0
$\begingroup$

Could anyone help me with this result?

To solve the equation $P(x)=\theta$, we can aply a modified Newton method such that the itarations are given by the formula $$x_{n+1}=x_n-[P'(x_1)]^{-1}P(x_n) \quad (1)$$ supposing that $P$ is Fréchet differentiable on a convex region $D$, and for some $x_1 \in D$, the inverse $[P'(x_1)]^{-1}$ exists.

Suppose that $x_2$ computed with $(1)$ belongs to $D$, also suppose that $$\rho=\lVert [P'(x_1)]^{-1}\rVert \cdot \sup_{x\in D}\lVert P'(x_1)-P'(x)\rVert<1$$ and the ball $$S=\{x:\lVert x-x_2\rVert<\frac{\rho}{1-\rho}\lVert x_1-x_2\rVert\}$$ is a subset of $D$. Show that this modified method converges to a solution $x_0\in S$.

Any help would be appreciated.

  • 0
    Show that the map given by the first formula maps $S \to S$ and is a contraction. I would work with the closure of $S$. The only tricky part is showing $S \to S$. – 2017-02-05
  • 0
    Could you please give me an idea how to prove the "tricky" part? I see that if i prove that the suggested map is a contraction, then aplying Banach's fixed point theorem the result holds, but I cannot see how to prove that this mapping go from $S$ to $S$. – 2017-02-06
  • 0
    Sorry, I should have said that the sequence of points lies in $S$ if it starts in $S$. – 2017-02-06
  • 0
    I think you need $S$ to be closed (complete) to conclude that $x_0 \in S$. – 2017-02-06

1 Answers 1

1

Let $\phi(x) = x - P'(x_1)^{-1} P(x)$, then $\phi'(x) = I - P'(x_1)^{-1} P'(x) = P'(x_1)^{-1} (P'(x_1)-P'(x))$ from which we can see that if $x,y \in D$ then $\|\phi(x)-\phi(y)\| \le \rho \|x-y\|$, with $\rho \in (0,1)$.

Let $S = \{ x | \|x-\phi(x_1)\| < {\rho \over 1-\rho } \|\phi(x_1) - x_1 \| \}$.

Then the iterates $x_{n+1} = \phi(x_n)$ lie in $S$ for $n >1$. First note that $x_2 = \phi(x_1) \in S$. Suppose $x_2,...,x_{n} \in S$. Note that $\|\phi(x_k) -\phi(x_{k-1}) \| \le \rho^{k-2} \| \phi(x_2) - \phi(x_1) \| $. Then $\|x_{n+1} - \phi(x_1) \| = \|\phi(x_{n}) - \phi(x_1) \| $ and \begin{eqnarray} \|\phi(x_{n}) - \phi(x_1) \| &\le& \sum_{k=2}^n \|\phi(x_{k}) - \phi(x_{k-1}) \| \\ &\le& (\rho^{n-2}+ \cdots + 1) \| \phi(x_2) - \phi(x_1) \| \\ &<& {1 \over 1-\rho} \| \phi(x_2) - \phi(x_1) \| \\ &\le& {\rho \over 1-\rho} \| \phi(x_1) -x_1 \| \end{eqnarray} hence $x_{n+1} \in S$.

A small modification of the above analysis shows that the sequence $x_n$ is Cauchy and hence converges to some $x_0 \in \overline{S}$ (the closure).

(I find Kantorovich & Akilov's "Functional analysis" useful for such things.)

  • 1
    Thank you a lot! @copper.hat – 2017-02-06