I think this function is strictly convex in the vector ${\bf x} = (x_1,x_2,x_3,x_4)$ but the fact that some terms are zero when variables take on the same values leaves me uncertain, i.e. when $x_1=x_2$ and $x_1=x_3$ then the argument of $c_1$ is zero.
$C({\bf x}) = c_0(a_0x_1^2)+c_1\left(a_1(x_1-x_2)^2+b_1(x_1-x_3)^2\right) + c_2\left(a_2(x_2-x_1)^2+b_2(x_2-x_4)^2\right)$
where each $c_i$ is strictly increasing and strictly convex, $c_i(0) =0 $ and all $a_i,b_i>0$. What is throwing me off is the fact that a function like $f(x,y) = (x-y)^2$ is weakly convex.
Thank you
First note a result which is straightforward to verify: Suppose $f$ is convex and $\phi: \mathbb{R} \to \mathbb{R}$ is convex and non-decreasing, then $\phi \circ f$ is convex. In addition, if $f$ is strictly convex, and $\phi$ is strictly increasing, then $\phi \circ f$ is strictly convex.
Let $A = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 1 & -1 & 0 & 0 \\ 1 & 0 & -1 & 0 \\ 0 & 1 & 0 & -1 \end{bmatrix}$, and note that $\det A = 1$.
Then $C = \kappa \circ A$, where $\kappa(y) = c_0(a_0 y_1^2)+c_1\left(a_1y_2^2+b_1 y_3^2\right) + c_2\left(a_2 y_2^2+b_2 y_4 ^2\right)$, and I am abusing notation slightly by using $A$ to represent the map $x \mapsto Ax$.
Since $A$ is an invertible linear map, we see that $C$ is strictly convex iff $\kappa$ is.
Let $f_0(y) = c_0(a_0 y_1^2)$, $f_1(x) = c_1\left(a_1y_2^2+b_1 y_3^2\right)$, $f_2(y) = c_2\left(a_2 y_2^2+b_2 y_4 ^2\right)$, and note that from the hypotheses, the $f_k$ are all convex. Also note that by the above result, the restricted maps $y_1 \mapsto c_0(a_0 y_1^2)$, $(y_2,y_3) \mapsto c_1\left(a_1y_2^2+b_1 y_3^2\right)$, $(y_2,y_4) \mapsto c_2\left(a_2 y_2^2+b_2 y_4 ^2\right)$ are strictly convex.
We have $\kappa = f_1+f_1+f_2$, and if $\lambda \in [0,1]$, we have $f_k(\lambda x + (1-\lambda ) y) \le \lambda f_k(x) + (1-\lambda) f_k (y) $, for each $k$, and so we see that $\kappa$ is convex. Furthermore, note that at any particular $x,y,\lambda$, if the inequality is strict for any of the $f_k$, then the inequality is strict for $\kappa$.
Now suppose $x\neq y$ and $\lambda \in (0,1)$. Then, since $x_i \neq y_i$ for at least one $i$, we see that the inequality is strict for at least one $f_k$, hence $\kappa$ is strictly convex.
Note that strict convexity depends on the domain in the sense that if you extend a strictly convex function to a larger domain, it may no longer be strictly convex. For example, the function $x_1 \mapsto x_1^2$ is strictly convex, but the function $(x_1 , x_2) \mapsto x_1^2$ is not.