I am considering a function $f_i : \mathcal{D}\times\mathcal{D} \to \mathbb{R}$, where $\mathcal{D}$ is a finite space in $\mathbb{R}^3$.
The function $f_i$ is differentiable with respective to $x_i\in\mathcal{D}$ and concavely decreasing with respect to $\Vert x_i-a \Vert$ with a fixedly given $a\in\mathcal{D}$.
For example, denoting the $\ell_2$-norm by $\Vert\cdot\Vert$, a function $f_i(x_i,a)$ can be defined as $100-\Vert x_i-a \Vert^2$, $500-e^{\Vert x_i-a \Vert}$, and etc.
In this case, I want to solve $$\max_{x_1,\ldots,x_N\in\mathcal{D}} f(x_1,\ldots,x_N) = \sum_{i=1}^{N} f_i(x_i,a_i)$$ where $a_i$ for $i=1,\ldots,N$ are all given.
I think the above problem is concave function with respect to $(x_1,\ldots,x_N)$. Hence, I first tried to solve the problem using the Gradient descent algorithm, and saw that it solved well by the Gradient descent algorithm. Now, I want to prove its convergence.
To prove the Gradient descent algorithm converges to a solution of the above problem, I tried to use the following theorem:
With any convex and differentiable function $f:\mathbb{R}^n\to\mathbb{R}$, if $$\Vert \nabla f(x) - \nabla f(y) \Vert \le L \Vert x-y\Vert$$ for any $x,y\in\mathbb{R}^n$, the Gradient descent with a fixed step size $t\le 1/L$ has convergence rate $O(1/k)$, where $O$ is big-O notation representing the complexity and $k$ stands for the $k$th iteration in the Gradient descent algorithm.
In this context, how can I prove that the gradient of the function $f$ with which I am concerned is Lipschitz continuous? That is, given $a_i$, how can I prove $$\Vert \nabla f_i(x,a_i) - \nabla f_i(y,a_i) \Vert \le L \Vert x-y\Vert?$$