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Let $H=\ell_2$, the real Hilbert space whose elements are the square-summable sequences of real scalars, i.e., $ H=\left\{u=(u_1,u_2,\ldots,u_i,\ldots): \sum_{i=1}^{\infty}|u_i|^2<+\infty\right\}\;. $ Let $F: H\rightarrow H$ be a mapping given by $ F(u)=(0, u_1, u_2, \ldots, u_n, \ldots ) \quad \forall u = (u_1, u_2, \ldots, u_n, \ldots)\in H. $ Finding a linear mapping $A: H\rightarrow H$ satisfying the following conditions:

  • There exists $L>0$ such that $\|Au\|\leq L \|u\|$ for all $u \in H;$

  • $\langle Au, u\rangle\geq 0 \quad \forall u \in H;$

  • There exists $\alpha \in (0, 1/L)$ such that $ I-\alpha A+\alpha^2 A^2=F, $ where $I:H\rightarrow H$ is an identity map.

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    @Nonliapunov: I hope that the answer is positive. This question takes a lot of your time. I am sorry about this. Thank you for your construction.2012-10-03

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The equation $I-\alpha A+\alpha^2 A^2=F$ cannot have a solution under the conditions stated. The left hand side has non-negative spectrum, while the right hand side doesn't.

Indeed, if $t\in\sigma(A)$, then we know that $|t|\leq L$, i.e. $\alpha|t|\leq1$. By the Spectral Mapping Theorem, the spectrum of $I-\alpha A+\alpha^2 A^2$ is of the form $\{1-\alpha t+\alpha^2 t^2:\ t\in\sigma(A)\}$; and $ 1-\alpha t+\alpha^2t^2\geq1-\alpha t\geq0, $ so $\sigma(I-\alpha A+\alpha^2 A^2)\subset[0,\infty)$. On the other hand it is easy to see that $\beta\in\sigma(F^T)$ for any $\beta\in[-1,0)$, so $F^T-\beta I$ is not invertible; then $F-\beta I=(F^T-\beta I)^T$ cannot be invertible, so $\beta\in\sigma(F)$.

Note that the reasoning above does not use the condition $\langle Au,u\rangle\geq0$, only the other two.