A non compact operator on $L^2[0,1]$
$begingroup$
Let $H$ be the Hilbert space $L^2[0,1]$.
and the operator $T : Hrightarrow H$, such as $T(f)(x)=x.f(x)$
Why $T$ isn't compact ?
functional-analysis operator-theory compact-operators
$endgroup$
add a comment |
$begingroup$
Let $H$ be the Hilbert space $L^2[0,1]$.
and the operator $T : Hrightarrow H$, such as $T(f)(x)=x.f(x)$
Why $T$ isn't compact ?
functional-analysis operator-theory compact-operators
$endgroup$
add a comment |
$begingroup$
Let $H$ be the Hilbert space $L^2[0,1]$.
and the operator $T : Hrightarrow H$, such as $T(f)(x)=x.f(x)$
Why $T$ isn't compact ?
functional-analysis operator-theory compact-operators
$endgroup$
Let $H$ be the Hilbert space $L^2[0,1]$.
and the operator $T : Hrightarrow H$, such as $T(f)(x)=x.f(x)$
Why $T$ isn't compact ?
functional-analysis operator-theory compact-operators
functional-analysis operator-theory compact-operators
asked Dec 22 '18 at 8:23
Anas BOUALIIAnas BOUALII
1438
1438
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $f_n(x) = sqrt{n}1_{[1-frac{1}{n},1]}(x)$. Then $||f_n||_2 = 1$ for each $n$ but $(Tf_n)_n$ has no convergent subsequence.
Indeed, suppose $Tf_{n_k} to g$ in $L^2$ for some $g in L^2$ and subsequence $(f_{n_k})_k$. Since $||f_{n_k}||_2 = 1$ for each $k$, $g$ cannot be $0$, so $int_0^1 |g(x)|^2dx > 0$. Then $int_0^a |g(x)|^2dx =: epsilon > 0$ for some $a in (0,1)$. But then for large $k$, $$int_0^1 |g(x)-xf_{n_k}(x)|^2dx = int_0^a |g(x)|^2 +int_a^1 |g(x)-xf_{n_k}(x)|^2dx ge int_0^a |g(x)|^2dx ge epsilon,$$ which contradicts $f_{n_k} to g$ in $L^2$.
$endgroup$
$begingroup$
I found dificulties in proving that $Tf_n=f_n$
$endgroup$
– Anas BOUALII
Dec 22 '18 at 9:08
$begingroup$
@AnasBOUALII I have added details
$endgroup$
– mathworker21
Dec 22 '18 at 9:18
$begingroup$
I see, thank you.
$endgroup$
– Anas BOUALII
Dec 22 '18 at 9:24
add a comment |
$begingroup$
Observe that $T$ is a self-adjoint operator. And also $$
ker (T-lambda)=(0)
$$ holds for all $lambdain mathbb{C}$. If $T$ is compact, then by the spectral theorem of compact operators, it follows that
$$
T=0.
$$ This leads to an obvious contradiction.
$endgroup$
$begingroup$
By the theorem it follows that the spectrum is $mathbb{C}$, which is a contradiction with the spectrum is compact. Right ?
$endgroup$
– Anas BOUALII
Dec 22 '18 at 8:56
$begingroup$
I'll elaborate on this. In fact, $ker(T-lambda)=(0)$ means that the point spectrum of $T$ is empty. For a compact operator $T$, it holds generally that $sigma_p(T)setminus{0} = sigma(T)setminus{0}$, where $sigma_p(T) := {lambda;|;ker(T-lambda) neq (0)}$ is the point spectrum of $T$. For the given $T$, we have $sigma_p(T) =varnothing$ and hence it follows $sigma(T) subset {0}$. However, we can show that $sigma(T) = [0,1]$ holds, leading to a contradiction!
$endgroup$
– Song
Dec 22 '18 at 9:03
$begingroup$
I agree with you. so since $T$ is self-adjoint operator then $sigma(T) subset [m,M]$ with $M=sup_{||f||=1}<Tf,f>$, $m=inf_{||f||=1}<Tf,f>$, but how to prove that $m=0$ and $M=1$?
$endgroup$
– Anas BOUALII
Dec 22 '18 at 9:22
$begingroup$
Well, $langle f,Tfranglesubset [0,1]$ is almost direct, and $f_n = sqrt{n}1_{[0,frac{1}{n}]}$ is an infimum-approaching sequence (note that $f_n$ is getting concentrated on $x=0$). In a similar way, $f_n =sqrt{n}1_{[1-frac{1}{n},1]}$ is a supremum-approaching sequence. But in fact, $sigma(T) =[0,1]$ can be seen in a more direct way. Since for $lambda in [0,1]$, $xmapsto frac{1}{x-lambda}$ is not bounded on $[0,1]$, it follows that $(T-lambda)^{-1}f(x) =frac{f(x)}{x-lambda}$ is not a bounded inverse in $L^2$ .
$endgroup$
– Song
Dec 22 '18 at 9:34
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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$begingroup$
Let $f_n(x) = sqrt{n}1_{[1-frac{1}{n},1]}(x)$. Then $||f_n||_2 = 1$ for each $n$ but $(Tf_n)_n$ has no convergent subsequence.
Indeed, suppose $Tf_{n_k} to g$ in $L^2$ for some $g in L^2$ and subsequence $(f_{n_k})_k$. Since $||f_{n_k}||_2 = 1$ for each $k$, $g$ cannot be $0$, so $int_0^1 |g(x)|^2dx > 0$. Then $int_0^a |g(x)|^2dx =: epsilon > 0$ for some $a in (0,1)$. But then for large $k$, $$int_0^1 |g(x)-xf_{n_k}(x)|^2dx = int_0^a |g(x)|^2 +int_a^1 |g(x)-xf_{n_k}(x)|^2dx ge int_0^a |g(x)|^2dx ge epsilon,$$ which contradicts $f_{n_k} to g$ in $L^2$.
$endgroup$
$begingroup$
I found dificulties in proving that $Tf_n=f_n$
$endgroup$
– Anas BOUALII
Dec 22 '18 at 9:08
$begingroup$
@AnasBOUALII I have added details
$endgroup$
– mathworker21
Dec 22 '18 at 9:18
$begingroup$
I see, thank you.
$endgroup$
– Anas BOUALII
Dec 22 '18 at 9:24
add a comment |
$begingroup$
Let $f_n(x) = sqrt{n}1_{[1-frac{1}{n},1]}(x)$. Then $||f_n||_2 = 1$ for each $n$ but $(Tf_n)_n$ has no convergent subsequence.
Indeed, suppose $Tf_{n_k} to g$ in $L^2$ for some $g in L^2$ and subsequence $(f_{n_k})_k$. Since $||f_{n_k}||_2 = 1$ for each $k$, $g$ cannot be $0$, so $int_0^1 |g(x)|^2dx > 0$. Then $int_0^a |g(x)|^2dx =: epsilon > 0$ for some $a in (0,1)$. But then for large $k$, $$int_0^1 |g(x)-xf_{n_k}(x)|^2dx = int_0^a |g(x)|^2 +int_a^1 |g(x)-xf_{n_k}(x)|^2dx ge int_0^a |g(x)|^2dx ge epsilon,$$ which contradicts $f_{n_k} to g$ in $L^2$.
$endgroup$
$begingroup$
I found dificulties in proving that $Tf_n=f_n$
$endgroup$
– Anas BOUALII
Dec 22 '18 at 9:08
$begingroup$
@AnasBOUALII I have added details
$endgroup$
– mathworker21
Dec 22 '18 at 9:18
$begingroup$
I see, thank you.
$endgroup$
– Anas BOUALII
Dec 22 '18 at 9:24
add a comment |
$begingroup$
Let $f_n(x) = sqrt{n}1_{[1-frac{1}{n},1]}(x)$. Then $||f_n||_2 = 1$ for each $n$ but $(Tf_n)_n$ has no convergent subsequence.
Indeed, suppose $Tf_{n_k} to g$ in $L^2$ for some $g in L^2$ and subsequence $(f_{n_k})_k$. Since $||f_{n_k}||_2 = 1$ for each $k$, $g$ cannot be $0$, so $int_0^1 |g(x)|^2dx > 0$. Then $int_0^a |g(x)|^2dx =: epsilon > 0$ for some $a in (0,1)$. But then for large $k$, $$int_0^1 |g(x)-xf_{n_k}(x)|^2dx = int_0^a |g(x)|^2 +int_a^1 |g(x)-xf_{n_k}(x)|^2dx ge int_0^a |g(x)|^2dx ge epsilon,$$ which contradicts $f_{n_k} to g$ in $L^2$.
$endgroup$
Let $f_n(x) = sqrt{n}1_{[1-frac{1}{n},1]}(x)$. Then $||f_n||_2 = 1$ for each $n$ but $(Tf_n)_n$ has no convergent subsequence.
Indeed, suppose $Tf_{n_k} to g$ in $L^2$ for some $g in L^2$ and subsequence $(f_{n_k})_k$. Since $||f_{n_k}||_2 = 1$ for each $k$, $g$ cannot be $0$, so $int_0^1 |g(x)|^2dx > 0$. Then $int_0^a |g(x)|^2dx =: epsilon > 0$ for some $a in (0,1)$. But then for large $k$, $$int_0^1 |g(x)-xf_{n_k}(x)|^2dx = int_0^a |g(x)|^2 +int_a^1 |g(x)-xf_{n_k}(x)|^2dx ge int_0^a |g(x)|^2dx ge epsilon,$$ which contradicts $f_{n_k} to g$ in $L^2$.
edited Dec 22 '18 at 9:18
answered Dec 22 '18 at 8:33
mathworker21mathworker21
8,9561928
8,9561928
$begingroup$
I found dificulties in proving that $Tf_n=f_n$
$endgroup$
– Anas BOUALII
Dec 22 '18 at 9:08
$begingroup$
@AnasBOUALII I have added details
$endgroup$
– mathworker21
Dec 22 '18 at 9:18
$begingroup$
I see, thank you.
$endgroup$
– Anas BOUALII
Dec 22 '18 at 9:24
add a comment |
$begingroup$
I found dificulties in proving that $Tf_n=f_n$
$endgroup$
– Anas BOUALII
Dec 22 '18 at 9:08
$begingroup$
@AnasBOUALII I have added details
$endgroup$
– mathworker21
Dec 22 '18 at 9:18
$begingroup$
I see, thank you.
$endgroup$
– Anas BOUALII
Dec 22 '18 at 9:24
$begingroup$
I found dificulties in proving that $Tf_n=f_n$
$endgroup$
– Anas BOUALII
Dec 22 '18 at 9:08
$begingroup$
I found dificulties in proving that $Tf_n=f_n$
$endgroup$
– Anas BOUALII
Dec 22 '18 at 9:08
$begingroup$
@AnasBOUALII I have added details
$endgroup$
– mathworker21
Dec 22 '18 at 9:18
$begingroup$
@AnasBOUALII I have added details
$endgroup$
– mathworker21
Dec 22 '18 at 9:18
$begingroup$
I see, thank you.
$endgroup$
– Anas BOUALII
Dec 22 '18 at 9:24
$begingroup$
I see, thank you.
$endgroup$
– Anas BOUALII
Dec 22 '18 at 9:24
add a comment |
$begingroup$
Observe that $T$ is a self-adjoint operator. And also $$
ker (T-lambda)=(0)
$$ holds for all $lambdain mathbb{C}$. If $T$ is compact, then by the spectral theorem of compact operators, it follows that
$$
T=0.
$$ This leads to an obvious contradiction.
$endgroup$
$begingroup$
By the theorem it follows that the spectrum is $mathbb{C}$, which is a contradiction with the spectrum is compact. Right ?
$endgroup$
– Anas BOUALII
Dec 22 '18 at 8:56
$begingroup$
I'll elaborate on this. In fact, $ker(T-lambda)=(0)$ means that the point spectrum of $T$ is empty. For a compact operator $T$, it holds generally that $sigma_p(T)setminus{0} = sigma(T)setminus{0}$, where $sigma_p(T) := {lambda;|;ker(T-lambda) neq (0)}$ is the point spectrum of $T$. For the given $T$, we have $sigma_p(T) =varnothing$ and hence it follows $sigma(T) subset {0}$. However, we can show that $sigma(T) = [0,1]$ holds, leading to a contradiction!
$endgroup$
– Song
Dec 22 '18 at 9:03
$begingroup$
I agree with you. so since $T$ is self-adjoint operator then $sigma(T) subset [m,M]$ with $M=sup_{||f||=1}<Tf,f>$, $m=inf_{||f||=1}<Tf,f>$, but how to prove that $m=0$ and $M=1$?
$endgroup$
– Anas BOUALII
Dec 22 '18 at 9:22
$begingroup$
Well, $langle f,Tfranglesubset [0,1]$ is almost direct, and $f_n = sqrt{n}1_{[0,frac{1}{n}]}$ is an infimum-approaching sequence (note that $f_n$ is getting concentrated on $x=0$). In a similar way, $f_n =sqrt{n}1_{[1-frac{1}{n},1]}$ is a supremum-approaching sequence. But in fact, $sigma(T) =[0,1]$ can be seen in a more direct way. Since for $lambda in [0,1]$, $xmapsto frac{1}{x-lambda}$ is not bounded on $[0,1]$, it follows that $(T-lambda)^{-1}f(x) =frac{f(x)}{x-lambda}$ is not a bounded inverse in $L^2$ .
$endgroup$
– Song
Dec 22 '18 at 9:34
add a comment |
$begingroup$
Observe that $T$ is a self-adjoint operator. And also $$
ker (T-lambda)=(0)
$$ holds for all $lambdain mathbb{C}$. If $T$ is compact, then by the spectral theorem of compact operators, it follows that
$$
T=0.
$$ This leads to an obvious contradiction.
$endgroup$
$begingroup$
By the theorem it follows that the spectrum is $mathbb{C}$, which is a contradiction with the spectrum is compact. Right ?
$endgroup$
– Anas BOUALII
Dec 22 '18 at 8:56
$begingroup$
I'll elaborate on this. In fact, $ker(T-lambda)=(0)$ means that the point spectrum of $T$ is empty. For a compact operator $T$, it holds generally that $sigma_p(T)setminus{0} = sigma(T)setminus{0}$, where $sigma_p(T) := {lambda;|;ker(T-lambda) neq (0)}$ is the point spectrum of $T$. For the given $T$, we have $sigma_p(T) =varnothing$ and hence it follows $sigma(T) subset {0}$. However, we can show that $sigma(T) = [0,1]$ holds, leading to a contradiction!
$endgroup$
– Song
Dec 22 '18 at 9:03
$begingroup$
I agree with you. so since $T$ is self-adjoint operator then $sigma(T) subset [m,M]$ with $M=sup_{||f||=1}<Tf,f>$, $m=inf_{||f||=1}<Tf,f>$, but how to prove that $m=0$ and $M=1$?
$endgroup$
– Anas BOUALII
Dec 22 '18 at 9:22
$begingroup$
Well, $langle f,Tfranglesubset [0,1]$ is almost direct, and $f_n = sqrt{n}1_{[0,frac{1}{n}]}$ is an infimum-approaching sequence (note that $f_n$ is getting concentrated on $x=0$). In a similar way, $f_n =sqrt{n}1_{[1-frac{1}{n},1]}$ is a supremum-approaching sequence. But in fact, $sigma(T) =[0,1]$ can be seen in a more direct way. Since for $lambda in [0,1]$, $xmapsto frac{1}{x-lambda}$ is not bounded on $[0,1]$, it follows that $(T-lambda)^{-1}f(x) =frac{f(x)}{x-lambda}$ is not a bounded inverse in $L^2$ .
$endgroup$
– Song
Dec 22 '18 at 9:34
add a comment |
$begingroup$
Observe that $T$ is a self-adjoint operator. And also $$
ker (T-lambda)=(0)
$$ holds for all $lambdain mathbb{C}$. If $T$ is compact, then by the spectral theorem of compact operators, it follows that
$$
T=0.
$$ This leads to an obvious contradiction.
$endgroup$
Observe that $T$ is a self-adjoint operator. And also $$
ker (T-lambda)=(0)
$$ holds for all $lambdain mathbb{C}$. If $T$ is compact, then by the spectral theorem of compact operators, it follows that
$$
T=0.
$$ This leads to an obvious contradiction.
answered Dec 22 '18 at 8:51
SongSong
14.1k1633
14.1k1633
$begingroup$
By the theorem it follows that the spectrum is $mathbb{C}$, which is a contradiction with the spectrum is compact. Right ?
$endgroup$
– Anas BOUALII
Dec 22 '18 at 8:56
$begingroup$
I'll elaborate on this. In fact, $ker(T-lambda)=(0)$ means that the point spectrum of $T$ is empty. For a compact operator $T$, it holds generally that $sigma_p(T)setminus{0} = sigma(T)setminus{0}$, where $sigma_p(T) := {lambda;|;ker(T-lambda) neq (0)}$ is the point spectrum of $T$. For the given $T$, we have $sigma_p(T) =varnothing$ and hence it follows $sigma(T) subset {0}$. However, we can show that $sigma(T) = [0,1]$ holds, leading to a contradiction!
$endgroup$
– Song
Dec 22 '18 at 9:03
$begingroup$
I agree with you. so since $T$ is self-adjoint operator then $sigma(T) subset [m,M]$ with $M=sup_{||f||=1}<Tf,f>$, $m=inf_{||f||=1}<Tf,f>$, but how to prove that $m=0$ and $M=1$?
$endgroup$
– Anas BOUALII
Dec 22 '18 at 9:22
$begingroup$
Well, $langle f,Tfranglesubset [0,1]$ is almost direct, and $f_n = sqrt{n}1_{[0,frac{1}{n}]}$ is an infimum-approaching sequence (note that $f_n$ is getting concentrated on $x=0$). In a similar way, $f_n =sqrt{n}1_{[1-frac{1}{n},1]}$ is a supremum-approaching sequence. But in fact, $sigma(T) =[0,1]$ can be seen in a more direct way. Since for $lambda in [0,1]$, $xmapsto frac{1}{x-lambda}$ is not bounded on $[0,1]$, it follows that $(T-lambda)^{-1}f(x) =frac{f(x)}{x-lambda}$ is not a bounded inverse in $L^2$ .
$endgroup$
– Song
Dec 22 '18 at 9:34
add a comment |
$begingroup$
By the theorem it follows that the spectrum is $mathbb{C}$, which is a contradiction with the spectrum is compact. Right ?
$endgroup$
– Anas BOUALII
Dec 22 '18 at 8:56
$begingroup$
I'll elaborate on this. In fact, $ker(T-lambda)=(0)$ means that the point spectrum of $T$ is empty. For a compact operator $T$, it holds generally that $sigma_p(T)setminus{0} = sigma(T)setminus{0}$, where $sigma_p(T) := {lambda;|;ker(T-lambda) neq (0)}$ is the point spectrum of $T$. For the given $T$, we have $sigma_p(T) =varnothing$ and hence it follows $sigma(T) subset {0}$. However, we can show that $sigma(T) = [0,1]$ holds, leading to a contradiction!
$endgroup$
– Song
Dec 22 '18 at 9:03
$begingroup$
I agree with you. so since $T$ is self-adjoint operator then $sigma(T) subset [m,M]$ with $M=sup_{||f||=1}<Tf,f>$, $m=inf_{||f||=1}<Tf,f>$, but how to prove that $m=0$ and $M=1$?
$endgroup$
– Anas BOUALII
Dec 22 '18 at 9:22
$begingroup$
Well, $langle f,Tfranglesubset [0,1]$ is almost direct, and $f_n = sqrt{n}1_{[0,frac{1}{n}]}$ is an infimum-approaching sequence (note that $f_n$ is getting concentrated on $x=0$). In a similar way, $f_n =sqrt{n}1_{[1-frac{1}{n},1]}$ is a supremum-approaching sequence. But in fact, $sigma(T) =[0,1]$ can be seen in a more direct way. Since for $lambda in [0,1]$, $xmapsto frac{1}{x-lambda}$ is not bounded on $[0,1]$, it follows that $(T-lambda)^{-1}f(x) =frac{f(x)}{x-lambda}$ is not a bounded inverse in $L^2$ .
$endgroup$
– Song
Dec 22 '18 at 9:34
$begingroup$
By the theorem it follows that the spectrum is $mathbb{C}$, which is a contradiction with the spectrum is compact. Right ?
$endgroup$
– Anas BOUALII
Dec 22 '18 at 8:56
$begingroup$
By the theorem it follows that the spectrum is $mathbb{C}$, which is a contradiction with the spectrum is compact. Right ?
$endgroup$
– Anas BOUALII
Dec 22 '18 at 8:56
$begingroup$
I'll elaborate on this. In fact, $ker(T-lambda)=(0)$ means that the point spectrum of $T$ is empty. For a compact operator $T$, it holds generally that $sigma_p(T)setminus{0} = sigma(T)setminus{0}$, where $sigma_p(T) := {lambda;|;ker(T-lambda) neq (0)}$ is the point spectrum of $T$. For the given $T$, we have $sigma_p(T) =varnothing$ and hence it follows $sigma(T) subset {0}$. However, we can show that $sigma(T) = [0,1]$ holds, leading to a contradiction!
$endgroup$
– Song
Dec 22 '18 at 9:03
$begingroup$
I'll elaborate on this. In fact, $ker(T-lambda)=(0)$ means that the point spectrum of $T$ is empty. For a compact operator $T$, it holds generally that $sigma_p(T)setminus{0} = sigma(T)setminus{0}$, where $sigma_p(T) := {lambda;|;ker(T-lambda) neq (0)}$ is the point spectrum of $T$. For the given $T$, we have $sigma_p(T) =varnothing$ and hence it follows $sigma(T) subset {0}$. However, we can show that $sigma(T) = [0,1]$ holds, leading to a contradiction!
$endgroup$
– Song
Dec 22 '18 at 9:03
$begingroup$
I agree with you. so since $T$ is self-adjoint operator then $sigma(T) subset [m,M]$ with $M=sup_{||f||=1}<Tf,f>$, $m=inf_{||f||=1}<Tf,f>$, but how to prove that $m=0$ and $M=1$?
$endgroup$
– Anas BOUALII
Dec 22 '18 at 9:22
$begingroup$
I agree with you. so since $T$ is self-adjoint operator then $sigma(T) subset [m,M]$ with $M=sup_{||f||=1}<Tf,f>$, $m=inf_{||f||=1}<Tf,f>$, but how to prove that $m=0$ and $M=1$?
$endgroup$
– Anas BOUALII
Dec 22 '18 at 9:22
$begingroup$
Well, $langle f,Tfranglesubset [0,1]$ is almost direct, and $f_n = sqrt{n}1_{[0,frac{1}{n}]}$ is an infimum-approaching sequence (note that $f_n$ is getting concentrated on $x=0$). In a similar way, $f_n =sqrt{n}1_{[1-frac{1}{n},1]}$ is a supremum-approaching sequence. But in fact, $sigma(T) =[0,1]$ can be seen in a more direct way. Since for $lambda in [0,1]$, $xmapsto frac{1}{x-lambda}$ is not bounded on $[0,1]$, it follows that $(T-lambda)^{-1}f(x) =frac{f(x)}{x-lambda}$ is not a bounded inverse in $L^2$ .
$endgroup$
– Song
Dec 22 '18 at 9:34
$begingroup$
Well, $langle f,Tfranglesubset [0,1]$ is almost direct, and $f_n = sqrt{n}1_{[0,frac{1}{n}]}$ is an infimum-approaching sequence (note that $f_n$ is getting concentrated on $x=0$). In a similar way, $f_n =sqrt{n}1_{[1-frac{1}{n},1]}$ is a supremum-approaching sequence. But in fact, $sigma(T) =[0,1]$ can be seen in a more direct way. Since for $lambda in [0,1]$, $xmapsto frac{1}{x-lambda}$ is not bounded on $[0,1]$, it follows that $(T-lambda)^{-1}f(x) =frac{f(x)}{x-lambda}$ is not a bounded inverse in $L^2$ .
$endgroup$
– Song
Dec 22 '18 at 9:34
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