Proving whether this function is convex

Question

I want to determine whether or not the following function is convex

\begin{equation} y(t) = \Big( \log A(t) \Big)^2 \end{equation}

where

\begin{equation} A(t) = \int_0^\infty f(x) \exp(-g(x)t) dx \end{equation}

where $f(x)$, $g(x)$, and $t$ are all non-negative. Furthermore, $f(x)$ satisfies

\begin{equation} \int_0^\infty f(x)dx = 1 \end{equation}

My approach has been as follows: first, calculating the second derivative:

\begin{equation} y''(t) = \frac{2}{A(t)^2} \left( A'(t)^2 + (A(t) A''(t) - A'(t)^2) \log A(t) \right) \end{equation}

where

\begin{align} A'(t) &= -\int_0^\infty f(x) g(x) \exp(-g(x)t) dx \\\\ A''(t) &= \int_0^\infty f(x) g(x)^2 \exp(-g(x)t) dx \end{align}

Setting the second derivative greater than or equal to zero gives us the following inequality:

\begin{equation} -\log A(t) \leq \frac{A'(t)^2}{A(t) A''(t) - A'(t)^2} \end{equation}

I now want to determine whether this inequality is true. A couple observations:

1. $A(t) A''(t) - A'(t)^2$ is non-negative, which can be shown using Cauchy-Schwarz $\left( \int_0^\infty h_1^2 dx \int_0^\infty h_2^2dx \geq \int_0^\infty h_1^2 h_2^2 dx\right)$ with $h_1 = \sqrt{f(x) \exp(-g(x)k)}$ and $h_2 = g(x) \sqrt{f(x) \exp(-g(x)k)}$.
2. $\log A(t)$ is convex. This follows from (1) after computing $\frac{d^2}{dt^2} \log A(t)$.
3. The inequality holds at $t=0$, since $A(0)=1$ and the RHS is non-negative.

Beyond this, I'm stuck. Any clues?

Edit: Ryszard Szwarc shows a counterexample. If we let $f(x) = e^{-x}$ and $g(x)=x$, we get $y(t) = [\log(1+t)]^2$, which is not convex for $t > e-1$.

Answer 1

Notice that we can write $A(t) = \mathbb E\left[e^{-g(X)t}\right]$ where $X$ is a real random variable with density $f$. Furthermore, we have the following (proof will be given at the end)

Lemma : If $f:I\to\mathbb R$ is convex, then $f^2$ is convex as well

Assuming the lemma is correct, it is thus enough to show that $z(t):=\log(A(t))$ is convex.

Now let $t_1,t_2\in \mathbb R$ and $0<\lambda<1$, we have $$\begin{align*}z(\lambda t_1 + (1-\lambda)t_2) &:= \log\mathbb E\bigg[\exp[-g(X)\lambda t_1 + -g(X)(1-\lambda)t_2]\bigg]\\ &=\log\mathbb E\bigg[\underbrace{\exp[-g(X)\lambda t_1 ]}_{A}\cdot\underbrace{\exp[ -g(X)(1-\lambda)t_2]}_{B}\bigg]\end{align*} $$ Now let $p:=\frac 1 \lambda$ and $q:= \frac{1}{1-\lambda}$. We clearly have $p,q\in[1,\infty]$, and it follows from Hölder's inequality that $$\begin{align*}z(\lambda t_1 + (1-\lambda)t_2) &= \log\mathbb E[A\cdot B]\\ &\le\log\bigg[\mathbb E[A^p]^{1/p}\mathbb E[B^q]^{1/q}\bigg]\\ &=\frac 1 p\log\mathbb E[A^p] + \frac 1 q \log\mathbb E[B^q]\\ &=\lambda z(t_1) + (1-\lambda)z(t_2)\end{align*} $$ As desired.

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Proof of Lemma : the proof can be found in this answer, but I reproduce it here for completeness. Let $x,y \in I$ and $0<\lambda<1$. By convexity of $f$ it holds $$f((1-t)x+ty) \leq (1-t) \, f(x) + t \, f(y) $$ Squaring both sides and expanding the RHS yields $$f^2((1-t)x+ty) \leq (1-t)^2 f^2(x) + t^2 f^2(y) + 2t(1-t)f(x)f(y)$$ Then by adding and substracting respectively $(1-t)\, f^2(x) $ and $ t \, f^2(y) $ we get $$\begin{align*}f^2((1-t)x+ty) &\leq \color{red}{(1-t)^2 f^2(x)} + \color{blue}{t^2 f^2(y)} + 2t(1-t)f(x)f(y)\\ & - \color{red}{(1-t)\, f^2(x)} - \color{blue}{t\, f^2(y)}\ + (1-t)\, f^2(x) + t \, f^2(y)\\ \\ &= \color{red}{(1-t)((1-t)-1)\ f^2(x)} + \color{blue}{t(t-1)f^2(y)}\\ &+\color{green}{2t(1-t)f(x)f(y)}+ (1-t)\, f^2(x) + t \, f^2(y)\\ \\ &=\color{brown}{-t(1-t)(f(x)-f(y))^2}+ (1-t)\, f^2(x) + t \, f^2(y)\\ &\le (1-t)\, f^2(x) + t \, f^2(y)\end{align*} $$ and we're done.