Convexity of Exponential Composite Function $f:\mathbb{R}_{+}^M\rightarrow\mathbb{R}_+$ is a convex analytic function. For $\mathbf{x}\in\mathbb{R}_{+}^M$ and $y\in\mathbb{R}_{+}$, consider the function $g:\mathbb{R}_{+}^{M+1}\rightarrow\mathbb{R}_+$ defined as
$g(\mathbf{x},y)=e^y[f(\mathbf{x})+c]$
where $c$ is a positive constant. I have a hunch that $g$ is a convex function but not able to prove it. Any hints, idea, or more importantly, counter example?
P. S. Forming the Hessian matrix for $g$ from that of $f$ is apparently a nice idea, but I am stuck to prove the final determinant will be positive. 
 A: In general, $g$ is not convex (everywhere).
For a counterexample, I pick $M = 1$, since that's easiest to write up, but of course a similar situation can arise in arbitrary dimensions. Let
$$f(x) = x^4 - \frac12 x + \frac{3}{16} + \varepsilon(x^2 + 1),$$
where $\varepsilon > 0$ is very small, it just serves to establish strict positivity and strict convexity and shall not interfere with the following. Let also $c > 0$ be sufficiently small. Then the Hessian (grr, that ought to be Hessean, the man was called Hesse) of $g$ is - modulo the factor $e^y$ -
$$\begin{pmatrix}
12x^2 + 2\varepsilon & 4x^3 -\frac12 + 2\varepsilon x\\
4x^3 -\frac12 + 2\varepsilon x & f(x)+c
\end{pmatrix}.$$
For $x = 0$, we obtain
$$\begin{pmatrix}
2\varepsilon & -\frac12\\
-\frac12 & \frac{3}{16} +\varepsilon + c
\end{pmatrix},$$
whose determinant is negative.
If $f$ is uniformly strictly convex (the smallest eigenvalue of the Hessian is strictly bounded away from zero), then choosing $c$ sufficiently large [if the values of $f$ are large enough, that is no constraint] guarantees convexity of $g$.
