In what case Hessian matrix off-diagnal values are zero In the case of off-diagnal values of Hessian matrix are zero, what will be the implication?
I found a related question here.
If a function $f$ has the from $$f(x_1,x_2,x_3)=g_1(x_1)+g_2(x_2)+g_3(x_3)$$ then the second order derivative $$\frac{\partial^2}{\partial x_i \partial x_j} f(x) = 0 \qquad \forall i\neq j$$.
In the case of Hessian matrix, does it only implies the function is linear as @ErdelvonMises mentioned?
 A: The other post doesn't actually solve the question as stated---let's do that. If you like you can appeal to the generalized Taylor expansion of $f$ to solve the problem immediately, or we can basically do the same thing with our "bare hands" below.
Suppose that $f$ is smooth. In showing that there are smooth functions $g_i$ on $\mathbb{R}$ such that $f(x_1, \ldots, x_n) = g_1(x_1) + \cdots + g_n(x_n)$, it will be slightly annoying that this decomposition is not unique if $n > 1$: e.g. we can always simultaneously replace $g_1$ with $g_1 - 1$ and $g_2$ with $g_2 + 1$ to get another distinct decomposition.
Instead let's impose the additional constraint that $g_i(0) = 0$ for all $i$ and that there exists a constant $c \in \mathbb{R}$ such that  $f(x_1, \ldots, x_n) = c + g_1(x_1) + \cdots + g_n(x_n)$. We will show that such a decomposition exists and is unique when the Hessian of $f$ is diagonal.
Let's first guess the functions $g_i$: we better have that
$$
g_i'(t) = \frac{d f}{d x_i}(0, \ldots, 0) + \int_0^t \frac{d^2 f}{d x_i^2}(0, \ldots, 0, t, 0, \ldots, 0) d t,
$$
i.e. so that
$$
g_i(t) := \frac{d f}{d x_i}(0, \ldots, 0) t + \int_0^t \int_0^s \frac{d^2 f}{d x_i^2}(0, \ldots, 0, s, 0, \ldots, 0) d s d t.
$$
(Recall that we want $g_i(0) = 0$.) Then we also need $c := f(0, \ldots, 0)$.
Ok, so now put $h(x_1, \ldots, x_n) = f(0, \ldots, 0) + g_1(x_1) + \cdots + g_n(x_n)$: we claim $f = h$. Indeed, for any $(x_1, \ldots, x_n) \in \mathbb{R}^n$ by the ordinary 1-dimensional fundamental theorem of calculus we have
$$
f(x_1, 0, \ldots, 0) - f(0, \ldots, 0)
= \int_0^{x_1} \frac{d f}{d x_1}(t, 0, \ldots, 0) d t,
$$
so that inductively we have
\begin{align*}
f(x_1, \ldots, x_n) - f(0, \ldots, 0)
&= \int_0^{x_1} \frac{d f}{d x_1}(t, 0, \ldots, 0) d t
+ \int_0^{x_2} \frac{d f}{d x_2}(x_1, t, \ldots, 0) d t\\
&\phantom{===}+ \cdots
+ \int_0^{x_n} \frac{d f}{d x_n}(x_1, \ldots, x_{n - 1}, t) d t.
\end{align*}
If we can show that $\frac{d f}{d x_k}(x_1, \ldots, x_{k - 1}, t, x_{k + 1}, \dots, x_n) = \frac{d f}{d x_k}(0, \ldots, 0, t, 0, \dots, 0)$ for all $k$ then we will be done, since then the previous equation just becomes
\begin{align*}
f(x_1, \ldots, x_n) - f(0, \ldots, 0)
&= \int_0^{x_1} g_1'(t) d t
+ \int_0^{x_2}  g_2'(t) d t 
+ \cdots
+ \int_0^{x_n}  g_n'(t) d t\\
&= g_1(x_1) + \cdots + g_n(x_n),
\end{align*}
as desired.
To check that $\frac{d f}{d x_k}(x_1, \ldots, x_{k - 1}, t, x_{k + 1}, \dots, x_n) = \frac{d f}{d x_k}(0, \ldots, 0, t, 0, \dots, 0)$ we finally use the hypothesis that off-diagonal entries of the Hessian of $f$ vanish: note just like before by the fundamental theorem of calculus we have
\begin{align*}
\frac{d f}{d x_k}(x_1, \ldots, 0, t, 0, \dots, 0) - \frac{d f}{d x_k}(0, \ldots, 0, t, 0, \dots, 0)
&= \int_0^{x_1} \frac{d}{d x_1} \frac{d f}{d x_k}(s, \ldots, 0, t, 0, \dots, 0) d s\\
&= 0,
\end{align*}
since $\frac{d}{d x_1} \frac{d f}{d x_k} = 0$ (because $k \not = 1$).
so that inductively we can conclude
\begin{align*}
\frac{d f}{d x_k}(x_1, \ldots, x_{k - 1}, t, x_{k + 1}, \dots, x_n) - \frac{d f}{d x_k}(0, \ldots, 0, t, 0, \dots, 0),
\end{align*}
as desired.
A: First recall the definition of the Hessian $H$ of a function $f$
$$ H_{i,j} = \partial_{x_i,x_j} f $$
For all the off-diagonal term to be zero (e.i. $H_{i,j} = 0 \;\;\;\; \forall ij : i \neq j$)
$$ 0 = \partial_{x_i,x_j} f \;\;\;\; \forall ij : i \neq j$$
must happen, suppose $f$ is a linear combination of functions with dependency in no more than a variable
$$ f = \sum_{m=1}^M a_m g_m(x_m) $$
so the $ 0 = \partial_{x_i,x_j} f \;\;\;\; \forall ij : i \neq j $ of $f$ is
$$ \partial_{x_i,x_j} f = \sum_{m=1}^M a_m \partial_{x_i,x_j} g_m(x_m)  = 0 $$
This trivially implies that if $f$ is a linear combination of functions with dependency in no more than a variable, then the off-diagonal entries will be zero. And I conjecture the before mentioned condition is nesesary and sufficient to the off-diagonal entriess of the Hessian be zero.
For gain some intuition of why it would no apply to other cases. Suppose $f$ is a multivariate polynomial of degree two
$$ f = a + \sum_{n = 1}^N \sum_{m=1}^2 x_n^m + \sum_{n = 1}^N \sum_{m = 1}^M c_{n,m} x_n x_m $$
We again we get the Hessian off-diagonal derivatives
$$ \partial_{x_i,x_j} f = \sum_{n = 1}^N \sum_{m = 1}^M c_{n,m} \partial_{x_i,x_j} x_n x_m \neq 0 \,\,\,\, \text{for} \,\, i \neq j$$
