It is known, that antigradient \(-\nabla f (x_0)\) is the direction of the steepest descent of the function \(f(x)\) at point \(x_0\). However, we can introduce another concept for choosing the best direction of function decreasing.

Given \(f(x)\) and a point \(x_0\). Define \(B_\varepsilon(x_0) = \{x \in \mathbb{R}^n : d(x, x_0) = \varepsilon^2 \}\) as the set of points with distance \(\varepsilon\) to \(x_0\). Here we presume the existence of a distance function \(d(x, x_0)\).

\[x^* = \text{arg}\min_{x \in B_\varepsilon(x_0)} f(x)\]

Then, we can define another steepest descent direction in terms of minimizer of function on a sphere:

\[s = \lim_{\varepsilon \to 0} \frac{x^* - x_0}{\varepsilon}\]

Let us assume that the distance is defined locally by some metric \(A\):

\[d(x, x_0) = (x-x_0)^\top A (x-x_0)\]

Let us also consider first order Taylor approximation of a function \(f(x)\) near the point \(x_0\):

\[\tag{A1} f(x_0 + \delta x) \approx f(x_0) + \nabla f(x_0)^\top \delta x\]

Now we can explicitly pose a problem of finding \(s\), as it was stated above.

\[\begin{split} &\min_{\delta x \in \mathbb{R^n}} f(x_0 + \delta x) \\ \text{s.t.}\;& \delta x^\top A \delta x = \varepsilon^2 \end{split}\]

Using \(\text{(A1)}\) it can be written as:

\[\begin{split} &\min_{\delta x \in \mathbb{R^n}} \nabla f(x_0)^\top \delta x \\ \text{s.t.}\;& \delta x^\top A \delta x = \varepsilon^2 \end{split}\]

Using Lagrange multipliers method, we can easily conclude, that the answer is:

\[\delta x = - \frac{2 \varepsilon^2}{\nabla f (x_0)^\top A^{-1} \nabla f (x_0)} A^{-1} \nabla f\]

Which means, that new direction of steepest descent is nothing else, but \(A^{-1} \nabla f(x_0)\).

Indeed, if the space is isotropic and \(A = I\), we immediately have gradient descent formula, while Newton method uses local Hessian as a metric matrix.

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