# Finding a linear mapping in a special Hilbert space

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. – blindman Oct 2 '12 at 7:35
@t.b.: Dear Sir. I would like to hear your advice and thoughts on this problem. – blindman Oct 2 '12 at 7:37
@LVK: Dear Sir. I would like to hear your advice and thoughts on this problem – blindman Oct 2 '12 at 7:40
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. – joriki Oct 2 '12 at 8:53
Have you try applying the continuous functional calculus to it? – Nonliapunov Oct 2 '12 at 8:57

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.