Consider a twice differentiable strongly convex function $f:\mathbb{R}^n \rightarrow \mathbb{R^+}$ that attains its minimum value at the point $x^\*$. I am wondering if one can compute a direction of *slowest* ascent $u$ from the point $x^\*$. 


Intuitively, the direction of slowest ascent would be a unit vector $u$ such that $f(x^\*+u)$ is as close as possible to $f(x^\*)$ (or at least, we would expect $f(x^\*+u)$ and $f(x^\*)$ to be close based on the information provided by the Hessian at $x^\*$).


I am having some difficulty formally defining $u$ since it depends on the size of the step that we take away from $x^\*$. That said, a poor man's definition would be:


$$ u = \text{argmin}\_{\|w\| \neq 0} ~~f(x^\* + w) - f(x^\*)$$



How to compute the direction of slowest ascent from the minimum of a strongly convex function?
There may not be a unique 'direction of slowest ascent', take for example the simple case of $f(\vec{x}) = x\_0^2 + x\_1^2$, in which case the minimum is clearly (0,0), and any direction is equally fast. That said, near enough to the minimum, the you can define a 'subspace of slowest ascent' which is basically just the eigenspace associated with the lowest eigenvalue of the Hessian. If the lowest eigenvalue is non-degenerate, then the direction you seek is well defined.


