\label{3114933:eq:1}\tag{1} j(2y) < 2j(y) \quad (y > 0).$$$$ If \eqref{3114933:eq:1} holds, then, for all $$x > 0$$, $$2x = 2j(h(x)) > j(2h(x)),$$ which proves: $$$$\label{3114933:eq:2}\tag{2} h(2x) > 2h(x) \quad (x > 0).$$$$ The necessity of \eqref{3114933:eq:2} is admittedly obvious. Its sufficiency seems less obvious, although it can probably be proved without going round the houses like this. Carrying on, anyway (so as not to edit my existing answer): the converse argument is formally identical. That is, if \eqref{3114933:eq:2} is satisfied, then, for all $$y > 0$$, $$2y = 2h(j(y)) < h(2j(y)),$$ which proves \eqref{3114933:eq:1}. Thus, \eqref{3114933:eq:1} and \eqref{3114933:eq:2} are equivalent; so \eqref{3114933:eq:2} is another necessary and sufficient condition for the strict convexity of $$f$$. $$\square$$

Perhaps this is tractable after all! (Have I made a silly mistake?)

By the foregoing, it is enough to prove that the function $$\frac{h(x)}{x} = \frac{e^{\sqrt{1 + x} - 1} - 1}{x} \quad (x > 0)$$ is strictly increasing. Therefore, it is enough to prove that the function $$\psi(u) = \frac{e^{u - 1} - 1}{u^2 - 1} \quad (u > 1)$$ is strictly increasing. But: \begin{align*} e^{u - 1} - 1 & = (u - 1) + \frac{(u - 1)^2}{2} + (u - 1)^3\rho(u) \\ & = \frac{u^2 - 1}{2} + (u - 1)^3\rho(u), \end{align*} where $$\rho(u) = \sum_{n=0}^\infty\frac{(u - 1)^n}{(n + 3)!} \text{ is strictly increasing for all } u \geqslant 1.$$ Therefore: $$\psi(u) = \frac{1}{2} + \frac{(u - 1)^2}{u + 1}\rho(u) \quad (u > 1),$$ whence it is enough to prove - without, of course, using differentiation! - that $$\frac{(u - 1)^2}{u + 1}$$ is strictly increasing, for all $$u \geqslant 1$$. But if $$a > b \geqslant 0$$, then $$\frac{a^2}{a + 2} - \frac{b^2}{b + 2} = \frac{(a - b)(ab + 2a + 2b)}{(a + 2)(b + 2)} > 0,$$ and this completes the proof.

• I wouldn't like to do the same thing with $e^{\sqrt[3]{x}}$, mind! So there's probably a better way. Commented Feb 18, 2019 at 3:53