China Economy Index

Last edited: February 2, 2026

convergence of self-concordant functions

Last edited: February 2, 2026

Functions is self-concordant if:

\begin{align} \mid f’’’\qty(x)\mid \leq 2f’’\qty(x)^{\frac{3}{2}}, \forall x \in \text{dom } f \end{align}

and \(f\) is self-concordant if \(g\qty(t) = f\qty(x+tv)\) is self concordant for all \(x \in \text{dom } f\).

Convergence analysis! There exists \(\eta \in (0, \frac{1}{4}]\), \(\gamma > 0\) such that:

if \(\lambda \qty(x) > \eta\), then \(f\qty(x^{(k+1)}) - f\qty(x^{(k)}) \leq -y\)
if \(\lambda \qty(x) \leq \eta\), then \(2\lambda \qty(x^{(k+1)}) \leq \qty(2 \lambda \qty(x^{(k)}))^{2}\)

and \(\eta, \gamma\) depends only on backtracking line search parameters. This gives bounds:

Last edited: February 2, 2026

EE364A.stanford.edu

Last edited: February 2, 2026

Descent methods are generally of shape:

\begin{align} x^{(k+1)} = x^{(k)} + t^{(k)} \delta x^{(k)} \end{align}

choosing these is a matter of which descent method you choose. The Hessian thus exposes:

\begin{align} \qty {x + v \mid v^{T} \nabla^{2} f\qty(x) v \leq 1} \end{align}

A first-order Taylor line in \(t\), centered about \(x\):

\begin{align} f\qty(x) + t \nabla f\qty(x)^{T} \Delta x \end{align}

We can degrade this to raise it to be slightly higher:

Last edited: February 2, 2026

Iterative methods require a starting point \(x^{(0)}\) such that:

\(f\) is strongly convex on \(S\) if there exists \(m > 0\) such that:

\begin{align} \nabla^{2}f\qty(x) \succeq mI, \forall x \in S \end{align}

if \(f\) is strongly convex for \(x, y \in S\), we have:

\begin{align} f\qty(y) \geq f\qty(x) + \nabla f\qty(x)^{T} \qty(y-x) + \frac{m}{2} \norm{x-y}_{2}^{2} \end{align}