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Let $\alpha,\beta\in\mathbb{R}^n$, $a\in \mathbb{R}$, and $A$ be an $n\times n$ nonsingular matrix. Use the Banach Fixed Point Theorem to prove that for a sufficiently small $\epsilon > 0$, $$\alpha - Ax + \epsilon(a - \beta^tx)x = 0_v \tag{*}$$ has a unique solution in $$D = \{x \in \mathbb{R}^n \,|\, \|x - A^{-1}\alpha\| \le 1\}.$$

I have a hunch that the solution to this problem is incredibly simple, but unfortunately my experience with the Banach Fixed Point Theorem is incredibly limited. I know that the Banach Fixed Point Theorem is as follows:

Let $S$ be a closed subset of a Banach space, and $T: S \to S$ be a contraction. Then there exists a unique $x^* \in S$ such that $$T(x^*) = x^*,$$

and that a contraction is defined as

A mapping $T: S \to S$ is called a contraction if there exists a $\rho \in (0,1)$ such that $$\|T(x) - T(y)\| \le \rho\|x - y\|$$ for all $x,y \in S.$

Here's what I think I need to do:

  1. Show that $D$ is a closed subset of $\mathbb{R}^n$
  2. Create a mapping $T: D\to D$ such that $T$ is a contraction
  3. Conclude by the Banach Fixed Point Theorem that $(*)$ has a unique solution in $D$.

I'm afraid that I need a few hints to part 2. Is there anything obvious about the problem as it's stated that gives us a hint as to how to construct a mapping $T$? Also from what I understand, the norm used is the Euclidean norm. I have skipped part 1 for the moment because 1), it seems trivial that $D \subseteq \mathbb{R}^n$ and 2), I'm not quite sure how to specifically show that $D$ is a closed subset, so if the solution to my question lies there, I apologize. Please let me know.

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1 Answer 1

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Hint

Note that the goal is $$\alpha - Ax + \epsilon(a - \beta^tx)x = 0_v \tag{*}.$$ Rearranging, it is the same as (since $A$ is invertible) $$ x = A^{-1} (\alpha + \epsilon(a - \beta^tx)x).$$ So if you let $T(x) = A^{-1} (\alpha + \epsilon(a - \beta^tx)x)$, you are looking for a fixed point of $T$.

Note that $D$ is a closed ball centered at $A^{-1}\alpha$ with radius $1$. Do you know how to show a closed ball is closed?

Actually, the harder part is to show $T :D\to D$ and $T$ is a contraction. This is where you need to choose a small $\epsilon$.

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  • $\begingroup$ I feel like a total idiot right now; I actually took a class on numerical analysis and we covered the topic of fixed-point iteration. Rearranging of $(*)$ is now completely obvious--I just unfortunately did not make the connection to the Banach Fixed Point Theorem in this case. And I know the general outline of how to show that $D$ is closed. We can find that information here: math.stackexchange.com/a/1149534/227902 I greatly appreciate your help! $\endgroup$
    – Decaf-Math
    Commented Sep 5, 2016 at 5:30
  • $\begingroup$ @pyrazolam You are welcome~ $\endgroup$
    – user99914
    Commented Sep 5, 2016 at 5:44

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