# Do convergence rates for (convex) gradient descent apply when domain is (convex) subset of reals?

I have a convex multi-variate optimization problem where each variable lies on the domain $$[x, \infty)$$ for some positive number $$x$$. I know the problem has a unique finite solution in the domain, and am interested in its convergence rate.

There are several convergence rates proven for Gradient Descent on convex problems (with Lipshitz gradient). However I always see such rates given assuming the variables range over all the real numbers.

Do these rates still apply when the domain is restricted to a convex subset like I stated earlier? In particular, does the $$O(\frac{1}{\epsilon})$$ rate of gradient descent apply? I assume one needs to modify the algorithm to ensure the value does not get less than $$x$$ for this to apply. I would assume it is something like: if $$y^{t+1} = y^t - \mu^t \nabla y^t$$ violates the $$y \geq x$$ condition then decrease $$\mu^t$$ until it no longer does, but would not be surprised if a more complicated sequence of step sizes is necessary for convergence.

If not, are there any conditions we could add to the problem to recover the convergence rate?

I assume that whatever the answer is will be a well-known (perhaps even trivial) result, but I have had a surprisingly difficult time finding a direct answer to this online, so any help would be appreciated.

• Since these results are asymptotic, it's hard to see why there would be any problem applying them to arbitrary convex subsets except when the convergence is to the boundary.
– whuber
Commented Jul 12 at 16:59

I think your assumption is mistaken. Most algorithms I worked with handle simple constrains on domains by automatically transforming to an unconstrained $$\mathbb{R}^n$$ under the hood. For your specific case, the typical solution would be to have:
$$y^\prime = \log (y - x)$$
Now $$y^\prime$$ ranges over the whole real line and the optimizer will work with $$y^\prime$$ and compute gradients w.r.t $$y^\prime$$. If needed, the algorithm will automatically adjust the target to account for this transformation. Other transformation functions from $$[x, \infty)$$ to $$\mathbb{R}$$ are obviously possible as long as they are sufficiently well behaved.