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A smooth function $u:(0,+\infty)\to \mathbb{R}$ is said to be a absolutely monotonic function if $u^{(k)}(x)\geq 0,\forall k\geq 0,x\in (0,+\infty),$ where $u^{(k)}$ is the k-th derivative of $u$ (we adapt the convention that $u^{(0)}(x)=u(x)$), now I want to show that $\ln u(e^r)$ is a convex function of $r:$

I want to use the fact that a smooth function is convex $\Longleftrightarrow f^{\prime\prime}(x)\geq 0,$ so I take derivative with respect to $r$ and compute ($g(r)=\ln u(e^r)$)

\begin{align*} g^{\prime}(r)=&~\frac{u^{\prime}(e^r)e^r}{u(e^r)}\qquad g^{\prime\prime}(r)=\frac{u^{\prime\prime}(e^r)u(e^r)e^{r}+u^{\prime}(e^r)u(e^r)-[u^{\prime}(e^r)]^2e^r}{[u(e^r)]^2}\cdot e^{r} \end{align*}

there is a negative term in the numerator of $g^{\prime\prime}(r),$ so I cannot determine its sign. Thus I let $t=e^{r}$ and consider $$\varphi(t)=u^{\prime\prime}(t)u(t)t+u^{\prime}(t)u(t)-[u^{\prime}(t)]^2t$$ note that $\varphi(0)\geq 0,$ so I would like to show that $\varphi$ is monotonically increasing, thus I compute

\begin{align*} \varphi^{\prime}=&~u^{(3)}ut+2u^{\prime\prime}u-u^{\prime\prime}u^{\prime}t \\ \varphi^{\prime\prime}=&~u^{(4)}ut+3u^{(3)}u+u^{\prime\prime}u^{\prime}-(u^{\prime\prime})^2t \end{align*}

but there are still negative terms that cannot be dropped out, so I am stuck here. Can someone help me?

Rodrigo de Azevedo
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Tiffany
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1 Answers1

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Since $f: \Bbb R \to \Bbb R$, $f(r) = \ln u(e^r)$ is continuous it suffices to show that $f$ is midpoint-convex, i.e. if $$ f \left( \frac{r+s}{2}\right) \le \frac 12 \bigl( f(r) + f(s) \bigr) $$ for all $r, s \in \Bbb R$. With the substitution $x=e^{r/2}$, $y = e^{s/2}$ this is equivalent to showing that $$ \tag{*} u (x y )^2 \le u(x^2) u(y^2) $$ for all $x, y > 0$.

For all $k$ is $u^{(k)}$ is nonnegative and nondecreasing, so that $\lim_{x \to 0+} u^{(k)}(x)$ exists. If we define $u(0) = \lim_{x \to 0+} u(x)$ then the extended function $u:[0, \infty) \to \Bbb R$ is continuous.

A repeated application of Prove that $f'(a)=\lim_{x\rightarrow a}f'(x)$. shows that the extended function is also infinitely often differentiable on $[0, \infty)$.

Now we can use a result of Bernstein, see for example Smooth Function with Derivative Nonnegative is Analytic : $u$ is analytic and can be represented as a power series $$ u(x) = \sum_{n=0}^\infty a_n x^n $$ with nonnegative coefficients $a_n \ge 0$.

So $(*)$ becomes $$ \tag{**} \left(\sum_{n=0}^\infty a_n x^n y^n\right)^2 \le \sum_{n=0}^\infty a_n x^{2n} \cdot \sum_{n=0}^\infty a_n y^{2n} $$ and that is just the Cauchy-Schwarz inequality.

Addendum: Equality holds in the Cauchy-Schwarz inequality $(**)$ only if the two “vectors” are a scalar multiple of each other, i.e. if $a_n x^{2n} = c a_n y^{2n}$ with a constant $c$ and all $n$. If at least two coefficients $a_n$ are nonzero then this is only possible for $x=y$. It follows that

$f$ is strictly convex unless $u$ is of the form $u(x) = a_n x^n$.

Martin R
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