Bounding a sequence involving minimizers of strictly convex quadratic functions

Question

Suppose that $H\succ 0$ and we have a sequence $\{x_j\}_{j\ge 1}$ such that each one is the unique solution of the following problem:

$$ x_{j} = \text{argmin}_{x\in \mathbb R^n}(x-x_{j-1})^T\nabla f(x_{j-1})+\frac{1}{2}(x-x_{j-1})^TH(x-x_{j-1})+\frac{1}{2}\rho_j\|x-y_j\|_2^2 .$$

Note that $\rho_j \to \infty$ as $j \to \infty$. Further, $x_0$ can be arbitrary and we know the sequence $\{y_j\}_j$ is bounded.

I want to show the sequence $$(x_j-x_{j-1})^T\nabla f(x_{j-1})+\frac{1}{2}(x_j-x_{j-1})^TH(x_j-x_{j-1}), \, j\in \mathbb N$$ is bounded below.

It turns out that if $f$ is convex and Lipshitz continues on the whole space, there is a lower bound (Existence of a lower-bound for an interesting function! and Lipschitz implies bounded gradient).

Unfortunately, I am not finding practical functions $f$ that satisfy these conditions (convexity and Lipshitz continuity on the entire space) together. So, I want to see if finding a lower bound is doable with less restrictive conditions or at least practical ones.

Thank you for your time!

Answer 1

I assume the $\mathrm{argmin}$ is supposed to define $x_j$ instead of $x_{j+1}$.

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In general, without further constraints, the sequence may be unbounded below. For instance, in dimension $1$, take $H = 1$, $x_0 = 0$, $f(x) = -x - (x+1)^4/4$, $y_j = 0$ and $\rho_j = j^2$ for all $j \ge 1$. One can show that $x_j = j$ for all $j \ge 0$, and $$(x_j - x_{j-1})\nabla f(x_{j-1}) + (1/2)H(x_j - x_{j-1})^2 = -(1 + j^3) + (1/2).$$

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In a positive direction, let's assume that there exists $K > 0$ such that $\| \nabla f(x)\| \le K$ for $x$ inside the closed ball of radius $3$ centred at $0$, and $\| \nabla f(x)\| \le K \|x\|$ for $x$ outside this ball. Then the sequence $(x_j - x_{j-1})^T \nabla f(x_{j-1}) + \frac{1}{2} (x_j - x_{j-1})^T H (x_j - x_{j-1})$ is bounded from below (by a possibly $x_0$-dependent constant).

We set $c := \mathrm{min}_{\|x\| = 1} x^t H x > 0$ and $C := \mathrm{max}_{\|x\| = 1} x^t H x > 0$. Without loss of generality (dilating/contracting $\mathbb{R}^n$ if necessary), we assume $\|y_j\| \le 1$ for all $j$. Since we are really only concerned with the case of large $j$ and $\rho_j \to \infty$, we may also assume that $\rho_j \ge 3(3C+K)$ for all $j$.

We compute that $$ (\clubsuit) \quad x_j = (H + \rho_j I)^{-1} \left( Hx_{j-1} + \rho_j y_j - \nabla f(x_{j-1}) \right) .$$ Note that $\|(H + \rho_j I)^{-1}\| \le (c + \rho_j)^{-1}$ and $\|(H + \rho_j I)^{-1}H\| \le (c + \rho_j)^{-1}C$.

• Suppose $\|x_{j-1}\| > 3$. From ($\clubsuit$), we get \begin{align} \|x_j\| \le (c + \rho_j)^{-1}(C+K)\|x_{j-1}\| + 1 \le (1/3) \|x_{j-1}\| + 1 \le (2/3) \|x_{j-1}\|. \end{align} Since $(2/3)^n \to 0$ as $n \to \infty$, we see that there exists $j_0$ such that $\|x_{j_0}\| \le 3$.

• Whenever $\|x_{j-1}\| \le 3$, from ($\clubsuit$), we get \begin{align} \|x_j\| \le (c + \rho_j)^{-1}(C \|x_{j-1}\| +K) + 1 \le 2 < 3. \end{align}

Therefore, we see that regardless of $x_0$, the sequence $x_j$ eventually falls and stays in the ball of radius $3$ centred at $0$. However, whenever $\|x_{j-1}\| \le 3$, we have the lower bound \begin{align} (x_j - x_{j-1})^T \nabla f(x_{j-1}) &+ \frac{1}{2} (x_j - x_{j-1})^T H (x_j - x_{j-1}) \\ &\ge - \|x_j - x_{j-1}\| K + (c/2) \|x_j - x_{j-1}\|^2 \\ &\ge - K^2/2c. \end{align}