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user69109
user69109

I am curious to know the answer to the following question:

Does there exist a continuous linear operator on some Banach space $X$ such that $\Vert T \Vert=1$, and $\sigma(T)\supset \{1\}$ is isolated in the spectrum of $T$ even though $\{1\}$ is not in the point spectrum? Or does an operator like that not exist?

It seems to me that having something that is isolated in the spectrum on a Banach space and of maximal absolute value ($=1$) is really hard to come by. My guess is that such an operator does not exist.

However, by combining a quasinilpotent operator $N$ with trivial kernel and adding to it the identity, we have an operator $N+1$ which has spectrum $\sigma(N+1)=\{1\}$ and empty point spectrum. This way, however, I cannot see how to ensure that $\Vert N+1 \Vert =1.$

I am curious to know the answer to the following question:

Does there exist a continuous linear operator on some Banach space $X$ such that $\Vert T \Vert=1$, and $\sigma(T)\supset \{1\}$ is isolated in the spectrum of $T$ even though $\{1\}$ is not in the point spectrum? Or does an operator like that not exist?

It seems to me that having something that is isolated in the spectrum on a Banach space and of maximal absolute value ($=1$) is really hard to come by. My guess is that such an operator does not exist.

However, by combining a quasinilpotent operator $N$ with trivial kernel and adding to it the identity, we have an operator $N+1$ which has spectrum $\sigma(N+1)=\{1\}$ and empty point spectrum. This way, however, I cannot see how to ensure that $\Vert N+1 \Vert =1.$

I am curious to know the answer to the following question:

Does there exist a continuous linear operator on some Banach space $X$ such that $\Vert T \Vert=1$, and $\sigma(T)\supset \{1\}$ is isolated in the spectrum of $T$ even though $\{1\}$ is not in the point spectrum? Or does an operator like that not exist?

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user69109
user69109

I would likeam curious to know the answer to the following question:

Does there exist a continuous linear operator on some Banach space $X$ such that $\Vert T \Vert=1$, and $\sigma(T)\supset \{1\}$ is isolated in the spectrum of $T$ even though $\{1\}$ is not in the point spectrum? Or does an operator like that not exist?

It seems to me that having something that is isolated in the spectrum on a Banach space and of maximal absolute value ($=1$) is really hard to come by. My guess is that such an operator does not exist.

However, by combining a quasinilpotent operator $N$ with trivial kernel and adding to it the identity, we have an operator $N+1$ which has spectrum $\sigma(N+1)=\{1\}$ and empty point spectrum. This way, however, I cannot see how to ensure that $\Vert N+1 \Vert =1.$

I would like to know the following:

Does there exist a continuous linear operator on some Banach space $X$ such that $\Vert T \Vert=1$, and $\sigma(T)\supset \{1\}$ is isolated in the spectrum of $T$ even though $\{1\}$ is not in the point spectrum? Or does an operator like that not exist?

It seems to me that having something that is isolated in the spectrum on a Banach space and of maximal absolute value ($=1$) is really hard to come by. My guess is that such an operator does not exist.

However, by combining a quasinilpotent operator $N$ with trivial kernel and adding to it the identity, we have an operator $N+1$ which has spectrum $\sigma(N+1)=\{1\}$ and empty point spectrum. This way, however, I cannot see how to ensure that $\Vert N+1 \Vert =1.$

I am curious to know the answer to the following question:

Does there exist a continuous linear operator on some Banach space $X$ such that $\Vert T \Vert=1$, and $\sigma(T)\supset \{1\}$ is isolated in the spectrum of $T$ even though $\{1\}$ is not in the point spectrum? Or does an operator like that not exist?

It seems to me that having something that is isolated in the spectrum on a Banach space and of maximal absolute value ($=1$) is really hard to come by. My guess is that such an operator does not exist.

However, by combining a quasinilpotent operator $N$ with trivial kernel and adding to it the identity, we have an operator $N+1$ which has spectrum $\sigma(N+1)=\{1\}$ and empty point spectrum. This way, however, I cannot see how to ensure that $\Vert N+1 \Vert =1.$

deleted 259 characters in body
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user69109
user69109

I would like to know the following:

Does there exist a continuous linear operator on some Banach space $X$ such that $\Vert T \Vert=1$, and $\sigma(T)\supset \{1\}$ is isolated in the spectrum of $T$ even though $\{1\}$ is not in the point spectrum? Or does an operator like that not exist?

It seems to me that having something that is isolated in the spectrum on a Banach space and of maximal absolute value ($=1$) is really hard to come by. My guess is that such an operator does not exist I would like to know the following:.

Does there existHowever, by combining a continuous linearquasinilpotent operator on some Banach space $X$ such that $\Vert T \Vert=1$,$N$ with trivial kernel and $\sigma(T)\supset \{1\}$ is isolated inadding to it the spectrum ofidentity, we have an operator $T$ even though$N+1$ which has spectrum $\{1\}$ is not in the$\sigma(N+1)=\{1\}$ and empty point spectrum? Or does an operator like that not exist?

It seems. This way, however, I cannot see how to me that having somethingensure that is isolated in the spectrum on a Banach space and of maximal absolute value ($=1$) is really hard to come by. My guess is that such an operator does not exist but this is only a vague feeling$\Vert N+1 \Vert =1.$

I would like to know the following:

Does there exist a continuous linear operator on some Banach space $X$ such that $\Vert T \Vert=1$, and $\sigma(T)\supset \{1\}$ is isolated in the spectrum of $T$ even though $\{1\}$ is not in the point spectrum? Or does an operator like that not exist?

It seems to me that having something that is isolated in the spectrum on a Banach space and of maximal absolute value ($=1$) is really hard to come by. My guess is that such an operator does not exist I would like to know the following:

Does there exist a continuous linear operator on some Banach space $X$ such that $\Vert T \Vert=1$, and $\sigma(T)\supset \{1\}$ is isolated in the spectrum of $T$ even though $\{1\}$ is not in the point spectrum? Or does an operator like that not exist?

It seems to me that having something that is isolated in the spectrum on a Banach space and of maximal absolute value ($=1$) is really hard to come by. My guess is that such an operator does not exist but this is only a vague feeling

I would like to know the following:

Does there exist a continuous linear operator on some Banach space $X$ such that $\Vert T \Vert=1$, and $\sigma(T)\supset \{1\}$ is isolated in the spectrum of $T$ even though $\{1\}$ is not in the point spectrum? Or does an operator like that not exist?

It seems to me that having something that is isolated in the spectrum on a Banach space and of maximal absolute value ($=1$) is really hard to come by. My guess is that such an operator does not exist.

However, by combining a quasinilpotent operator $N$ with trivial kernel and adding to it the identity, we have an operator $N+1$ which has spectrum $\sigma(N+1)=\{1\}$ and empty point spectrum. This way, however, I cannot see how to ensure that $\Vert N+1 \Vert =1.$

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user69109
user69109