Prove a multivariate function is convex Show that the following multi-dimensional function is (strictly?) convex $$ f(x) =  \sum_{j=1}^m log\left(\frac{1}{2}+e^{2 x_j}\right) - \sum_{j=1}^m x_j  $$
Does it simply suffice to show that if the function is convex in one dimensional space, then it is in multivariate case too? e.g. the second derivative is clearly non-negative:
$$\frac{\partial^2}{\partial x} = \frac{4e^{2 x}}{(0.5+e^{2 x})^2}>0$$
But if not, I am unsure how to compute for when $\{x_j\}_{j=1}^p$ is multivariate, I think I need to show the $p\times p$ Hessian Matrix is positive definite, so i calculated the Hessian, but am not sure what to do now.
$$H = \pmatrix{\frac{4e^{2 x_1}}{(0.5+e^{2 x_1})^2}, \frac{2(x_1x_2e^{2 (x_1x_2)}+0.5e^{2x_1x_2}+e^{4x_1x_2})}{(0.5+e^{2 x_1x_2})^2},..., \frac{2(x_1x_pe^{2 (x_1x_p)}+0.5e^{2x_1x_p}+e^{4x_1x_p})}{(0.5+e^{2 x_1x_p})^2}  \\
    \frac{2(x_1x_2e^{2 (x_1x_2)}+0.5e^{2x_1x_2}+e^{4xy})}{(0.5+e^{2 x_1x_2})^2},  \frac{4e^{2 x_2}}{(0.5+e^{2 x_2})^2},...,  \frac{2(x_2x_pe^{2 (x_1x_p)}+0.5e^{2x_2x_p}+e^{4x_2x_p})}{(0.5+e^{2 x_1x_p})^2} \\
.\\
.\\
.    }$$
Alternatively, looking at this answer
 A: If $f: \Bbb R \to \Bbb R$ is (strictly) convex then
$$
 F: \Bbb R^m \to \Bbb R, \, F(x) = \sum_{j=1}^m f(x_j)
$$
is (strictly) convex as well.
This can be verified directly:  For $x \ne y \in \Bbb R^m$ and $0 < \lambda < 1$ is
$$
F((1-\lambda)x + \lambda y) = \sum_{j=1}^m f((1-\lambda)x_j + \lambda y_j) \\ \le \sum_{j=1}^m (1-\lambda) f(x_j) + \lambda f(y_j) = (1-\lambda) F(x) + \lambda F(y)
$$
because $f$ is convex. If $f$ is strictly convex then strict inequality holds here, since $x_j \ne y_j$ for at least one index $j$.
Alternatively (if $f$ is twice differentiable) this can be verified by computing the Hessian matrix $H$. The entries on the  diagonal are $f''(x_j)$, $j=1, \ldots, m$, and all mixed derivatives are zero. $H$ is a diagonal matrix with non-negative entries on the diagonal, and therefore positive semi-definite. If $f$ is strictly convex then the diagonal entries are positive and $H$ is positive definite.
In your case,
$$
 f: \Bbb R \to \Bbb R, f(t) = \log\left(\frac{1}{2}+e^{2 t}\right) - t
$$
is a strictly convex function, so that  $F$ is strictly convex.
Slightly more general: If $f_j: \Bbb R \to \Bbb R$, $j=1\, \ldots, m$ are $m$ (strictly) convex functions then
$$
 F: \Bbb R^m \to \Bbb R, \, F(x) = \sum_{j=1}^m f_j(x_j)
$$
is (strictly) convex as well.
