In Delphi 7 when I try to add a application icon (project->options->application), it says "Duplicate resource type 3 name 1". Does anyone know how to fix this?


Question: A question on unbounded operators
Answer: 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 

$$ \begin{pmatrix} 0 &amp; 1 &amp;   &amp;   &amp;   &amp;   &amp; \cdots \\
                   0 &amp; 0 &amp;   &amp;   &amp;   &amp;   &amp; \cdots \\
                     &amp;   &amp; 0 &amp; 2 &amp;   &amp;   &amp; \cdots \\
                     &amp;   &amp; 0 &amp; 0 &amp;   &amp;   &amp; \cdots \\
                     &amp;   &amp;   &amp;   &amp; 0 &amp; 3 &amp; \cdots \\
                     &amp;   &amp;   &amp;   &amp; 0 &amp; 0 &amp; \ddots \\
        \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \ddots &amp; \ddots \end{pmatrix},$$
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 

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

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: $\deg p &gt; 2$.

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

$$ \begin{pmatrix} 0 &amp; a_1 &amp; &amp; &amp; \cdots  &amp; &amp; \\
                   &amp; 0 &amp; a_2 &amp; &amp; \cdots  &amp; &amp; \\
                   &amp; &amp; 0 &amp; a_3 &amp; \cdots  &amp; &amp; \\
              \vdots &amp; \vdots &amp; \vdots &amp;  \ddots &amp; \ddots  &amp; &amp;\\
\vdots &amp; \vdots &amp; \vdots &amp;  \vdots &amp; \ddots  &amp; \ddots  &amp; \\
   &amp; &amp; &amp; &amp; &amp;  0 &amp; a_{n-1} \\
a_n &amp; 0 &amp; 0 &amp; &amp; \dotsc &amp; 0 &amp; 0 \end{pmatrix} $$
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).