Compact operator on $L_2([0,1],m)$
$begingroup$
Consider the Hilbert space $H=L_2([0,1],m)$ where $m$ is the Lebesgue measure on the interval $[0,1]$. Let $T in mathcal{L}(H,H)$ given by
begin{equation*}
T f(x)=x f(x) f in H, x in[0,1]
end{equation*}
Is $T$ compact? I know that one way to prove that an operator is compact is to find a sequence of compact operator that converges to $T$. On the other side, to prove that $T$ is not compact, I should find a sequence such that its image doesn't admit a convergent subsequence.
functional-analysis hilbert-spaces compact-operators
asked Jan 17 at 10:00
user289143user289143
913313
$endgroup$
add a comment |
$begingroup$
Consider the Hilbert space $H=L_2([0,1],m)$ where $m$ is the Lebesgue measure on the interval $[0,1]$. Let $T in mathcal{L}(H,H)$ given by
begin{equation*}
T f(x)=x f(x) f in H, x in[0,1]
end{equation*}
Is $T$ compact? I know that one way to prove that an operator is compact is to find a sequence of compact operator that converges to $T$. On the other side, to prove that $T$ is not compact, I should find a sequence such that its image doesn't admit a convergent subsequence.
functional-analysis hilbert-spaces compact-operators
asked Jan 17 at 10:00
user289143user289143
913313
$endgroup$
add a comment |
$begingroup$
Consider the Hilbert space $H=L_2([0,1],m)$ where $m$ is the Lebesgue measure on the interval $[0,1]$. Let $T in mathcal{L}(H,H)$ given by
begin{equation*}
T f(x)=x f(x) f in H, x in[0,1]
end{equation*}
Is $T$ compact? I know that one way to prove that an operator is compact is to find a sequence of compact operator that converges to $T$. On the other side, to prove that $T$ is not compact, I should find a sequence such that its image doesn't admit a convergent subsequence.
functional-analysis hilbert-spaces compact-operators
asked Jan 17 at 10:00
user289143user289143
913313
$endgroup$
Consider the Hilbert space $H=L_2([0,1],m)$ where $m$ is the Lebesgue measure on the interval $[0,1]$. Let $T in mathcal{L}(H,H)$ given by
begin{equation*}
T f(x)=x f(x) f in H, x in[0,1]
end{equation*}
Is $T$ compact? I know that one way to prove that an operator is compact is to find a sequence of compact operator that converges to $T$. On the other side, to prove that $T$ is not compact, I should find a sequence such that its image doesn't admit a convergent subsequence.
functional-analysis hilbert-spaces compact-operators
functional-analysis hilbert-spaces compact-operators
asked Jan 17 at 10:00
user289143user289143
913313
asked Jan 17 at 10:00
user289143user289143
913313
asked Jan 17 at 10:00
user289143user289143
913313
asked Jan 17 at 10:00
user289143user289143
913313
asked Jan 17 at 10:00
user289143user289143
913313
913313
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
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You can easily check that $T$ has no eigenvalues. However, $lambda=frac 1 2$ is in the spectrum but not in the point spectrum. Just try to solve for
$$left( frac 1 2 - xright) f(x) = 1$$
and see that no solution exists.
But we know that the spectrum of a compact operator except zero consists only of eigenvalues. Hence the operator $T$ can not be compact.
answered Jan 17 at 10:51
ShashiShashi
7,1781628
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add a comment |
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It is trivial to check that $T$ is a positive operator. Hence its norm is same as its spectral radius. If it is compact then the norm would be $0$ because it has no non-zero eigen values. Hence $T$ is not compact.
answered Jan 17 at 10:14
Kavi Rama MurthyKavi Rama Murthy
59.2k42161
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add a comment |
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Let $$s = sup_{|f|_2 leq 1} ,langle Tf,,frangle=1.$$
It is quite easy to see that $s$ is not a maximum, so there exists some sequence $f_n$ such that $|f_n| leq 1$, and $langle Tf_n ,, f_n rangle rightarrow 1$.
Assume that there is some converging subsequence $Tf_{varphi(n)} rightarrow g$. Then you easily get $langle g ,,f_{varphi(n)}rangle rightarrow 1$.
We know there is a weakly converging subsequence $f_{varphi(psi(n))} rightarrow h$ thus $|g| leq 1$ and $|h| leq 1$ and their inner product is $1$, thus $g=h$ and their norm is $1$.
Now, this entails (using the fact that $T$ is self-adjoint) that $Th=h$, which is impossible.
answered Jan 17 at 10:27
MindlackMindlack
3,86018
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can easily check that $T$ has no eigenvalues. However, $lambda=frac 1 2$ is in the spectrum but not in the point spectrum. Just try to solve for
$$left( frac 1 2 - xright) f(x) = 1$$
and see that no solution exists.
But we know that the spectrum of a compact operator except zero consists only of eigenvalues. Hence the operator $T$ can not be compact.
answered Jan 17 at 10:51
ShashiShashi
7,1781628
$endgroup$
add a comment |
$begingroup$
You can easily check that $T$ has no eigenvalues. However, $lambda=frac 1 2$ is in the spectrum but not in the point spectrum. Just try to solve for
$$left( frac 1 2 - xright) f(x) = 1$$
and see that no solution exists.
But we know that the spectrum of a compact operator except zero consists only of eigenvalues. Hence the operator $T$ can not be compact.
answered Jan 17 at 10:51
ShashiShashi
7,1781628
$endgroup$
add a comment |
$begingroup$
You can easily check that $T$ has no eigenvalues. However, $lambda=frac 1 2$ is in the spectrum but not in the point spectrum. Just try to solve for
$$left( frac 1 2 - xright) f(x) = 1$$
and see that no solution exists.
But we know that the spectrum of a compact operator except zero consists only of eigenvalues. Hence the operator $T$ can not be compact.
answered Jan 17 at 10:51
ShashiShashi
7,1781628
$endgroup$
You can easily check that $T$ has no eigenvalues. However, $lambda=frac 1 2$ is in the spectrum but not in the point spectrum. Just try to solve for
$$left( frac 1 2 - xright) f(x) = 1$$
and see that no solution exists.
But we know that the spectrum of a compact operator except zero consists only of eigenvalues. Hence the operator $T$ can not be compact.
answered Jan 17 at 10:51
ShashiShashi
7,1781628
edited Jan 17 at 11:22
answered Jan 17 at 10:51
ShashiShashi
7,1781628
answered Jan 17 at 10:51
ShashiShashi
7,1781628
answered Jan 17 at 10:51
ShashiShashi
7,1781628
7,1781628
add a comment |
add a comment |
$begingroup$
It is trivial to check that $T$ is a positive operator. Hence its norm is same as its spectral radius. If it is compact then the norm would be $0$ because it has no non-zero eigen values. Hence $T$ is not compact.
answered Jan 17 at 10:14
Kavi Rama MurthyKavi Rama Murthy
59.2k42161
$endgroup$
add a comment |
$begingroup$
It is trivial to check that $T$ is a positive operator. Hence its norm is same as its spectral radius. If it is compact then the norm would be $0$ because it has no non-zero eigen values. Hence $T$ is not compact.
answered Jan 17 at 10:14
Kavi Rama MurthyKavi Rama Murthy
59.2k42161
$endgroup$
add a comment |
$begingroup$
It is trivial to check that $T$ is a positive operator. Hence its norm is same as its spectral radius. If it is compact then the norm would be $0$ because it has no non-zero eigen values. Hence $T$ is not compact.
answered Jan 17 at 10:14
Kavi Rama MurthyKavi Rama Murthy
59.2k42161
$endgroup$
It is trivial to check that $T$ is a positive operator. Hence its norm is same as its spectral radius. If it is compact then the norm would be $0$ because it has no non-zero eigen values. Hence $T$ is not compact.
answered Jan 17 at 10:14
Kavi Rama MurthyKavi Rama Murthy
59.2k42161
answered Jan 17 at 10:14
Kavi Rama MurthyKavi Rama Murthy
59.2k42161
answered Jan 17 at 10:14
Kavi Rama MurthyKavi Rama Murthy
59.2k42161
answered Jan 17 at 10:14
Kavi Rama MurthyKavi Rama Murthy
59.2k42161
59.2k42161
add a comment |
add a comment |
$begingroup$
Let $$s = sup_{|f|_2 leq 1} ,langle Tf,,frangle=1.$$
It is quite easy to see that $s$ is not a maximum, so there exists some sequence $f_n$ such that $|f_n| leq 1$, and $langle Tf_n ,, f_n rangle rightarrow 1$.
Assume that there is some converging subsequence $Tf_{varphi(n)} rightarrow g$. Then you easily get $langle g ,,f_{varphi(n)}rangle rightarrow 1$.
We know there is a weakly converging subsequence $f_{varphi(psi(n))} rightarrow h$ thus $|g| leq 1$ and $|h| leq 1$ and their inner product is $1$, thus $g=h$ and their norm is $1$.
Now, this entails (using the fact that $T$ is self-adjoint) that $Th=h$, which is impossible.
answered Jan 17 at 10:27
MindlackMindlack
3,86018
$endgroup$
add a comment |
$begingroup$
Let $$s = sup_{|f|_2 leq 1} ,langle Tf,,frangle=1.$$
It is quite easy to see that $s$ is not a maximum, so there exists some sequence $f_n$ such that $|f_n| leq 1$, and $langle Tf_n ,, f_n rangle rightarrow 1$.
Assume that there is some converging subsequence $Tf_{varphi(n)} rightarrow g$. Then you easily get $langle g ,,f_{varphi(n)}rangle rightarrow 1$.
We know there is a weakly converging subsequence $f_{varphi(psi(n))} rightarrow h$ thus $|g| leq 1$ and $|h| leq 1$ and their inner product is $1$, thus $g=h$ and their norm is $1$.
Now, this entails (using the fact that $T$ is self-adjoint) that $Th=h$, which is impossible.
answered Jan 17 at 10:27
MindlackMindlack
3,86018
$endgroup$
add a comment |
$begingroup$
Let $$s = sup_{|f|_2 leq 1} ,langle Tf,,frangle=1.$$
It is quite easy to see that $s$ is not a maximum, so there exists some sequence $f_n$ such that $|f_n| leq 1$, and $langle Tf_n ,, f_n rangle rightarrow 1$.
Assume that there is some converging subsequence $Tf_{varphi(n)} rightarrow g$. Then you easily get $langle g ,,f_{varphi(n)}rangle rightarrow 1$.
We know there is a weakly converging subsequence $f_{varphi(psi(n))} rightarrow h$ thus $|g| leq 1$ and $|h| leq 1$ and their inner product is $1$, thus $g=h$ and their norm is $1$.
Now, this entails (using the fact that $T$ is self-adjoint) that $Th=h$, which is impossible.
answered Jan 17 at 10:27
MindlackMindlack
3,86018
$endgroup$
Let $$s = sup_{|f|_2 leq 1} ,langle Tf,,frangle=1.$$
It is quite easy to see that $s$ is not a maximum, so there exists some sequence $f_n$ such that $|f_n| leq 1$, and $langle Tf_n ,, f_n rangle rightarrow 1$.
Assume that there is some converging subsequence $Tf_{varphi(n)} rightarrow g$. Then you easily get $langle g ,,f_{varphi(n)}rangle rightarrow 1$.
We know there is a weakly converging subsequence $f_{varphi(psi(n))} rightarrow h$ thus $|g| leq 1$ and $|h| leq 1$ and their inner product is $1$, thus $g=h$ and their norm is $1$.
Now, this entails (using the fact that $T$ is self-adjoint) that $Th=h$, which is impossible.
answered Jan 17 at 10:27
MindlackMindlack
3,86018
answered Jan 17 at 10:27
MindlackMindlack
3,86018
answered Jan 17 at 10:27
MindlackMindlack
3,86018
answered Jan 17 at 10:27
MindlackMindlack
3,86018
3,86018
add a comment |
add a comment |
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