A minimization problem in function fitting setup

Let $\Omega$ be a convex, closed, compact set in $\mathbb{R}^d$ with a smooth boundary.

Given a data $(x_i,d_i)$, $x_i \in \Omega$,$d_i \in \mathbb{R}$, $i = 1,2,3...N$, $N>d$ and $\sum\limits_{i=1}^N d_i = 0$. Also given that, there are always $d$ vectors in $\{x_i\}$ which are linearly independent.

Let $A = \int_{\Omega}dx$

I want to find a continuous function $f:\Omega \to \mathbb{R}$, such that,

1. $\int_{\Omega}f(x)dx = 0$ and
2. $C(f)$ is minimum, where $$C(f) = \frac{A^{1/d}}{N}\bigg(\sum\limits_{i=1}^N |f(x_i)-d_i|^d\bigg)^{1/d} +\|f\|_{L^d}+ A^{1/d} \||\nabla f|\|_{L^d}$$

Does the solution exist? Is the solution unique? I mean any two solutions are equal almost everywhere?

The quick answer to your question (for $d > 1$) is no, there is no continuous solution. This can be rectified, however, by using a different $L^p$ norm.

Starting with the case of the $L^d$ norm as asked in the question, where $d$ is also the dimension of the Euclidean space, consider the following. For any $x_0\in\Omega$ and real numbers $r_0,A > 0$, define $f\colon\Omega\to\mathbb R$ by $$ f(x) = \begin{cases} 0,&{\rm if\ }r\ge r_0,\\ 1,&{\rm if\ }r\le e^{-A}r_0,\\ A^{-1}\log(r_0/r)&{\rm if\ }e^{-A}r_0 < r < r_0 \end{cases} $$ where $r=\lVert x-x_0\rVert_2$. The norm $\lVert f\rVert_{L^d}$ is bounded by $cr_0$ for constant $c$, so can be made small by taking $r_0$ to be small. Also, $\nabla f$ is of magnitude $1/(Ar)$ for $e^{-A}r_0 < r < r_0$. So (for some other constant $c$), \begin{align} \lVert\nabla f\rVert_{L^d}^d&=c\int_{e^{-A}r_0}^{r_0}(Ar)^{-d}\,(r^{d-1}dr)\\ &=cA^{-d}\left[\log r\right]_{e^{-A}r_0}^{r_0}=cA^{-(d-1)} \end{align} So, by taking $A$ large and $r_0$ small, we can make $\lVert f\rVert_{L^d}$ and $\lVert\nabla f\rVert_{L^d}$ as small as we like, with $f(x_0)=1$. By taking linear combinations of such functions (with $x_0$ replaced by the $x_i$ in the question, $C(f)$ can be made arbitrarily small. Clearly, the limit $C(f)=0$ is not obtained by a continuous function.

Instead, consider using the following for $C(f)$, $$ C(f)=a\left(\sum_{i=1}^N\lvert f(x_i)-d_i\rvert^p\right)^{1/p}+b\lVert f\rVert_{L^p}+c\lVert\nabla f\rVert_{L^p} \tag1 $$ for any positive constants $a,b,c$ and $p > d$. The Sobolev embedding theorem and Morrey's inequality bounds the $C^{0,\alpha}$-norm of $f$ by a positive multiple of $C(f)$, with $\alpha=1-\frac dp$. This guarantees that the limit of any sequence of continuous $f_n\colon\Omega\to\mathbb R$ with $C(f_n)$ bounded is, at the very least, $\alpha$-Hölder continuous. Any sequence $f_n$ with $C(f_n)$ converging to $\inf\{C(f)\colon \int f=0\}$ satisfies $$ a\left(\sum_{i=1}^N\lvert f_n(x_i)-f(x_i)\rvert^p\right)^{1/p}+b\lVert f_n-f\rVert_{L^p}+c\lVert\nabla(f_n-f)\rVert_{L^p}\to0 \tag2 $$ where $f\colon\Omega\to\mathbb R$ is the unique continuous function minimising $C(f)$. The convergence of $f_n$ and the existence and uniqueness of the limit $f$ follows from uniform convexity of the $L^p$ norm for $1 < p < \infty$.