# How to compute the direction of slowest ascent from the minimum of a strongly convex function?

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:

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

• Near the minimum $f-f(x^*)$ is equivalent to a positive quadratic form, given by the Hessian. This defines a family of hyper-ellipsoids, with coinciding center and axis. The corresponding directions form a diagonalizing orthonormal basis. One candidate vector is probably a unit eigenvector corresponding to smallest eigenvalues. Sep 15, 2015 at 19:51

## 1 Answer

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.