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Example of a noncompact operator on $L^2([0,1])$
Clash Royale CLAN TAG#URR8PPP
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5
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For $f in L^2([0,1])$, define operator $Tf: x mapsto frac1xint_0^x f(y)dy$. Show that $T$ is not a compact operator on $L^2([0,1])$ and that $T$ is bounded.
For the second part, I can show $T$ is bounded by looking at $|Tf|_2$ and rewriting it by integration by parts and then apply the Cauchy-Schwartz inequality. However, I was not able to find a bounded sequence of $L^2$ functions so that its image under $T$ is not precompact in $L^2$.
Any help is tremendously appreciated.
real-analysis functional-analysis compact-operators
edited Jul 23 at 12:40
mechanodroid
22.2k52041
asked Jul 22 at 23:14
S_j
198110
add a comment |Â
up vote
5
down vote
favorite
For $f in L^2([0,1])$, define operator $Tf: x mapsto frac1xint_0^x f(y)dy$. Show that $T$ is not a compact operator on $L^2([0,1])$ and that $T$ is bounded.
For the second part, I can show $T$ is bounded by looking at $|Tf|_2$ and rewriting it by integration by parts and then apply the Cauchy-Schwartz inequality. However, I was not able to find a bounded sequence of $L^2$ functions so that its image under $T$ is not precompact in $L^2$.
Any help is tremendously appreciated.
real-analysis functional-analysis compact-operators
edited Jul 23 at 12:40
mechanodroid
22.2k52041
asked Jul 22 at 23:14
S_j
198110
1
Hint: try $f_n(x) = sqrtn 1_[0,1/n](x)$. A helpful fact is that if a sequence converges in $L^2$ then it has a subsequence which converges almost surely, so you can identify the possible limit points of $T f_n$.
– Chris Janjigian
Jul 23 at 0:22
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
For $f in L^2([0,1])$, define operator $Tf: x mapsto frac1xint_0^x f(y)dy$. Show that $T$ is not a compact operator on $L^2([0,1])$ and that $T$ is bounded.
For the second part, I can show $T$ is bounded by looking at $|Tf|_2$ and rewriting it by integration by parts and then apply the Cauchy-Schwartz inequality. However, I was not able to find a bounded sequence of $L^2$ functions so that its image under $T$ is not precompact in $L^2$.
Any help is tremendously appreciated.
real-analysis functional-analysis compact-operators
edited Jul 23 at 12:40
mechanodroid
22.2k52041
asked Jul 22 at 23:14
S_j
198110
For $f in L^2([0,1])$, define operator $Tf: x mapsto frac1xint_0^x f(y)dy$. Show that $T$ is not a compact operator on $L^2([0,1])$ and that $T$ is bounded.
For the second part, I can show $T$ is bounded by looking at $|Tf|_2$ and rewriting it by integration by parts and then apply the Cauchy-Schwartz inequality. However, I was not able to find a bounded sequence of $L^2$ functions so that its image under $T$ is not precompact in $L^2$.
Any help is tremendously appreciated.
real-analysis functional-analysis compact-operators
edited Jul 23 at 12:40
mechanodroid
22.2k52041
asked Jul 22 at 23:14
S_j
198110
edited Jul 23 at 12:40
mechanodroid
22.2k52041
edited Jul 23 at 12:40
mechanodroid
22.2k52041
edited Jul 23 at 12:40
mechanodroid
22.2k52041
22.2k52041
asked Jul 22 at 23:14
S_j
198110
asked Jul 22 at 23:14
S_j
198110
asked Jul 22 at 23:14
S_j
198110
198110
1
Hint: try $f_n(x) = sqrtn 1_[0,1/n](x)$. A helpful fact is that if a sequence converges in $L^2$ then it has a subsequence which converges almost surely, so you can identify the possible limit points of $T f_n$.
– Chris Janjigian
Jul 23 at 0:22
add a comment |Â
1
Hint: try $f_n(x) = sqrtn 1_[0,1/n](x)$. A helpful fact is that if a sequence converges in $L^2$ then it has a subsequence which converges almost surely, so you can identify the possible limit points of $T f_n$.
– Chris Janjigian
Jul 23 at 0:22
1
1
Hint: try $f_n(x) = sqrtn 1_[0,1/n](x)$. A helpful fact is that if a sequence converges in $L^2$ then it has a subsequence which converges almost surely, so you can identify the possible limit points of $T f_n$.
– Chris Janjigian
Jul 23 at 0:22
Hint: try $f_n(x) = sqrtn 1_[0,1/n](x)$. A helpful fact is that if a sequence converges in $L^2$ then it has a subsequence which converges almost surely, so you can identify the possible limit points of $T f_n$.
– Chris Janjigian
Jul 23 at 0:22
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
For $f_n(x) = sqrt n chi_(0, frac 1 n](x)$ we have
begineqnarray*
Tf_n(x)
& = &
frac 1 x int_0^xsqrt nchi_(0, frac 1 n](y); rm d y\
& = &
begincases
sqrt n & text if 0< x leq frac 1 n\
frac1xsqrt n & text if frac 1 n< x< 1.
endcases
endeqnarray*
From this formula we see that $lim_ntoinftyTf_n(x)to 0$ pointwise for all $0< x< 1$. Therefore, if there were some $L^2$-convergent subsequence of $Tf_n$, say $(Tf_n_k)_k = 1^infty$, then we must have $Tf_n_kto0$ in $L^2([0, 1])$. However, the following computation shows that no member of the sequence $(Tf_n)_n = 1^infty$ can get anywhere near its potential subsequential limit. For every $ngeq 1$,
begineqnarray*
|Tf_n|_L^2([0, 1])^2
& = & int_0^1/n(sqrt n)^2; rm dx + int_1/n^1left(frac 1xsqrt nright)^2; rm dx
\
& = &
1 + frac 1 nint_1/n^1x^-2; rm d x
\
& = &
2 - frac 1 n\
& > &
1.
endeqnarray*
answered Jul 23 at 1:27
BindersFull
498110
add a comment |Â
up vote
2
down vote
Another approach using the same sequence $f_n = sqrtnchi_[0, 1/n]$.
Here it is shown that $(f_n)_n$ converges to $0$ weakly:
$|f_n|_2 = 1, forall ninmathbbN$ so $(f_n)_n$ is bounded
for $g in L^infty[0,1] cap L^2[0,1]$ we have
$$|langle f_n, grangle| le | g |_infty cdot sqrtn int_0^1/n! dx = fracsqrtn to 0$$
and $L^infty[0,1] cap L^2[0,1]$ is dense in $L^2[0,1]$.
If $T$ were compact, it would map weakly convergent sequences to strongly convergent sequences so we would have $Tf_n to 0$ strongly.
But $|Tf_n|_2 = sqrt2-frac1n$ which doesn't converge to $0$.
answered Jul 23 at 12:40
mechanodroid
22.2k52041
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
For $f_n(x) = sqrt n chi_(0, frac 1 n](x)$ we have
begineqnarray*
Tf_n(x)
& = &
frac 1 x int_0^xsqrt nchi_(0, frac 1 n](y); rm d y\
& = &
begincases
sqrt n & text if 0< x leq frac 1 n\
frac1xsqrt n & text if frac 1 n< x< 1.
endcases
endeqnarray*
From this formula we see that $lim_ntoinftyTf_n(x)to 0$ pointwise for all $0< x< 1$. Therefore, if there were some $L^2$-convergent subsequence of $Tf_n$, say $(Tf_n_k)_k = 1^infty$, then we must have $Tf_n_kto0$ in $L^2([0, 1])$. However, the following computation shows that no member of the sequence $(Tf_n)_n = 1^infty$ can get anywhere near its potential subsequential limit. For every $ngeq 1$,
begineqnarray*
|Tf_n|_L^2([0, 1])^2
& = & int_0^1/n(sqrt n)^2; rm dx + int_1/n^1left(frac 1xsqrt nright)^2; rm dx
\
& = &
1 + frac 1 nint_1/n^1x^-2; rm d x
\
& = &
2 - frac 1 n\
& > &
1.
endeqnarray*
answered Jul 23 at 1:27
BindersFull
498110
add a comment |Â
up vote
2
down vote
accepted
For $f_n(x) = sqrt n chi_(0, frac 1 n](x)$ we have
begineqnarray*
Tf_n(x)
& = &
frac 1 x int_0^xsqrt nchi_(0, frac 1 n](y); rm d y\
& = &
begincases
sqrt n & text if 0< x leq frac 1 n\
frac1xsqrt n & text if frac 1 n< x< 1.
endcases
endeqnarray*
From this formula we see that $lim_ntoinftyTf_n(x)to 0$ pointwise for all $0< x< 1$. Therefore, if there were some $L^2$-convergent subsequence of $Tf_n$, say $(Tf_n_k)_k = 1^infty$, then we must have $Tf_n_kto0$ in $L^2([0, 1])$. However, the following computation shows that no member of the sequence $(Tf_n)_n = 1^infty$ can get anywhere near its potential subsequential limit. For every $ngeq 1$,
begineqnarray*
|Tf_n|_L^2([0, 1])^2
& = & int_0^1/n(sqrt n)^2; rm dx + int_1/n^1left(frac 1xsqrt nright)^2; rm dx
\
& = &
1 + frac 1 nint_1/n^1x^-2; rm d x
\
& = &
2 - frac 1 n\
& > &
1.
endeqnarray*
answered Jul 23 at 1:27
BindersFull
498110
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
For $f_n(x) = sqrt n chi_(0, frac 1 n](x)$ we have
begineqnarray*
Tf_n(x)
& = &
frac 1 x int_0^xsqrt nchi_(0, frac 1 n](y); rm d y\
& = &
begincases
sqrt n & text if 0< x leq frac 1 n\
frac1xsqrt n & text if frac 1 n< x< 1.
endcases
endeqnarray*
From this formula we see that $lim_ntoinftyTf_n(x)to 0$ pointwise for all $0< x< 1$. Therefore, if there were some $L^2$-convergent subsequence of $Tf_n$, say $(Tf_n_k)_k = 1^infty$, then we must have $Tf_n_kto0$ in $L^2([0, 1])$. However, the following computation shows that no member of the sequence $(Tf_n)_n = 1^infty$ can get anywhere near its potential subsequential limit. For every $ngeq 1$,
begineqnarray*
|Tf_n|_L^2([0, 1])^2
& = & int_0^1/n(sqrt n)^2; rm dx + int_1/n^1left(frac 1xsqrt nright)^2; rm dx
\
& = &
1 + frac 1 nint_1/n^1x^-2; rm d x
\
& = &
2 - frac 1 n\
& > &
1.
endeqnarray*
answered Jul 23 at 1:27
BindersFull
498110
For $f_n(x) = sqrt n chi_(0, frac 1 n](x)$ we have
begineqnarray*
Tf_n(x)
& = &
frac 1 x int_0^xsqrt nchi_(0, frac 1 n](y); rm d y\
& = &
begincases
sqrt n & text if 0< x leq frac 1 n\
frac1xsqrt n & text if frac 1 n< x< 1.
endcases
endeqnarray*
From this formula we see that $lim_ntoinftyTf_n(x)to 0$ pointwise for all $0< x< 1$. Therefore, if there were some $L^2$-convergent subsequence of $Tf_n$, say $(Tf_n_k)_k = 1^infty$, then we must have $Tf_n_kto0$ in $L^2([0, 1])$. However, the following computation shows that no member of the sequence $(Tf_n)_n = 1^infty$ can get anywhere near its potential subsequential limit. For every $ngeq 1$,
begineqnarray*
|Tf_n|_L^2([0, 1])^2
& = & int_0^1/n(sqrt n)^2; rm dx + int_1/n^1left(frac 1xsqrt nright)^2; rm dx
\
& = &
1 + frac 1 nint_1/n^1x^-2; rm d x
\
& = &
2 - frac 1 n\
& > &
1.
endeqnarray*
answered Jul 23 at 1:27
BindersFull
498110
answered Jul 23 at 1:27
BindersFull
498110
answered Jul 23 at 1:27
BindersFull
498110
answered Jul 23 at 1:27
BindersFull
498110
498110
add a comment |Â
add a comment |Â
up vote
2
down vote
Another approach using the same sequence $f_n = sqrtnchi_[0, 1/n]$.
Here it is shown that $(f_n)_n$ converges to $0$ weakly:
$|f_n|_2 = 1, forall ninmathbbN$ so $(f_n)_n$ is bounded
for $g in L^infty[0,1] cap L^2[0,1]$ we have
$$|langle f_n, grangle| le | g |_infty cdot sqrtn int_0^1/n! dx = fracsqrtn to 0$$
and $L^infty[0,1] cap L^2[0,1]$ is dense in $L^2[0,1]$.
If $T$ were compact, it would map weakly convergent sequences to strongly convergent sequences so we would have $Tf_n to 0$ strongly.
But $|Tf_n|_2 = sqrt2-frac1n$ which doesn't converge to $0$.
answered Jul 23 at 12:40
mechanodroid
22.2k52041
add a comment |Â
up vote
2
down vote
Another approach using the same sequence $f_n = sqrtnchi_[0, 1/n]$.
Here it is shown that $(f_n)_n$ converges to $0$ weakly:
$|f_n|_2 = 1, forall ninmathbbN$ so $(f_n)_n$ is bounded
for $g in L^infty[0,1] cap L^2[0,1]$ we have
$$|langle f_n, grangle| le | g |_infty cdot sqrtn int_0^1/n! dx = fracsqrtn to 0$$
and $L^infty[0,1] cap L^2[0,1]$ is dense in $L^2[0,1]$.
If $T$ were compact, it would map weakly convergent sequences to strongly convergent sequences so we would have $Tf_n to 0$ strongly.
But $|Tf_n|_2 = sqrt2-frac1n$ which doesn't converge to $0$.
answered Jul 23 at 12:40
mechanodroid
22.2k52041
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Another approach using the same sequence $f_n = sqrtnchi_[0, 1/n]$.
Here it is shown that $(f_n)_n$ converges to $0$ weakly:
$|f_n|_2 = 1, forall ninmathbbN$ so $(f_n)_n$ is bounded
for $g in L^infty[0,1] cap L^2[0,1]$ we have
$$|langle f_n, grangle| le | g |_infty cdot sqrtn int_0^1/n! dx = fracsqrtn to 0$$
and $L^infty[0,1] cap L^2[0,1]$ is dense in $L^2[0,1]$.
If $T$ were compact, it would map weakly convergent sequences to strongly convergent sequences so we would have $Tf_n to 0$ strongly.
But $|Tf_n|_2 = sqrt2-frac1n$ which doesn't converge to $0$.
answered Jul 23 at 12:40
mechanodroid
22.2k52041
Another approach using the same sequence $f_n = sqrtnchi_[0, 1/n]$.
Here it is shown that $(f_n)_n$ converges to $0$ weakly:
$|f_n|_2 = 1, forall ninmathbbN$ so $(f_n)_n$ is bounded
for $g in L^infty[0,1] cap L^2[0,1]$ we have
$$|langle f_n, grangle| le | g |_infty cdot sqrtn int_0^1/n! dx = fracsqrtn to 0$$
and $L^infty[0,1] cap L^2[0,1]$ is dense in $L^2[0,1]$.
If $T$ were compact, it would map weakly convergent sequences to strongly convergent sequences so we would have $Tf_n to 0$ strongly.
But $|Tf_n|_2 = sqrt2-frac1n$ which doesn't converge to $0$.
answered Jul 23 at 12:40
mechanodroid
22.2k52041
answered Jul 23 at 12:40
mechanodroid
22.2k52041
answered Jul 23 at 12:40
mechanodroid
22.2k52041
answered Jul 23 at 12:40
mechanodroid
22.2k52041
22.2k52041
add a comment |Â
add a comment |Â
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1
Hint: try $f_n(x) = sqrtn 1_[0,1/n](x)$. A helpful fact is that if a sequence converges in $L^2$ then it has a subsequence which converges almost surely, so you can identify the possible limit points of $T f_n$.
– Chris Janjigian
Jul 23 at 0:22