Let $\mathbb{X}$ be a real Banach space.

The subject will be the linear operator equation $$f = \mathbf{A} f + b , \qquad f \in \mathbb{X}. $$ Given are the continuous linear operator $\mathbf{A} \colon \mathbb{X} \to \mathbb{X}$, and the fixed element $b \in \mathbb{X}$.

Suppose that $0< \| \mathbf{A} \| <1$. Then, the Neumann series theorem implies that $\mathbf {I} - \mathbf{A}$ is a bijection on $\mathbb{X}$, and hence the inverse exists as a continuous linear operator on $\mathbb{X}$ ( i.e., $(\mathbf {I} - \mathbf{A})^{-1} = \sum^{\infty}_{n=0}\mathbf{A}^{n}$. )

Thus, we know that the unique solution of the linear operator equation $f = \mathbf{A} f + b $ is $f^{\ast} = (\mathbf {I} - \mathbf{A})^{-1} b$.

Under this circumstance, could we get any algebraically simple relation between the fixed point $f^{\ast}$ and the fixed element $b$ ? ( i.e., $f^{\ast} = \lambda b$ for some real number $\lambda$ ? )

To answer this question, I was thinking to calculate the norm of the operator $(\mathbf {I} - \mathbf{A})^{-1}$ or the spectral radius $\rho \left( (\mathbf {I} - \mathbf{A})^{-1} \right)$ of this operator. But I get suck in the computing of this spectral radius.

Compared with the simplest case when let $\mathbb{X} = \mathbb{R}$, we have $ \rho(\mathbf{A}) = \| \mathbf{A} \| = |\mathbf{A}| <1 $ and hence, $\rho \left( (\mathbf {I} - \mathbf{A})^{-1} \right) = \left( 1 - \rho(\mathbf{A}) \right)^{-1}$.

Now I'm thinking and wondering that is there any similar (or generalized) result about $ \| (\mathbf {I} - \mathbf{A})^{-1} \| = \left( 1 - \| \mathbf{A} \| \right)^{-1}$ ( or, $\rho \left( (\mathbf {I} - \mathbf{A})^{-1} \right) = \left( 1 - \rho (\mathbf{A}) \right)^{-1}$ ) for some general Banach space (such as $L_p$ space)? Or, there might be some inequality relation between $ \| (\mathbf {I} - \mathbf{A})^{-1} \|$ and $\left( 1 - \| \mathbf{A} \| \right)^{-1}$?

Any idea or suggestion would be much appreciated! Thanks:)

  • $\begingroup$ What do you mean by "...which is defined by $f = Af + b$ for some fixed $b$"...? That certainly does not define $A$; it may define $f$ (if the solution is unique). $\endgroup$ – Robert Israel Jul 24 '17 at 5:00
  • $\begingroup$ @RobertIsrael Thanks for pointing out this issue. I've already revised it:) $\endgroup$ – Paradiesvogel Jul 24 '17 at 5:20

$$\|(I - A)^{-1}\| \le \frac{1}{1-\|A\|} \text{ if $\|A\| < 1$} $$ again using the Neumann series. But $I - A$ can be invertible for any $\|A\|$.

In the other direction, $$\|(I-A)^{-1}\| \ge \frac{1}{\|I - A\|} \ge \frac{1}{1+\|A\|}$$

And no, there is no algebraically simple relation between $f$ and $b$. In particular, there is no reason for them to be linearly dependent.

  • $\begingroup$ Is it possible to make them linearly dependent? I thought one case in which $b$ is an eigenvector of $(\mathbf{I} - \mathbf{A})^{-1}$, then for every eigenvector there must be a unique eigenvalue corresponding it. Are there any other cases? Many thanks again:) $\endgroup$ – Paradiesvogel Jul 24 '17 at 5:52

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