True/False: Self-adjoint compact operator 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). 
 A: 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
$$\begin{align}
\lVert x-u_\varepsilon\rVert = \sqrt{\sum_{n=n_0}^\infty \left\lvert x_n - \frac{x_0}{n}\right\rvert^2} \leqslant \sqrt{\sum_{n=n_0}^\infty \lvert x_n\rvert^2} + \lvert x_0\rvert\sqrt{\sum_{n=n_0}^\infty \frac{1}{n^2}} < \varepsilon.
\end{align}$$
