# A bounded operator on a non-separable Hilbert space has $0$ in its point spectrum

If $H$ is a non-separable Hilbert space and $T:H \to H$ be compact , then show that $0 \in \sigma_{p}(T)$.

I have no idea about working on a non-separable Hilbert space. It is obvious that it won't have a countable orthonormal basis. Apart from that I don't know anything about non-separble Hilbert Space. Thanks in advance !!!

• Use (and prove, if unfamiliar) the fact that the range of a compact operator is always separable. – Aweygan May 9 '17 at 15:05
• I know this fact. But then you can say that T is not onto. Then 0 cannot be in the resolvent. Hence 0 is in the spectrum. But what I need to prove is that 0 is an eigen value, that is T is not one-one.. But can we easily conclude that T is not one-one? – Riju May 9 '17 at 15:25
• Have you looked at $S=T^*T$, which is selfadjoint and compact? – DisintegratingByParts May 9 '17 at 16:19
• @DisintegrationByParts Is it correct? If wrong kindly clarify. Since $T^*T$ is compact $0\in \sigma(T^*T)$. Moreover $\exists x_n\in S_H$ such that $lim_{n\to infty}<T^*Tx_n,x_n>=0$ i.e, $||Tx_n||\to 0$. Hence $Tx_n\to 0$ as $n\to \infty$. Since $T$ is compact $\exists y\in S_H$ such that $Ty=0$. Hence $T$ is not injective. Then we can get the conclusion. – Happy May 9 '17 at 16:41
• @happy how do you get sequence $X_{n}$ such that $\langle T^{*}T(x_{n}), x_{n} \rangle$ tend to 0 as n tends to infinity. – Riju May 9 '17 at 16:52

The point is to show that $\ker T\ne\{0\}$.
Assume first that $T$ is selfadjoint. Since $T$ is compact, every nonzero element of the spectrum is an eigenvalue with finite multiplicity, and there are at most countably many nonzero eigenvalues. That is, $$T=\sum_{k=1}^\infty\lambda_kP_k,$$ where $\{P_j\}$ are pairwise orthogonal finite-rank projections.
As $Tx=\sum_{k=1}^\infty\lambda_kP_kx$, we deduce that the image of $T$ is (at most) countably-dimensional. Now, since $$H=\ker T\oplus \overline{\text{ran}\,T},$$ it follows that $\ker T$ is nonzero; actually, it has uncountable dimension.
For non-selfadjoint $T$, use the fact that $\ker T=\ker T^*T$.