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?



Convergence of a continuous time stochastic gradient descent algorithm
here "[An SDE perspective on stochastic convex optimization](https://arxiv.org/abs/2207.02750#:%7E:text=We%20analyze%20the%20global%20and,problems%20with%20noisy%20gradient%20input.)" 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](https://math.stackexchange.com/questions/2842884/c1-convex-function-but-not-c1-1) 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](https://math.stackexchange.com/questions/1507138/examples-where-constant-step-size-gradient-descent-fails-everywhere?rq=1) without this Lipschitz condition, we don't even have convergence for the deterministic gradient descent.


