Suppose $f:\mathbb{R^n}\rightarrow\mathbb{R}$, $f(x) = x^TAx$, where $x = (x_1, \cdots,x_n)^T$, and $A$ is a symmetric positive definite matrix (thus $f(x)$ is strictly convex). And suppose $g:\mathbb{R}^n\rightarrow\mathbb{R}^n$, $g(x) = (\exp(x_1), x_2, \dots,x_n)^T$. Is $$h(x) = f\circ g(x) = g(x)^TAg(x)$$ strictly convex?
I tried to use Hessian, but found Hessian is not positive definite. Thus I cannot use Hessian to prove strict convexity. Now, I am trying to use the definition to prove it, but I got stuck. I tried the simplest case where $A$ is two by two matrix \begin{matrix} a_{11} & a_{12}\\ a_{12} & a_{22} \end{matrix}
and got $$h(x) = a_{11}\exp(2x_1)+2a_{12}\exp(x_1)x_2+a_{22}x_2^2$$ So at least it is strictly convex on each axis (marginally). But I am not able to show its strictly convexity jointly. Thanks!
We have $$ h(x) = e^{2x_1}a_{11} + 2\left(\sum_{i=2}^n a_{i1} e^{x_1}x_i \right)+ \left(\sum_{i,j=2}^n a_{ij} x_ix_j\right) $$ and $$ \frac{\partial^2}{\partial x_1^2} h(x) = 4a_{11} e^{2x_1} + 2\sum_{i=2}^n a_{i1} e^{x_1}x_i . $$ If one of the numbers $a_{i1}$ is non-zero then set $x_j=0$ for $j\ne i$, $x_i > -\frac{4a_{11}}{a_{i1}}$. For such $x$, we have $\frac{\partial^2}{\partial x_1^2} h(x)<0$.
This implies that $h$ is not convex! If $h$ would have been convex, then $\frac{\partial^2}{\partial x_1^2} h(x)\ge0$ for all $x$.