One of these two operators is not invertible I have a Hilbert space $H$ and a bounded self-adjoint operator $T$ on it with $||T||=1$. I've been trying to show that at least one of $I+T, I-T$ are not invertible, but I haven't been able to make any progress yet. Any suggestions would be appreciated.
 A: As $ T $ is bounded and self-adjoint with norm $ 1 $, its spectrum $ \sigma(T) $ is a compact subset of $ [-1,1] $, and its spectral radius $ r(T) $ equals $ 1 $. Hence, either $ -1 \in \sigma(T) $ or $ 1 \in \sigma(T) $.


*

*If $ -1 \in \sigma(T) $, then $ 0 \in \sigma(I + T) $, and so $ I + T $ is not invertible.

*If $ 1 \in \sigma(T) $, then $ 0 \in \sigma(I - T) $, and so $ I - T $ is not invertible.

A: Here is another answer that is suitable for the poster’s background. We make use of the fact that for a self-adjoint bounded operator $ T $ on a Hilbert space $ \mathcal{H} $, we have
$$
  \| T \|_{B(\mathcal{H})}
= \sup(\{
  |\langle T(h),h \rangle_{\mathcal{H}}| \mid h \in \mathbb{S}(\mathcal{H})
  \}),
$$
where $ \mathbb{S}(\mathcal{H}) $ denotes the unit sphere of $ \mathcal{H} $.
Suppose that $ \| T \|_{B(\mathcal{H})} = 1 $. Then there is a sequence $ (h_{n})_{n \in \mathbb{N}} $ in $ \mathbb{S}(\mathcal{H}) $ such that
$$
\lim_{n \to \infty} |\langle T(h_{n}),h_{n} \rangle_{\mathcal{H}}| = 1.
$$
As $ T $ is self-adjoint, we have
$$
\forall h \in \mathcal{H}: \quad
  \langle T(h),h \rangle_{\mathcal{H}}
= \langle h,T(h) \rangle_{\mathcal{H}}
= \overline{\langle T(h),h \rangle_{\mathcal{H}}},
$$
which means that $ \langle T(h),h \rangle_{\mathcal{H}} \in \mathbb{R} $ for any $ h \in \mathcal{H} $. Hence, we can find a subsequence $ (h_{n_{k}})_{k \in \mathbb{N}} $ of $ (h_{n})_{n \in \mathbb{N}} $ such that either


*

*$ \displaystyle \lim_{k \to \infty} \langle T(h_{n_{k}}),h_{n_{k}} \rangle_{\mathcal{H}} = 1 $, or

*$ \displaystyle \lim_{k \to \infty} \langle T(h_{n_{k}}),h_{n_{k}} \rangle_{\mathcal{H}} = -1 $.


Suppose that Case (1) occurs. Observe that
\begin{align}
\forall h \in \mathcal{H}: \quad
       \left\langle {(T - I)^{2}}(h),h \right\rangle_{\mathcal{H}}
& =    \langle (T - I)(h),(T - I)(h) \rangle_{\mathcal{H}} \\
& =    \langle T(h),T(h) \rangle_{\mathcal{H}} +
       \langle h,h \rangle_{\mathcal{H}} -
       \langle T(h),h \rangle_{\mathcal{H}} -
       \langle h,T(h) \rangle_{\mathcal{H}} \\
& =    \langle T(h),T(h) \rangle_{\mathcal{H}} +
       \langle h,h \rangle_{\mathcal{H}} -
       2 \langle T(h),h \rangle_{\mathcal{H}} \qquad (\text{As $ T = T^{*} $.}) \\
& =    \| T(h) \|_{\mathcal{H}}^{2} + \| h \|_{\mathcal{H}}^{2} -
       2 \langle T(h),h \rangle_{\mathcal{H}} \\
& \leq \| h \|_{\mathcal{H}}^{2} + \| h \|_{\mathcal{H}}^{2} -
       2 \langle T(h),h \rangle_{\mathcal{H}} \qquad
       (\text{As $ \| T \|_{B(\mathcal{H})} = 1 $.}) \\
& =    2 \| h \|_{\mathcal{H}}^{2} - 2 \langle T(h),h \rangle_{\mathcal{H}}.
\end{align}
As
\begin{align}
    \lim_{k \to \infty}
    \left(
    2 \| h_{n_{k}} \|_{\mathcal{H}}^{2} -
    2 \langle T(h_{n_{k}}),h_{n_{k}} \rangle_{\mathcal{H}}
    \right)
& = \lim_{k \to \infty}
    (2 - 2 \langle T(h_{n_{k}}),h_{n_{k}} \rangle_{\mathcal{H}}) \qquad
    (\text{As $ h_{n_{k}} \in \mathbb{S}(\mathcal{H}) $.}) \\
& = 2 - 2(1) \qquad (\text{By the assumption of Case (1).}) \\
& = 0,
\end{align}
it follows from the Squeeze Theorem that
$$
  \lim_{k \to \infty}
  \left\langle {(T - I)^{2}}(h_{n_{k}}),h_{n_{k}} \right\rangle_{\mathcal{H}}
= 0.
$$
Hence, there cannot exist a $ \delta > 0 $ such that $ \left\langle {(T - I)^{2}}(h),h \right\rangle_{\mathcal{H}} \geq \delta \| h \|_{\mathcal{H}}^{2} $ for all $ h \in \mathcal{H} $, which immediately implies that $ T - I $ is not invertible.
If Case (2) occurs, then by a similar argument, $ T + I $ is not invertible.
