Check if linear operator is bounded
$begingroup$
I have an operator $A: C_{L_1[0,1]} rightarrow C_{L_2[0,1]}, Ax(t)=t^3int_0^1x(tau)dtau $.
Notation can be bizzare, so
$ C_{L_1[0,1]} $ is a set of functions which metric is $ rho(f,g)= int_a^b |f(x)-g(x)|dx $
and
$ C_{L_2[0,1]} $ is the same, but it's metric is $ rho(f,g)=(int_a^b(f(x)-g(x))^2 dx)^{1/2} $.
As the topic says, I want to prove that it's bounded, i.e.
$ exists c: ||Ax||_{C_{L_2[0,1]}} leqslant c||x||_{C_{L_1[0,1]}} $.
Ok, here's the scariest part. H o w. I guess I missed something really important or just having trouble to do the first step. It seems ok for $ C_{L_1[0,1]} $ since it a somewhat classic integral transform (is it?) and I honestly looked up in a book about this case, but what to do in my case is, er, an open question.
functional-analysis
asked Dec 13 '18 at 17:33
MajileeMajilee
82
$endgroup$
add a comment |
$begingroup$
I have an operator $A: C_{L_1[0,1]} rightarrow C_{L_2[0,1]}, Ax(t)=t^3int_0^1x(tau)dtau $.
Notation can be bizzare, so
$ C_{L_1[0,1]} $ is a set of functions which metric is $ rho(f,g)= int_a^b |f(x)-g(x)|dx $
and
$ C_{L_2[0,1]} $ is the same, but it's metric is $ rho(f,g)=(int_a^b(f(x)-g(x))^2 dx)^{1/2} $.
As the topic says, I want to prove that it's bounded, i.e.
$ exists c: ||Ax||_{C_{L_2[0,1]}} leqslant c||x||_{C_{L_1[0,1]}} $.
Ok, here's the scariest part. H o w. I guess I missed something really important or just having trouble to do the first step. It seems ok for $ C_{L_1[0,1]} $ since it a somewhat classic integral transform (is it?) and I honestly looked up in a book about this case, but what to do in my case is, er, an open question.
functional-analysis
asked Dec 13 '18 at 17:33
MajileeMajilee
82
$endgroup$
add a comment |
$begingroup$
I have an operator $A: C_{L_1[0,1]} rightarrow C_{L_2[0,1]}, Ax(t)=t^3int_0^1x(tau)dtau $.
Notation can be bizzare, so
$ C_{L_1[0,1]} $ is a set of functions which metric is $ rho(f,g)= int_a^b |f(x)-g(x)|dx $
and
$ C_{L_2[0,1]} $ is the same, but it's metric is $ rho(f,g)=(int_a^b(f(x)-g(x))^2 dx)^{1/2} $.
As the topic says, I want to prove that it's bounded, i.e.
$ exists c: ||Ax||_{C_{L_2[0,1]}} leqslant c||x||_{C_{L_1[0,1]}} $.
Ok, here's the scariest part. H o w. I guess I missed something really important or just having trouble to do the first step. It seems ok for $ C_{L_1[0,1]} $ since it a somewhat classic integral transform (is it?) and I honestly looked up in a book about this case, but what to do in my case is, er, an open question.
functional-analysis
asked Dec 13 '18 at 17:33
MajileeMajilee
82
$endgroup$
I have an operator $A: C_{L_1[0,1]} rightarrow C_{L_2[0,1]}, Ax(t)=t^3int_0^1x(tau)dtau $.
Notation can be bizzare, so
$ C_{L_1[0,1]} $ is a set of functions which metric is $ rho(f,g)= int_a^b |f(x)-g(x)|dx $
and
$ C_{L_2[0,1]} $ is the same, but it's metric is $ rho(f,g)=(int_a^b(f(x)-g(x))^2 dx)^{1/2} $.
As the topic says, I want to prove that it's bounded, i.e.
$ exists c: ||Ax||_{C_{L_2[0,1]}} leqslant c||x||_{C_{L_1[0,1]}} $.
Ok, here's the scariest part. H o w. I guess I missed something really important or just having trouble to do the first step. It seems ok for $ C_{L_1[0,1]} $ since it a somewhat classic integral transform (is it?) and I honestly looked up in a book about this case, but what to do in my case is, er, an open question.
functional-analysis
functional-analysis
asked Dec 13 '18 at 17:33
MajileeMajilee
82
asked Dec 13 '18 at 17:33
MajileeMajilee
82
asked Dec 13 '18 at 17:33
MajileeMajilee
82
asked Dec 13 '18 at 17:33
MajileeMajilee
82
asked Dec 13 '18 at 17:33
MajileeMajilee
82
82
add a comment |
add a comment |
1 Answer
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The proof is quite simple, once you start to write down the definitions step by step. What you want to prove is that $| A x |_{L^2[0,1]} leq C | x |_{L^1[0,1]}$ for some positive constant $C >0$. Note that I am using the more common notation $L^2$ and $L^1$, rather than $C_{L^2[0,1]}$ as you wrote. So let's prove this:
begin{equation}
| A x |_{L^2[0,1]}^2 = int_0^1 (Ax)(t)^2 : dt = left( int_0^1 x(t) : dt right)^2 int_0^1 t^6 : dt
end{equation}
Now taking square roots on both sides, we see
begin{equation}
| A x |_{L^2[0,1]} = sqrt{int_0^1 t^6 : dt} ; left| int_0^1 x(t) : dt right| leq C int_0^1 |x(t)| : dt = C |x |_{L^1[0,1]}
end{equation}
if we set $C=sqrt{int_0^1 t^6 : dt} = sqrt{frac{1}{7}}$ and by pulling the absolute value inside the integral. This shows what we wanted to show!
answered Dec 13 '18 at 17:48
MaxMax
1287
$endgroup$
$begingroup$
And that's when I feel very stupid. Thanks a lot, now this seems obvious (as do most of the problems with a trick but anyway).
$endgroup$
– Majilee
Dec 13 '18 at 18:19
$begingroup$
No need to feel stupid ;-) You're welcome.
$endgroup$
– Max
Dec 14 '18 at 7:18
add a comment |
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1 Answer
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$begingroup$
The proof is quite simple, once you start to write down the definitions step by step. What you want to prove is that $| A x |_{L^2[0,1]} leq C | x |_{L^1[0,1]}$ for some positive constant $C >0$. Note that I am using the more common notation $L^2$ and $L^1$, rather than $C_{L^2[0,1]}$ as you wrote. So let's prove this:
begin{equation}
| A x |_{L^2[0,1]}^2 = int_0^1 (Ax)(t)^2 : dt = left( int_0^1 x(t) : dt right)^2 int_0^1 t^6 : dt
end{equation}
Now taking square roots on both sides, we see
begin{equation}
| A x |_{L^2[0,1]} = sqrt{int_0^1 t^6 : dt} ; left| int_0^1 x(t) : dt right| leq C int_0^1 |x(t)| : dt = C |x |_{L^1[0,1]}
end{equation}
if we set $C=sqrt{int_0^1 t^6 : dt} = sqrt{frac{1}{7}}$ and by pulling the absolute value inside the integral. This shows what we wanted to show!
answered Dec 13 '18 at 17:48
MaxMax
1287
$endgroup$
$begingroup$
And that's when I feel very stupid. Thanks a lot, now this seems obvious (as do most of the problems with a trick but anyway).
$endgroup$
– Majilee
Dec 13 '18 at 18:19
$begingroup$
No need to feel stupid ;-) You're welcome.
$endgroup$
– Max
Dec 14 '18 at 7:18
add a comment |
$begingroup$
The proof is quite simple, once you start to write down the definitions step by step. What you want to prove is that $| A x |_{L^2[0,1]} leq C | x |_{L^1[0,1]}$ for some positive constant $C >0$. Note that I am using the more common notation $L^2$ and $L^1$, rather than $C_{L^2[0,1]}$ as you wrote. So let's prove this:
begin{equation}
| A x |_{L^2[0,1]}^2 = int_0^1 (Ax)(t)^2 : dt = left( int_0^1 x(t) : dt right)^2 int_0^1 t^6 : dt
end{equation}
Now taking square roots on both sides, we see
begin{equation}
| A x |_{L^2[0,1]} = sqrt{int_0^1 t^6 : dt} ; left| int_0^1 x(t) : dt right| leq C int_0^1 |x(t)| : dt = C |x |_{L^1[0,1]}
end{equation}
if we set $C=sqrt{int_0^1 t^6 : dt} = sqrt{frac{1}{7}}$ and by pulling the absolute value inside the integral. This shows what we wanted to show!
answered Dec 13 '18 at 17:48
MaxMax
1287
$endgroup$
$begingroup$
And that's when I feel very stupid. Thanks a lot, now this seems obvious (as do most of the problems with a trick but anyway).
$endgroup$
– Majilee
Dec 13 '18 at 18:19
$begingroup$
No need to feel stupid ;-) You're welcome.
$endgroup$
– Max
Dec 14 '18 at 7:18
add a comment |
$begingroup$
The proof is quite simple, once you start to write down the definitions step by step. What you want to prove is that $| A x |_{L^2[0,1]} leq C | x |_{L^1[0,1]}$ for some positive constant $C >0$. Note that I am using the more common notation $L^2$ and $L^1$, rather than $C_{L^2[0,1]}$ as you wrote. So let's prove this:
begin{equation}
| A x |_{L^2[0,1]}^2 = int_0^1 (Ax)(t)^2 : dt = left( int_0^1 x(t) : dt right)^2 int_0^1 t^6 : dt
end{equation}
Now taking square roots on both sides, we see
begin{equation}
| A x |_{L^2[0,1]} = sqrt{int_0^1 t^6 : dt} ; left| int_0^1 x(t) : dt right| leq C int_0^1 |x(t)| : dt = C |x |_{L^1[0,1]}
end{equation}
if we set $C=sqrt{int_0^1 t^6 : dt} = sqrt{frac{1}{7}}$ and by pulling the absolute value inside the integral. This shows what we wanted to show!
answered Dec 13 '18 at 17:48
MaxMax
1287
$endgroup$
The proof is quite simple, once you start to write down the definitions step by step. What you want to prove is that $| A x |_{L^2[0,1]} leq C | x |_{L^1[0,1]}$ for some positive constant $C >0$. Note that I am using the more common notation $L^2$ and $L^1$, rather than $C_{L^2[0,1]}$ as you wrote. So let's prove this:
begin{equation}
| A x |_{L^2[0,1]}^2 = int_0^1 (Ax)(t)^2 : dt = left( int_0^1 x(t) : dt right)^2 int_0^1 t^6 : dt
end{equation}
Now taking square roots on both sides, we see
begin{equation}
| A x |_{L^2[0,1]} = sqrt{int_0^1 t^6 : dt} ; left| int_0^1 x(t) : dt right| leq C int_0^1 |x(t)| : dt = C |x |_{L^1[0,1]}
end{equation}
if we set $C=sqrt{int_0^1 t^6 : dt} = sqrt{frac{1}{7}}$ and by pulling the absolute value inside the integral. This shows what we wanted to show!
answered Dec 13 '18 at 17:48
MaxMax
1287
answered Dec 13 '18 at 17:48
MaxMax
1287
answered Dec 13 '18 at 17:48
MaxMax
1287
answered Dec 13 '18 at 17:48
MaxMax
1287
1287
$begingroup$
And that's when I feel very stupid. Thanks a lot, now this seems obvious (as do most of the problems with a trick but anyway).
$endgroup$
– Majilee
Dec 13 '18 at 18:19
$begingroup$
No need to feel stupid ;-) You're welcome.
$endgroup$
– Max
Dec 14 '18 at 7:18
add a comment |
$begingroup$
And that's when I feel very stupid. Thanks a lot, now this seems obvious (as do most of the problems with a trick but anyway).
$endgroup$
– Majilee
Dec 13 '18 at 18:19
$begingroup$
No need to feel stupid ;-) You're welcome.
$endgroup$
– Max
Dec 14 '18 at 7:18
$begingroup$
And that's when I feel very stupid. Thanks a lot, now this seems obvious (as do most of the problems with a trick but anyway).
$endgroup$
– Majilee
Dec 13 '18 at 18:19
$begingroup$
And that's when I feel very stupid. Thanks a lot, now this seems obvious (as do most of the problems with a trick but anyway).
$endgroup$
– Majilee
Dec 13 '18 at 18:19
$begingroup$
No need to feel stupid ;-) You're welcome.
$endgroup$
– Max
Dec 14 '18 at 7:18
$begingroup$
No need to feel stupid ;-) You're welcome.
$endgroup$
– Max
Dec 14 '18 at 7:18
add a comment |
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