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Notes: This is pre-gelfand naimark so can't invoke that

My thoughts: If $||1-T||<1$ then T has an inverse and the inverse is the limit of polynomials in T so in the algebra. But what about invertible T with $||1-T||>1$?

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Notes: This is pre-gelfand naimark so can't invoke that

My thoughts: If $||1-T||<1$ then T has an inverse and the inverse is the limit of polynomials in T so in the algebra. But what about invertible T with $||1-T||>1$?

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- Feb 7, 2012

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In a general Banach algebra it is possible for a non-invertible element to have an inverse in a larger algebra. That cannot happen for selfadjoint subalgebras of $B(H)$. It follows that any proof of this fact cannot rely on general Banach algebra properties but must make some use of the special structure of $B(H)$. The usual proof uses the Gelfand–Naimark theorem.

Notes: This is pre-gelfand naimark so can't invoke that

My thoughts: If $||1-T||<1$ then T has an inverse and the inverse is the limit of polynomials in T so in the algebra. But what about invertible T with $||1-T||>1$?

However, you can avoid Gelfand–Naimark theory if you know a theorem about Banach subalgebras. Suppose that $A$ is a unital Banach subalgebra of a Banach algebra $B$. Let $a\in A$ and let $\sigma_A(a),\sigma_B(a)$ denote the spectrum of $a$ in $A$ and $B$ respectively. Then $\sigma_B(a) \subseteq \sigma_A(a)$, but if we look at the boundary of the spectrum then the theorem says that the inclusion goes the other way, $\partial\sigma_A(a) \subseteq \partial\sigma_B(a)$.

Given $T\in A\subseteq B(H)$ with an inverse in $B(H)$, the operator $T^*T$ is selfadjoint in $A$, and invertible in $B(H)$. If it is not invertible in $A$ then $0$ will be in its spectrum. But the spectrum of a selfadjoint operator is real, so every point in $\sigma_A(T^*T)$ is in the boundary of the spectrum. Therefore $0$ is in $\sigma_{B(H)}(T^*T)$. But this means that $T^*T$ is not invertible in $B(H)$. That contradiction shows that $T^*T$ has an inverse in $A$. There is then a little trick (Proposition 4.1.5 in Kadison and Ringrose) to show that $T$ also has an inverse in $A$.

(That proof works for any $T\in B(H)$. The condition that $T$ should be normal is not needed.)

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I'm not familiar with the boundary. From wikipedia, it is the set of points p such that every neighbourhood of p contains at least one point of the spectrum (in this case). What has that to do with the spectrum being real? Also, a certain irony here, but I cannot read chapter 4 in that preview!In a general Banach algebra it is possible for a non-invertible element to have an inverse in a larger algebra. That cannot happen for selfadjoint subalgebras of $B(H)$. It follows that any proof of this fact cannot rely on general Banach algebra properties but must make some use of the special structure of $B(H)$. The usual proof uses the Gelfand–Naimark theorem.

However, you can avoid Gelfand–Naimark theory if you know a theorem about Banach subalgebras. Suppose that $A$ is a unital Banach subalgebra of a Banach algebra $B$. Let $a\in A$ and let $\sigma_A(a),\sigma_B(a)$ denote the spectrum of $a$ in $A$ and $B$ respectively. Then $\sigma_B(a) \subseteq \sigma_A(a)$, but if we look at the boundary of the spectrum then the theorem says that the inclusion goes the other way, $\partial\sigma_A(a) \subseteq \partial\sigma_B(a)$.

Given $T\in A\subseteq B(H)$ with an inverse in $B(H)$, the operator $T^*T$ is selfadjoint in $A$, and invertible in $B(H)$. If it is not invertible in $A$ then $0$ will be in its spectrum. But the spectrum of a selfadjoint operator is real, so every point in $\sigma_A(T^*T)$ is in the boundary of the spectrum. Therefore $0$ is in $\sigma_{B(H)}(T^*T)$. But this means that $T^*T$ is not invertible in $B(H)$. That contradiction shows that $T^*T$ has an inverse in $A$. There is then a little trick (Proposition 4.1.5 in Kadison and Ringrose) to show that $T$ also has an inverse in $A$.

(That proof works for any $T\in B(H)$. The condition that $T$ should be normal is not needed.)

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Is it th set of limit points?In a general Banach algebra it is possible for a non-invertible element to have an inverse in a larger algebra. That cannot happen for selfadjoint subalgebras of $B(H)$. It follows that any proof of this fact cannot rely on general Banach algebra properties but must make some use of the special structure of $B(H)$. The usual proof uses the Gelfand–Naimark theorem.

However, you can avoid Gelfand–Naimark theory if you know a theorem about Banach subalgebras. Suppose that $A$ is a unital Banach subalgebra of a Banach algebra $B$. Let $a\in A$ and let $\sigma_A(a),\sigma_B(a)$ denote the spectrum of $a$ in $A$ and $B$ respectively. Then $\sigma_B(a) \subseteq \sigma_A(a)$, but if we look at the boundary of the spectrum then the theorem says that the inclusion goes the other way, $\partial\sigma_A(a) \subseteq \partial\sigma_B(a)$.

Given $T\in A\subseteq B(H)$ with an inverse in $B(H)$, the operator $T^*T$ is selfadjoint in $A$, and invertible in $B(H)$. If it is not invertible in $A$ then $0$ will be in its spectrum. But the spectrum of a selfadjoint operator is real, so every point in $\sigma_A(T^*T)$ is in the boundary of the spectrum. Therefore $0$ is in $\sigma_{B(H)}(T^*T)$. But this means that $T^*T$ is not invertible in $B(H)$. That contradiction shows that $T^*T$ has an inverse in $A$. There is then a little trick (Proposition 4.1.5 in Kadison and Ringrose) to show that $T$ also has an inverse in $A$.

(That proof works for any $T\in B(H)$. The condition that $T$ should be normal is not needed.)

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- Feb 7, 2012

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The boundary of a closed subset S of a space (in this case the space of complex numbers) is the set of all points in $S$ that are arbitrarily close to points outside $S$. The spectrum of a selfadjoint operator consists of real numbers, and any real number has nonreal complex numbers arbitrarily close to it. So in this case the whole of the spectrum consists of boundary points.Is it th set of limit points?

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Have I got this right so far? the precise deifinition is that every open ball centred at a point contains both points of S and not of S. That is true for real numbers so all the spectrum is in the boundary which implies it is in the boundary of the spectrum considered as an element of B(H). But the spectrum is closed and all points in the boundary are limiting points, hence in the spectrum. So T*T is not invertible.The boundary of a closed subset S of a space (in this case the space of complex numbers) is the set of all points in $S$ that are arbitrarily close to points outside $S$. The spectrum of a selfadjoint operator consists of real numbers, and any real number has nonreal complex numbers arbitrarily close to it. So in this case the whole of the spectrum consists of boundary points.

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How to prove 'Therefore $0$ is in $\sigma_{B(H)}(T^*T)$'? ThanksIn a general Banach algebra it is possible for a non-invertible element to have an inverse in a larger algebra. That cannot happen for selfadjoint subalgebras of $B(H)$. It follows that any proof of this fact cannot rely on general Banach algebra properties but must make some use of the special structure of $B(H)$. The usual proof uses the Gelfand–Naimark theorem.

However, you can avoid Gelfand–Naimark theory if you know a theorem about Banach subalgebras. Suppose that $A$ is a unital Banach subalgebra of a Banach algebra $B$. Let $a\in A$ and let $\sigma_A(a),\sigma_B(a)$ denote the spectrum of $a$ in $A$ and $B$ respectively. Then $\sigma_B(a) \subseteq \sigma_A(a)$, but if we look at the boundary of the spectrum then the theorem says that the inclusion goes the other way, $\partial\sigma_A(a) \subseteq \partial\sigma_B(a)$.

Given $T\in A\subseteq B(H)$ with an inverse in $B(H)$, the operator $T^*T$ is selfadjoint in $A$, and invertible in $B(H)$. If it is not invertible in $A$ then $0$ will be in its spectrum. But the spectrum of a selfadjoint operator is real, so every point in $\sigma_A(T^*T)$ is in the boundary of the spectrum. Therefore $0$ is in $\sigma_{B(H)}(T^*T)$. But this means that $T^*T$ is not invertible in $B(H)$. That contradiction shows that $T^*T$ has an inverse in $A$. There is then a little trick (Proposition 4.1.5 in Kadison and Ringrose) to show that $T$ also has an inverse in $A$.

(That proof works for any $T\in B(H)$. The condition that $T$ should be normal is not needed.)

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That is because of the theorem I quoted, that if a number is in the boundary of the spectrum of an element in an algebra $A$, then it is also in the spectrum of that element in any algebra that contains $A$.How to prove 'Therefore $0$ is in $\sigma_{B(H)}(T^*T)$'? Thanks

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if x is in the boundary, it is a limiting point, thus in the spectrum since it is closed. But just because x is not invertible in A does not mean there is no inverse in B, so it is not immediately clear. Do you have a proof of this?That is because of the theorem I quoted, that if a number is in the boundary of the spectrum of an element in an algebra $A$, then it is also in the spectrum of that element in any algebra that contains $A$.

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It is a standard theorem in Banach algebra theory, based on the fact that a boundary point of the spectrum must be a topological zero-divisor. I learned it from Rickart (Theorem 1.5.7). But that was over 50 years ago, so I expect that there are more recent sources available.if x is in the boundary, it is a limiting point, thus in the spectrum since it is closed. But just because x is not invertible in A does not mean there is no inverse in B, so it is not immediately clear. Do you have a proof of this?

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