Let $T$ be a nonzero bounded linear operator in $B(H)$, where $H$ is an infinite dimensional Hilbert space. If the norm of $T$ is known, how to compute the norm $\|I–T\|$,where $I$ is the identity operator.
Compute the norm of a bounded linear operator
banach-algebrasbanach-spacesc-star-algebrasfunctional-analysisoperator-theory
Related Solutions
It is obvious that $\Vert Tx\Vert_\infty\leq\Vert x\Vert_\infty$, hence $\Vert T\Vert\leq 1$. Since $\Vert T(1,0,0,\ldots)\Vert_\infty=\Vert (1,0,\ldots)\Vert_\infty$, we conclude that $\Vert T\Vert=1$.
One can easily verify that $$T(l^\infty)=\left\{(\xi_j):\exists C>0 \text{ such that } |\xi_j|\leq C/j\text{ for all j}\right\}$$
$T(l^\infty)$ is not closed in $l^\infty$: Let $x_n=(1,\dfrac{1}{\sqrt{2}},\ldots,\dfrac{1}{\sqrt{n}},0,0,\ldots)=T(1,\sqrt{2},\ldots,\sqrt{n},0,\ldots)$. The sequence $(x_n)$ obviously converges to $x=\left(1/\sqrt{j}\right)_{j=1}^\infty\in l^\infty$, which does not lie in $T(l^\infty)$ (if it did, what would be it's pre-image?).
The inverse operator $T^{-1}$ is not bounded: Consider the sequence $(x_n)\subseteq T(l^\infty)$ as above. This sequence is bounded but the image $\left\{T^{-1}x_n\right\}$ is not, since $\Vert T^{-1}x_n\Vert_\infty=\sqrt{n}$.
You could also argue in this way: If $T^{-1}$ were bounded, then for every Cauchy sequence $(y_n)\in T(l^\infty)$, the sequence $(T^{-1} y_n)$ would also be Cauchy, and hence would converge to some $x\in l^\infty$ since $l^\infty$ is complete. But then, since $T$ is bounded, $y_n$ would converge to $Tx$. Then $T(l^\infty)$ would be complete, contradicting the fact that it is not closed in $l^\infty$ (this argument can actually be used to show that if $T:X\rightarrow Y$ is an isomorphism between a Banach space $X$ and some normed space $Y$ such that $T$ and $T^{-1}$ are bounded, then $Y$ is Banach).
There are many counterexamples. Before I give one I want to give some context. Recall that for any normal matrix $A$ there exists a basis of eigenvectors. Let $\sigma(A)=\{\lambda_1,\ldots,\lambda_n\}$ be the eigenvalues with eigenspaces $\{V_{\lambda_1}, \ldots, V_{\lambda_n}\}$. Let $P_k$ be the orthogonal projection onto $V_{\lambda_k}$. Rewriting the spectral theorem for finite dimensional spaces a bit shows that
$$ A = \sum_{k=1}^n \lambda_k P_k. $$
Note that sums can be written as integrals over atomic measures and we can loosely write this as
$$ A = \int_{\sigma(A)} \lambda dP(\lambda). $$
More precisely, we do not integrate w.r.t. a measure in the usual sense. But integrate w.r.t a "projection-valued measure". I won't go to deep however.
Why did I write the spectral theorem for finite dimensional operators in such a convoluted way? Well there is a general spectral theorem that says that for any normal bounded operator $A$ on a Hilbert space $\mathcal{H}$ there exists a so called spectral measure $P$ such that we can write
$$ A = \int_{\sigma(A)} \lambda(A) dP(\lambda). $$
However, while previously, the set $\sigma(A)$ consisted of the eigenvalues of $A$ (i.e. the values $\lambda \in \mathbb{C}$ such that $A - \lambda 1$ is not injective), we now need to take the following definition:
$$ \sigma(A) = \{ \lambda \in \mathbb{C} \mid A-\lambda 1 \text{ is not invertible}. \} $$
In the finite dimensional case (and even when $A$ is compact) this boils down to $\sigma(A)$ being the eigenvalues of $A$. However in general, $A-\lambda 1$ might not be invertible while there exists no non-zero vector $v \in \mathcal{H}$ such that $A v =\lambda v$. While the generalized spectrum $\sigma(A)$ may be defined for any bounded operator (and having some really nice properties popping-up from complex analysis), it can be split up (for normal operators) into a discrete part (the real eigenvalues) and a continuous part. The continuous part makes general normal operators so different than the finite dimensional ones. This also makes the domain of the integral above over a non-discrete set.
Now to give a counterexample, take the operator $$A: L^2([0,1]) \rightarrow L^2([0,1]), A(f)(x) = x f(x) \quad \text{(multiplication by }x). $$ $A$ has no eigenvalues (for every $z \in \mathbb{C}$, there is no non-zero function $f$ with $A(f) = z f$, this would mean that $x = z$ for all $x$ where $f(x)$ is non-zero, so $f$ is zero a.e.). In particular, there is no basis of eigenvectors. However the spectrum $\sigma(A)$ can be shown to be equal to $[0,1]$ (the image of the function $x$ on $[0,1])$.
Edit: To show that $\sigma(A) = [0,1]$ for the example above, assume that $\lambda \notin \sigma(A) $. This means that there exists some $T \in \mathcal{B}(L^2([0,1])$ such that
$$T \circ (A-\lambda \text{Id}) = (A-\lambda \text{Id}) \circ T = \text{Id}. $$
Put $g = T(1) \in L^2([0,1])$. It follows that$(x-\lambda 1) g(x) = 1$ for all $x \in [0,1]$ a.e. or written differently that
$$ g(x) = \frac{1}{x-\lambda} \quad \forall x \in [0,1] \text{ a.e.} $$
However $g$ is not square integrable if $\lambda \in [0,1]$. This shows that $[0,1] \subset \sigma(A)$. For other inclusion, it is easy to see that whenever $\lambda \notin [0,1]$, then multiplication by the function $g(x) = \frac{1}{x-\lambda}$ is an inverse to $A-\lambda 1$. Because $g$ is uniformly bounded, the multiplication operator by $g$ is a bounded operator.
As a side note, $A$ from above is not a compact operator, as you can tell from the fact that is has a continuous spectrum (compact operators are diagonisable in the classical sense (having a countable orthonormal basis of eigenvectors), hence they have a discrete spectrum).
Best Answer
Not possible. You have the estimate $$ 0\leq\|I-T\|\leq\|I\|+\|T\|, $$ and both inequalities can be made sharp: take $T=I$ for the first one, and $T=-I$ for the second one; in both cases, $\|T\|=1$.