Mixing
Show that $(Tu)(x)=int_{alpha(x)}^{beta(x)} u(t)dt$ is Compact linear operator on $C([0,1])$
$begingroup$
Show that
begin{equation}
(Tu)(x)=int_{alpha(x)}^{beta(x)} u(t)dt
end{equation}
is Compact linear operator on $C([0,1],R)$ where $alpha, beta:[0,1]rightarrow [0,1]$ are continuous.
My Attempt
$T$ is obviously linear.
Let $M=max (alpha(x)-beta(x))$. Then
$|Tu(x)|leq ||u||_{infty}M$ and $||T||=M$.
For compactness, let $B={uin (C[0,1],R):||u||_{infty}leq 1}$. We show $T(B)$ is equicontinuous family so that by Arzela Ascoli it is relatively compact.
begin{equation}
|Tu(x)-Tu(y)|=bigg|int_{alpha(x)}^{beta(x)} u(t)dt-int_{alpha(y)}^{beta(y)} u(t)dtbigg|
end{equation}
Please how do I subtract this integrals. I'm thinking to make the assumption $alpha(y)<beta(y)<alpha(x)<beta(x)$.
functional-analysis operator-theory compact-operators
asked Jan 12 at 20:17
Muhammad MubarakMuhammad Mubarak
969
$endgroup$
add a comment |
$begingroup$
Show that
begin{equation}
(Tu)(x)=int_{alpha(x)}^{beta(x)} u(t)dt
end{equation}
is Compact linear operator on $C([0,1],R)$ where $alpha, beta:[0,1]rightarrow [0,1]$ are continuous.
My Attempt
$T$ is obviously linear.
Let $M=max (alpha(x)-beta(x))$. Then
$|Tu(x)|leq ||u||_{infty}M$ and $||T||=M$.
For compactness, let $B={uin (C[0,1],R):||u||_{infty}leq 1}$. We show $T(B)$ is equicontinuous family so that by Arzela Ascoli it is relatively compact.
begin{equation}
|Tu(x)-Tu(y)|=bigg|int_{alpha(x)}^{beta(x)} u(t)dt-int_{alpha(y)}^{beta(y)} u(t)dtbigg|
end{equation}
Please how do I subtract this integrals. I'm thinking to make the assumption $alpha(y)<beta(y)<alpha(x)<beta(x)$.
functional-analysis operator-theory compact-operators
asked Jan 12 at 20:17
Muhammad MubarakMuhammad Mubarak
969
$endgroup$
$begingroup$
Is there any assumption on the regularity of $alpha$ and $beta$ at all?
$endgroup$
– BigbearZzz
Jan 12 at 20:24
$begingroup$
They are continuous. I will edit now.
$endgroup$
– Muhammad Mubarak
Jan 12 at 20:25
$begingroup$
My answer below can be rewritten without using the characteristic function $chi$ by arguing case by case. However, I prefer using $chi$ as it makes some idea clearer.
$endgroup$
– BigbearZzz
Jan 12 at 22:39
$begingroup$
Thanks a lot. Its clear now
$endgroup$
– Muhammad Mubarak
Jan 13 at 9:56
add a comment |
$begingroup$
Show that
begin{equation}
(Tu)(x)=int_{alpha(x)}^{beta(x)} u(t)dt
end{equation}
is Compact linear operator on $C([0,1],R)$ where $alpha, beta:[0,1]rightarrow [0,1]$ are continuous.
My Attempt
$T$ is obviously linear.
Let $M=max (alpha(x)-beta(x))$. Then
$|Tu(x)|leq ||u||_{infty}M$ and $||T||=M$.
For compactness, let $B={uin (C[0,1],R):||u||_{infty}leq 1}$. We show $T(B)$ is equicontinuous family so that by Arzela Ascoli it is relatively compact.
begin{equation}
|Tu(x)-Tu(y)|=bigg|int_{alpha(x)}^{beta(x)} u(t)dt-int_{alpha(y)}^{beta(y)} u(t)dtbigg|
end{equation}
Please how do I subtract this integrals. I'm thinking to make the assumption $alpha(y)<beta(y)<alpha(x)<beta(x)$.
functional-analysis operator-theory compact-operators
asked Jan 12 at 20:17
Muhammad MubarakMuhammad Mubarak
969
$endgroup$
Show that
begin{equation}
(Tu)(x)=int_{alpha(x)}^{beta(x)} u(t)dt
end{equation}
is Compact linear operator on $C([0,1],R)$ where $alpha, beta:[0,1]rightarrow [0,1]$ are continuous.
My Attempt
$T$ is obviously linear.
Let $M=max (alpha(x)-beta(x))$. Then
$|Tu(x)|leq ||u||_{infty}M$ and $||T||=M$.
For compactness, let $B={uin (C[0,1],R):||u||_{infty}leq 1}$. We show $T(B)$ is equicontinuous family so that by Arzela Ascoli it is relatively compact.
begin{equation}
|Tu(x)-Tu(y)|=bigg|int_{alpha(x)}^{beta(x)} u(t)dt-int_{alpha(y)}^{beta(y)} u(t)dtbigg|
end{equation}
Please how do I subtract this integrals. I'm thinking to make the assumption $alpha(y)<beta(y)<alpha(x)<beta(x)$.
functional-analysis operator-theory compact-operators
functional-analysis operator-theory compact-operators
asked Jan 12 at 20:17
Muhammad MubarakMuhammad Mubarak
969
asked Jan 12 at 20:17
Muhammad MubarakMuhammad Mubarak
969
edited Jan 12 at 20:52
Muhammad Mubarak
asked Jan 12 at 20:17
Muhammad MubarakMuhammad Mubarak
969
asked Jan 12 at 20:17
Muhammad MubarakMuhammad Mubarak
969
asked Jan 12 at 20:17
Muhammad MubarakMuhammad Mubarak
969
969
$begingroup$
Is there any assumption on the regularity of $alpha$ and $beta$ at all?
$endgroup$
– BigbearZzz
Jan 12 at 20:24
$begingroup$
They are continuous. I will edit now.
$endgroup$
– Muhammad Mubarak
Jan 12 at 20:25
$begingroup$
My answer below can be rewritten without using the characteristic function $chi$ by arguing case by case. However, I prefer using $chi$ as it makes some idea clearer.
$endgroup$
– BigbearZzz
Jan 12 at 22:39
$begingroup$
Thanks a lot. Its clear now
$endgroup$
– Muhammad Mubarak
Jan 13 at 9:56
add a comment |
$begingroup$
Is there any assumption on the regularity of $alpha$ and $beta$ at all?
$endgroup$
– BigbearZzz
Jan 12 at 20:24
$begingroup$
They are continuous. I will edit now.
$endgroup$
– Muhammad Mubarak
Jan 12 at 20:25
$begingroup$
My answer below can be rewritten without using the characteristic function $chi$ by arguing case by case. However, I prefer using $chi$ as it makes some idea clearer.
$endgroup$
– BigbearZzz
Jan 12 at 22:39
$begingroup$
Thanks a lot. Its clear now
$endgroup$
– Muhammad Mubarak
Jan 13 at 9:56
$begingroup$
Is there any assumption on the regularity of $alpha$ and $beta$ at all?
$endgroup$
– BigbearZzz
Jan 12 at 20:24
$begingroup$
Is there any assumption on the regularity of $alpha$ and $beta$ at all?
$endgroup$
– BigbearZzz
Jan 12 at 20:24
$begingroup$
They are continuous. I will edit now.
$endgroup$
– Muhammad Mubarak
Jan 12 at 20:25
$begingroup$
They are continuous. I will edit now.
$endgroup$
– Muhammad Mubarak
Jan 12 at 20:25
$begingroup$
My answer below can be rewritten without using the characteristic function $chi$ by arguing case by case. However, I prefer using $chi$ as it makes some idea clearer.
$endgroup$
– BigbearZzz
Jan 12 at 22:39
$begingroup$
My answer below can be rewritten without using the characteristic function $chi$ by arguing case by case. However, I prefer using $chi$ as it makes some idea clearer.
$endgroup$
– BigbearZzz
Jan 12 at 22:39
$begingroup$
Thanks a lot. Its clear now
$endgroup$
– Muhammad Mubarak
Jan 13 at 9:56
$begingroup$
Thanks a lot. Its clear now
$endgroup$
– Muhammad Mubarak
Jan 13 at 9:56
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The integral from $a$ to $b$ can be viewed as $int_a^bf=int_{0}^1chi_{[a,b]}f$. Recall the relation
$$
|chi_A-chi_B| = chi_{A Delta B}
$$
where $ADelta B$ is the symmetric difference of $A$ and $B$. It is also easy to verify that
$$
[a,b] Delta [c,d] subset [a,c] cup[c,a]cup[b,d]cup[d,b]
$$
where $[x,y]=emptyset$ if $y<x$, so half of the terms of the right hand side would disappear, depending on the order of $a,b,c,d$. By symmetry of the right hand side, we may assume $ale c$ and $ble d$.
Using the above, we can compute that
$$begin{align}
left|int_a^bf-int_c^df right| &le int_0^1 |chi_{[a,b]} - chi_{[c,d]}| ,|f|\
&= int_0^1 chi_{[a,b] Delta [c,d]} |f|\
&leint_0^1 chi_{[a,c]} |f| + int_0^1 chi_{[b,d]}|f| \
&le (|c-a|+|d-b|)sup_{xin[0,1]} |f(x)|.
end{align}$$
Back to your question, we can deduce that
$$begin{align}
|Tu(x)-Tu(y)| &le big(|alpha(x)-alpha(y)|+ |beta(x)-beta(y)| big), ||u||_infty \
&le |alpha(x)-alpha(y)|+ |beta(x)-beta(y)|
end{align}$$
for any $uin B$. By continuity of $alpha$ and $beta$, for any given $varepsilon>0$ we may find $delta$ such that
$$
|x-y|<delta implies |alpha(x)-alpha(y)|+ |beta(x)-beta(y)|<varepsilon,
$$
which proves what you want.
answered Jan 12 at 21:14
BigbearZzzBigbearZzz
8,69121652
$endgroup$
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The integral from $a$ to $b$ can be viewed as $int_a^bf=int_{0}^1chi_{[a,b]}f$. Recall the relation
$$
|chi_A-chi_B| = chi_{A Delta B}
$$
where $ADelta B$ is the symmetric difference of $A$ and $B$. It is also easy to verify that
$$
[a,b] Delta [c,d] subset [a,c] cup[c,a]cup[b,d]cup[d,b]
$$
where $[x,y]=emptyset$ if $y<x$, so half of the terms of the right hand side would disappear, depending on the order of $a,b,c,d$. By symmetry of the right hand side, we may assume $ale c$ and $ble d$.
Using the above, we can compute that
$$begin{align}
left|int_a^bf-int_c^df right| &le int_0^1 |chi_{[a,b]} - chi_{[c,d]}| ,|f|\
&= int_0^1 chi_{[a,b] Delta [c,d]} |f|\
&leint_0^1 chi_{[a,c]} |f| + int_0^1 chi_{[b,d]}|f| \
&le (|c-a|+|d-b|)sup_{xin[0,1]} |f(x)|.
end{align}$$
Back to your question, we can deduce that
$$begin{align}
|Tu(x)-Tu(y)| &le big(|alpha(x)-alpha(y)|+ |beta(x)-beta(y)| big), ||u||_infty \
&le |alpha(x)-alpha(y)|+ |beta(x)-beta(y)|
end{align}$$
for any $uin B$. By continuity of $alpha$ and $beta$, for any given $varepsilon>0$ we may find $delta$ such that
$$
|x-y|<delta implies |alpha(x)-alpha(y)|+ |beta(x)-beta(y)|<varepsilon,
$$
which proves what you want.
answered Jan 12 at 21:14
BigbearZzzBigbearZzz
8,69121652
$endgroup$
add a comment |
$begingroup$
The integral from $a$ to $b$ can be viewed as $int_a^bf=int_{0}^1chi_{[a,b]}f$. Recall the relation
$$
|chi_A-chi_B| = chi_{A Delta B}
$$
where $ADelta B$ is the symmetric difference of $A$ and $B$. It is also easy to verify that
$$
[a,b] Delta [c,d] subset [a,c] cup[c,a]cup[b,d]cup[d,b]
$$
where $[x,y]=emptyset$ if $y<x$, so half of the terms of the right hand side would disappear, depending on the order of $a,b,c,d$. By symmetry of the right hand side, we may assume $ale c$ and $ble d$.
Using the above, we can compute that
$$begin{align}
left|int_a^bf-int_c^df right| &le int_0^1 |chi_{[a,b]} - chi_{[c,d]}| ,|f|\
&= int_0^1 chi_{[a,b] Delta [c,d]} |f|\
&leint_0^1 chi_{[a,c]} |f| + int_0^1 chi_{[b,d]}|f| \
&le (|c-a|+|d-b|)sup_{xin[0,1]} |f(x)|.
end{align}$$
Back to your question, we can deduce that
$$begin{align}
|Tu(x)-Tu(y)| &le big(|alpha(x)-alpha(y)|+ |beta(x)-beta(y)| big), ||u||_infty \
&le |alpha(x)-alpha(y)|+ |beta(x)-beta(y)|
end{align}$$
for any $uin B$. By continuity of $alpha$ and $beta$, for any given $varepsilon>0$ we may find $delta$ such that
$$
|x-y|<delta implies |alpha(x)-alpha(y)|+ |beta(x)-beta(y)|<varepsilon,
$$
which proves what you want.
answered Jan 12 at 21:14
BigbearZzzBigbearZzz
8,69121652
$endgroup$
add a comment |
$begingroup$
The integral from $a$ to $b$ can be viewed as $int_a^bf=int_{0}^1chi_{[a,b]}f$. Recall the relation
$$
|chi_A-chi_B| = chi_{A Delta B}
$$
where $ADelta B$ is the symmetric difference of $A$ and $B$. It is also easy to verify that
$$
[a,b] Delta [c,d] subset [a,c] cup[c,a]cup[b,d]cup[d,b]
$$
where $[x,y]=emptyset$ if $y<x$, so half of the terms of the right hand side would disappear, depending on the order of $a,b,c,d$. By symmetry of the right hand side, we may assume $ale c$ and $ble d$.
Using the above, we can compute that
$$begin{align}
left|int_a^bf-int_c^df right| &le int_0^1 |chi_{[a,b]} - chi_{[c,d]}| ,|f|\
&= int_0^1 chi_{[a,b] Delta [c,d]} |f|\
&leint_0^1 chi_{[a,c]} |f| + int_0^1 chi_{[b,d]}|f| \
&le (|c-a|+|d-b|)sup_{xin[0,1]} |f(x)|.
end{align}$$
Back to your question, we can deduce that
$$begin{align}
|Tu(x)-Tu(y)| &le big(|alpha(x)-alpha(y)|+ |beta(x)-beta(y)| big), ||u||_infty \
&le |alpha(x)-alpha(y)|+ |beta(x)-beta(y)|
end{align}$$
for any $uin B$. By continuity of $alpha$ and $beta$, for any given $varepsilon>0$ we may find $delta$ such that
$$
|x-y|<delta implies |alpha(x)-alpha(y)|+ |beta(x)-beta(y)|<varepsilon,
$$
which proves what you want.
answered Jan 12 at 21:14
BigbearZzzBigbearZzz
8,69121652
$endgroup$
The integral from $a$ to $b$ can be viewed as $int_a^bf=int_{0}^1chi_{[a,b]}f$. Recall the relation
$$
|chi_A-chi_B| = chi_{A Delta B}
$$
where $ADelta B$ is the symmetric difference of $A$ and $B$. It is also easy to verify that
$$
[a,b] Delta [c,d] subset [a,c] cup[c,a]cup[b,d]cup[d,b]
$$
where $[x,y]=emptyset$ if $y<x$, so half of the terms of the right hand side would disappear, depending on the order of $a,b,c,d$. By symmetry of the right hand side, we may assume $ale c$ and $ble d$.
Using the above, we can compute that
$$begin{align}
left|int_a^bf-int_c^df right| &le int_0^1 |chi_{[a,b]} - chi_{[c,d]}| ,|f|\
&= int_0^1 chi_{[a,b] Delta [c,d]} |f|\
&leint_0^1 chi_{[a,c]} |f| + int_0^1 chi_{[b,d]}|f| \
&le (|c-a|+|d-b|)sup_{xin[0,1]} |f(x)|.
end{align}$$
Back to your question, we can deduce that
$$begin{align}
|Tu(x)-Tu(y)| &le big(|alpha(x)-alpha(y)|+ |beta(x)-beta(y)| big), ||u||_infty \
&le |alpha(x)-alpha(y)|+ |beta(x)-beta(y)|
end{align}$$
for any $uin B$. By continuity of $alpha$ and $beta$, for any given $varepsilon>0$ we may find $delta$ such that
$$
|x-y|<delta implies |alpha(x)-alpha(y)|+ |beta(x)-beta(y)|<varepsilon,
$$
which proves what you want.
answered Jan 12 at 21:14
BigbearZzzBigbearZzz
8,69121652
edited Jan 12 at 23:39
answered Jan 12 at 21:14
BigbearZzzBigbearZzz
8,69121652
answered Jan 12 at 21:14
BigbearZzzBigbearZzz
8,69121652
answered Jan 12 at 21:14
BigbearZzzBigbearZzz
8,69121652
8,69121652
add a comment |
add a comment |
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$begingroup$
Is there any assumption on the regularity of $alpha$ and $beta$ at all?
$endgroup$
– BigbearZzz
Jan 12 at 20:24
$begingroup$
They are continuous. I will edit now.
$endgroup$
– Muhammad Mubarak
Jan 12 at 20:25
$begingroup$
My answer below can be rewritten without using the characteristic function $chi$ by arguing case by case. However, I prefer using $chi$ as it makes some idea clearer.
$endgroup$
– BigbearZzz
Jan 12 at 22:39
$begingroup$
Thanks a lot. Its clear now
$endgroup$
– Muhammad Mubarak
Jan 13 at 9:56