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Assume that $H$ is a separable Hilbert space. Is there a polynomial $p(z)\in \mathbb{C}[x]$ with $deg(p)>1$ with the following property?:

Every densely defined operator $A:D(A)\to D(A),\;D(A)\subset H$ with $p(A)=0$ is necessarily a bounded operator on $H$.

That is the polynomial-operator equation $p(A)=0$ has only bounded solution.

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2 Answers 2

up vote 2 down vote accepted

I do not think so.

Observation: Without loss of generality, $p(x)$ can be taken to be monic (constant multiples won't affect either $p(A) = 0$ or boundedness).

Case 1: $p$ is degree $2$.

By the above reduction, $p(x) = (x - \lambda)(x - \mu)$ for some $\lambda$ and $\mu$ in $\mathbb{C}$ (since $A$ clearly commutes with itself and with $I$, and since $A$ maps $D(A)$ to itself, this decomposition is reasonable). Yet if $p(A) = 0$, take the operator $\displaystyle B = A - \frac{\lambda + \mu}{2} I$ (with the same domain), and we see that for $\displaystyle \nu = \frac{\lambda - \mu}{2}$, $B$ satisfies $(B + \nu)(B - \nu) = 0$, or $B^2 - \nu^2 = 0$. We will show that an unbounded choice of $B$ exists, satisfying $B: D(B) \to D(B)$, hence an unbounded choice of $A$ exists, with $A: D(A) \to D(A)$.

Subcase 1: $\nu = 0$. Then take $H = \ell^2(\mathbb{N})$, let $H_0 = D(B)$ be the sequences with only finitely many nonzero elements, and let $B$ be the operator represented by the infinite matrix

$$ \begin{pmatrix} 0 & 1 & & & & & \cdots \\ 0 & 0 & & & & & \cdots \\ & & 0 & 2 & & & \cdots \\ & & 0 & 0 & & & \cdots \\ & & & & 0 & 3 & \cdots \\ & & & & 0 & 0 & \ddots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \ddots \end{pmatrix},$$ which is clearly well-defined on $H_0$ and clearly maps $H_0$ to itself. Then $B^2 = 0$, but letting $e_j$ be the $j$th basis vector, $B e_{2j} = j e_{2j - 1}$, so clearly $B$ is unbounded.

Subcase 2: $\nu \neq 0$. Then again take $H = \ell^2(\mathbb{N})$, and $H_0$ the almost-everywhere-$0$ sequences. We now define $B$ by the matrix

$$ \begin{pmatrix} 0 & \nu & & & & & \cdots \\ -\nu & 0 & & & & & \cdots \\ & & 0 & 2\nu & & & \cdots \\ & & -\frac{1}{2}\nu & 0 & & & \cdots \\ & & & & 0 & 3\nu & \cdots \\ & & & & -\frac{1}{3}\nu & 0 & \ddots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \ddots \end{pmatrix}.$$

Again, $B$ maps $H_0$ to itself, and $B^2 e_j = \nu^2 e_j$ for all $j$, so $B^2 = \nu^2$ on $H_0$. Yet $Be_{2j} = j e_{2j - 1}$, so $B$ is unbounded.

Case 2: $\deg p > 2$.

Well, then $p(x) = q_1(x) q_2(x)$, with $\deg q_1 = 2$, $\deg q_2 \geq 1$. Again take $H$ and $H_0$ as above, and take $A$ to be an unbounded solution to $q_1(A) = 0$, satisfying $A: D(A) \to D(A)$. Then $p(A) = q_1(A) q_2(A) = 0 q_2(A) = 0$, and $A$ is unbounded. [Again, my naive factoring really depends on $D(A) \subseteq D(A^n)$, hence I am strongly using the $A: D(A) \to D(A)$ fact here.] QED.

Of course, Case 2 is sort of a cheat. I think there should be a "natural" example in general, since you can construct unbounded examples to $A^n = 0$ by just increasing the order of the nilpotency, and $A^n = I$ by taking positive real numbers $a_1, \dotsc, a_n$ with $a_1 a_2 \dotsc a_n = 1$ and letting the building-block matrix be

$$ \begin{pmatrix} 0 & a_1 & & & \cdots & & \\ & 0 & a_2 & & \cdots & & \\ & & 0 & a_3 & \cdots & & \\ \vdots & \vdots & \vdots & \ddots & \ddots & &\\ \vdots & \vdots & \vdots & \vdots & \ddots & \ddots & \\ & & & & & 0 & a_{n-1} \\ a_n & 0 & 0 & & \dotsc & 0 & 0 \end{pmatrix} $$ Then again make a block-diagonal infinite matrix such that as we repeat the blocks, $a_3, \dotsc, a_n$ are constant, and $a_1 \to \infty$ and $a_2 \to 0$ (or somesuch).

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You mean an infinite-dimensional separable Hilbert space. The answer is no.

Suppose $p(z)$ has distinct roots $\alpha_1, \alpha_2$. Define a sequence $x_1, x_2, \ldots$ in the unit sphere of $H$ such that

  1. $x_1,\ldots, x_n$ are linearly independent for all $n$.
  2. $\|x_i - x_{i+1}\| \to 0$ as $i \to \infty$.
  3. the sequence is dense in the unit sphere of $H$.

Define $A$ on the linear span of the sequence so that $A x_i = \alpha_1 x_i$ if $i$ is odd, $\alpha_2 x_i$ otherwise.

On the other hand, if $p$ has only one root, say $p(z) = (z - \alpha)^d$, then with the same sequence as above take $A x_i = \alpha x_i + x_{i+1}$ for $i$ not divisible by $d$, $\alpha x_i$ otherwise.

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I thank you very much for your interewsting answer. I apologize if I did not accept it. meta.mathoverflow.net/questions/1491/… –  Ali Taghavi Aug 12 at 9:12

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