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Let $H$ be a hilbert space and $T$ a compact self-adjoint operator on it. T is also injective on a dense subspace $U \subset H$ and we also have that $T(H) \subset U$. Now I am asked whether it is true that $T$ has to be injective on the whole Hilbert space?

Until now, I don't know whether this is true or false. Maybe I could add that the only compact self-adjoint operator we treated in class was a Fredholm operator with a symmetric kernel function.(maybe this could serve as a counterexample).

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  • $\begingroup$ Is $U$ a subspace or a subset? $\endgroup$
    – gerw
    Jan 29, 2014 at 19:26
  • $\begingroup$ $U$ is a subspace that is dense in $H$ $\endgroup$
    – user66906
    Jan 29, 2014 at 19:28
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    $\begingroup$ That makes a difference ;) $\endgroup$
    – gerw
    Jan 29, 2014 at 19:30
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    $\begingroup$ I found that my answer was misleading and did not provide a solution to your problem, sorry. $\endgroup$
    – gerw
    Jan 30, 2014 at 11:28
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    $\begingroup$ Check math.stackexchange.com/questions/470240/… also. The answer is the same. $\endgroup$ Jan 31, 2014 at 11:56

1 Answer 1

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The conditions do not imply that $T$ is injective. As usual, let $H = \ell^2(\mathbb{N})$ the space of square-summable complex sequences. Let $\mu_0 = 0$, and $\mu_n = \frac{1}{n}$ for $n > 0$, and $T$ the multiplication operator for the sequence $\mu$,

$$T(x_n) = (\mu_n\cdot x_n).$$

Since $\mu$ is real, $T$ is self-adjoint, and since $\mu_n \to 0$, $T$ is compact. Furthermore $T$ is not injective, as $\ker T = \operatorname{span} (e_0)$.

Now let us construct a dense subspace $U$ on which $T$ is injective and that contains $\mathcal{R}(T)$. We have

$$\mathcal{R}(T) = \left\lbrace x \in H : x_0 = 0, \sum_{n=1}^\infty n^2\lvert x_n\rvert^2 < \infty\right\rbrace,$$

in particular $e_n \in \mathcal{R}(T)$ for all $n > 0$. Let $\xi = e_0 + \sum\limits_{n=1}^\infty \frac{1}{n}\cdot e_n$, and

$$U = \mathcal{R}(T) \oplus \mathbb{C}\cdot\xi.$$

Then $U \cap \ker T = \{0\}$, since $\sum\limits_{n=1}^\infty \frac{1}{n}\cdot e_n \notin \mathcal{R}(T)$, so $T\lvert_U$ is injective.

It remains to see that $U$ is dense.

Let $x \in H$, and $\varepsilon > 0$ arbitrary. Choose $n_0 > 0$ so large that

$$\begin{gather} \sum_{n=n_0}^\infty \lvert x_n\rvert^2 < \frac{\varepsilon^2}{4},\\ \lvert x_0\rvert^2\sum_{n=n_0}^\infty \frac{1}{n^2} < \frac{\varepsilon^2}{4}. \end{gather}$$

Then

$$u_\varepsilon := x_0\cdot \xi + \sum_{n=1}^{n_0-1} \left(x_n - \frac{x_0}{n}\right)\cdot e_n \in U,$$

and

$$\begin{align} \lVert x-u_\varepsilon\rVert = \sqrt{\sum_{n=n_0}^\infty \left\lvert x_n - \frac{x_0}{n}\right\rvert^2} \leqslant \sqrt{\sum_{n=n_0}^\infty \lvert x_n\rvert^2} + \lvert x_0\rvert\sqrt{\sum_{n=n_0}^\infty \frac{1}{n^2}} < \varepsilon. \end{align}$$

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