Convergence of a continuous time stochastic gradient descent algorithm

Question

Let $f: \mathbb R \to \mathbb R$ be a $C^1$ convex function, satisfying the growth conditions

$$\lim_{x \to -\infty} \nabla f(x) = -\infty, \lim_{x \to \infty} \nabla f(x) = \infty.$$

and let $\gamma_t: [0, \infty) \to \mathbb R$ be a deterministic, Borel measurable process satisfying the following conditions:

1. $\gamma_t > 0$ for all $t \in [0, \infty)$.
2. $\gamma_t \to 0$ as $t \to \infty$.
3. $\int_0^\infty \gamma_t^2 \, dt < \infty$.

Consider the solution $X$ to the one dimensional SDE

$$dX_t = -\nabla f(X_t) \, dt + \gamma_t \, dW_t, \, \, X_0 = x_0$$

with $W$ a standard Brownian motion, and $x_0 \in \mathbb R$ an arbitrary initial condition.

Question: Is it true that for all initial conditions $x_0$, we have $\nabla f(X_t) \to 0$ as $t \to \infty$ almost surely?

Answer 1

here "An SDE perspective on stochastic convex optimization" they study these type of SDEs (Stochastic gradient descent) and basically ask bounded and an L2 condition for gamma_t in theorem 3.1.

Here in (H0) they also have extra constraints to $f$ that might be sharp.

As shown here a function can be convex and C1 but its gradient not be Lipschitz

The map $f:\Bbb R\to \Bbb R$, $f(x)=\frac23\lvert x\rvert^{3/2}=\int_0^x \lvert t\rvert^{1/2}\operatorname{sgn}t\,dt$ is convex and $C^1$, but $f'(x)=\lvert x\rvert^{1/2}\operatorname{sgn}x$ is not Lipschitz continuous in any neighbourhood of $0$. More generally, integrate your favourite monotone increasing continuous function which is not locally Lipschitz and you'll obtain a counterexample in $\Bbb R$.

And as shown here without this Lipschitz condition, we don't even have convergence for the deterministic gradient descent.