It is well know that if one wants to perform gradient descent on a function $f(x)$ the ideal conditions would be those of strong convexity and Lipschitz continuous gradient, i. e., $$ 0 <\ell \le \nabla^2 f(x) \le L $$ for $\ell, L \in \mathbb{R}$.
Assume Instead we have a function $g(x)$ that does not have Lipschitz continuous gradient nor is strongly convex but we know its gradient $\nabla g(x)$ and that it has a (unique) minimum in $x^*$.
Could we perform gradient descent successfully utilizing as a gradient $h(x) \nabla g(x)$, where $h(x)$ is a function that shares the support of $g(x)$ and is s.t. $h(x^*) \nabla g(x^*) = 0$ (only at $x^*$) and
$$ 0 <\ell \le \nabla (h(x) \nabla g(x)) \le L ,$$
instead of just $\nabla g(x)$.
That is utilizing as an iterative optimization scheme
$$ x_{k+1} = x_k + \alpha h(x_k) \nabla g(x_k) $$
where $\alpha < 2 / L$ instead of
$$ x_{k+1} = x_k + \alpha \nabla g(x_k) $$
I do know believe we can find such a function $h$ to make $h(x)\nabla g(x)$ be a gradient of a strongly convex function. Here's my argument.
We can find an optimization problem which is not possible to solve. For examples, solving an optimization problem minimizing a general $10$th-order polynomial with $10$ variables is impossible. We know that it has (unique) global optimum $x^\ast$, but do not know how to solve it (in a reasonable amount of time). If we COULD find such $h(x)$, the original problem $g(x)$ WOULD be solvable, which is a contradiction!