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Is the following statement true?

A self-adjoint non-negative compact operator on a Hilbert space $(\mathcal{H}, \langle \cdot, \cdot \rangle)$ is injective if and only if the closure of its range is infinite-dimensional.

Taking the contraposite, it would mean that $T$ is non injective if and only if it is finite rank.

My attempt of proof is as follows. Since $T$ is self-adjoint, $\mathcal{H}$ can be decomposed into the direct sum $$ \mathcal{H} = \operatorname{ker}(T) \oplus \overline{\operatorname{range}(T)} $$ Let us assume that $T$ is injective which is equivalent to $\operatorname{ker}(T) = \{0\}$. Therefore, by the direct sum decomposition, it implies $\overline{\operatorname{range}(T)}=\mathcal{H}$ which is infinite dimensional.

For the other direction, let us assume that $T$ is not injective. We now use the assumption that $T$ is self-adjoint, non-negative and compact. By the spectral theorem, $T$ admits a countable sequence of non-negative eigenvalues $\{\lambda_i\}_{i \geq 1} $ with limit $0$ and such that the positive eigenvalues are associated with an orthonormal family that forms a basis of $\overline{\operatorname{range}(T)}$. Since $T$ is not injective, $\operatorname{ker}(T) \ne \{0\}$ and one of the eigenvalues is exactly $0$. Therefore, since $\lambda_i \to 0_+$, the number of positive eigenvalues is finite which implies that $\overline{\operatorname{range}(T)}$ is finite dimensional (and $\overline{\operatorname{range}(T)}=\operatorname{range}(T)$).

Is this proof correct? I could not find trace of this statement in standard functional analysis books. If the proof is correct, I think we can drop the non-negative assumption using singular values instead of eigenvalues.

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  • $\begingroup$ The conclusion is not true in an obvious way. As you have observed a self adjoint operator $T$ is injective iff its range is dense. So infinite dimensional range is not sufficient for $T$ to be injective. $\endgroup$ Commented Jun 11 at 14:21

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Here is a counterexample (I leave it to you to use it to locate the flaw in your proof).

Let $\mathcal H:=\ell^2(\Bbb N_0)$. The operator $T:\mathcal H\to\mathcal H$ defined by $$T(x_0,x_1,x_2,\dots,x_n,\dots):=\left(0,x_1,\frac{x_2}2,\dots,\frac{x_n}n,\dots\right)$$ is compact, self-adjoint, non-negative, non-injective, but its range is infinite-dimensional.

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  • $\begingroup$ Thank you, I could check all the properties given for this operator, including the fact that the eigenvalues are $(1, 1/2, \ldots, 1/n, \ldots)$ and $0$. The flaw of the proof is the following: when applying the spectral theorem, the fact that $\lambda_i \to 0_+$ is only about the positive eigenvalues, the eigenvalue $0$ is not included in this statement. I can now see that $\operatorname{ker}(T) \ne \{0\}$ does not tell us anything about the number of positive eigenvalues. Please let me know if the reasoning is wrong, $\endgroup$
    – DimSum
    Commented Jun 11 at 13:12
  • $\begingroup$ The flaw is indeed in "since $\lambda_i \to 0_+$, the number of positive eigenvalues is finite" (even if you include the eigenvalue $0$. For instance here, the sequence $\lambda_0=0,\lambda_1=1,\dots,\lambda_n=\frac1n,\dots$ includes $0$ and does tend to $0_+$). $\endgroup$ Commented Jun 11 at 13:19
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    $\begingroup$ Indeed. I implicitly assumed that the eigenvalues were ordered by decreasing order. Which does not change that the proof is false. $\endgroup$
    – DimSum
    Commented Jun 11 at 13:24

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