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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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    @Martin Sleziak: Dear Sir. I would like to hear your advice and thoughts on this problem.2012-10-02
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    @t.b.: Dear Sir. I would like to hear your advice and thoughts on this problem.2012-10-02
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    @LVK: Dear Sir. I would like to hear your advice and thoughts on this problem2012-10-02
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    I suspect that the answer must be a solution of the quadratic equation, $A=\left(1\pm\sqrt{1+4(F-1)}\right)/(2\alpha)$, interpreted as a power series in $F-1$, which is the difference operator, but proving that one of these fulfils the other two conditions seems tricky.2012-10-02
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    Have you try applying the continuous functional calculus to it?2012-10-02
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    @joriki: Thank you for your consideration and comment. I am really stuck in this question.2012-10-02
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    @Nonliapunov: Thank you for your consideration. I have tried. But I find it difficult to overcome.2012-10-02
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    If you consider $H=\ell^2$ as a complex Hilbert space, the answer is that there is no such a map. The spectrum of $F$ is the closed unit disk, the spectrum of $A$ is non-negative, so $\sigma(A)=[0,r)$ and by the spectral mapping theorem $f(\sigma(A))=\sigma(f(A))\neq\sigma(F)$ with $f(x)=I-\alpha x+\alpha^2 x^2$. But in a real Hilbert space, the spectral mapping theorem holds with inclusions and not equalities, I don't know how to get over that. Hope it helps.2012-10-02
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    @Nonliapunov: Dear Sir. Thank you for your helping. I confess that I want the answer is positive.2012-10-02
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    I'm afraid it might not be true. Since $F$ is the right shift, $-1\in \sigma(F)$, hence $I+F$ is not "invertible". On the other hand, $(I-\alpha A)+(I+\alpha^2A^2)$ is invertible: $\alpha\in (0, 1/\|A\|)$ so that $\|\alpha A\|=r<1$, then by the spectral radius result, $1\notin \sigma(\alpha A)$, similarly $-1\notin \sigma(\alpha^2 A^2)$. If you have a source where it says its true, I'd like to check it. Hope it helps.2012-10-03
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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.