unbounded differential operator in Lp
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By semigroup theory the following are well known
Suppose that $A:X longrightarrow X $ is a bounded (or unbounded) operator on some Banach's space $X$, and given the Cauchy problem: $$frac{d}{dt}u(t)=Au(t), quad u(0)=u_0$$
the solution is $u(t)=e^{tA}u_0$ for any $u_0in X$ is unique. The set generated by ${e^{tA}:tgeq0}$ is bounded even if $A$ is unbounded.
My question
Consider an example:
Given $$partial_tu=partial_xu, quad u(0,x)=u_0(x)$$ and we let $A=frac{partial}{partial x}$ be the differential operator, on $X=L^p$ for $pgeq1$ then $A$ is unbounded.
How do you show that $A$ is unbounded? Thanks in advance
functional-analysis analysis pde
asked Nov 15 at 16:00
Kwame Atta
61
add a comment |
up vote
0
down vote
favorite
By semigroup theory the following are well known
Suppose that $A:X longrightarrow X $ is a bounded (or unbounded) operator on some Banach's space $X$, and given the Cauchy problem: $$frac{d}{dt}u(t)=Au(t), quad u(0)=u_0$$
the solution is $u(t)=e^{tA}u_0$ for any $u_0in X$ is unique. The set generated by ${e^{tA}:tgeq0}$ is bounded even if $A$ is unbounded.
My question
Consider an example:
Given $$partial_tu=partial_xu, quad u(0,x)=u_0(x)$$ and we let $A=frac{partial}{partial x}$ be the differential operator, on $X=L^p$ for $pgeq1$ then $A$ is unbounded.
How do you show that $A$ is unbounded? Thanks in advance
functional-analysis analysis pde
asked Nov 15 at 16:00
Kwame Atta
61
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
By semigroup theory the following are well known
Suppose that $A:X longrightarrow X $ is a bounded (or unbounded) operator on some Banach's space $X$, and given the Cauchy problem: $$frac{d}{dt}u(t)=Au(t), quad u(0)=u_0$$
the solution is $u(t)=e^{tA}u_0$ for any $u_0in X$ is unique. The set generated by ${e^{tA}:tgeq0}$ is bounded even if $A$ is unbounded.
My question
Consider an example:
Given $$partial_tu=partial_xu, quad u(0,x)=u_0(x)$$ and we let $A=frac{partial}{partial x}$ be the differential operator, on $X=L^p$ for $pgeq1$ then $A$ is unbounded.
How do you show that $A$ is unbounded? Thanks in advance
functional-analysis analysis pde
asked Nov 15 at 16:00
Kwame Atta
61
By semigroup theory the following are well known
Suppose that $A:X longrightarrow X $ is a bounded (or unbounded) operator on some Banach's space $X$, and given the Cauchy problem: $$frac{d}{dt}u(t)=Au(t), quad u(0)=u_0$$
the solution is $u(t)=e^{tA}u_0$ for any $u_0in X$ is unique. The set generated by ${e^{tA}:tgeq0}$ is bounded even if $A$ is unbounded.
My question
Consider an example:
Given $$partial_tu=partial_xu, quad u(0,x)=u_0(x)$$ and we let $A=frac{partial}{partial x}$ be the differential operator, on $X=L^p$ for $pgeq1$ then $A$ is unbounded.
How do you show that $A$ is unbounded? Thanks in advance
functional-analysis analysis pde
functional-analysis analysis pde
asked Nov 15 at 16:00
Kwame Atta
61
asked Nov 15 at 16:00
Kwame Atta
61
asked Nov 15 at 16:00
Kwame Atta
61
asked Nov 15 at 16:00
Kwame Atta
61
asked Nov 15 at 16:00
Kwame Atta
61
61
add a comment |
add a comment |
1 Answer
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One way is to take a sequence ${f_k}$ of functions such that $|f_k|$ is some fixed constant but $|f_k'|$ becomes large as $ntoinfty$. You may take $f_k(x)=sqrt ke^{-k^2x^2}$. This is a standard example. Verify that $f_k(x)in L^2(mathbb{R})$ but $|f_k'(x)|_{L^2}^2toinfty$ as $ntoinfty$.
answered Nov 15 at 16:21
Hirak
2,54911336
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
One way is to take a sequence ${f_k}$ of functions such that $|f_k|$ is some fixed constant but $|f_k'|$ becomes large as $ntoinfty$. You may take $f_k(x)=sqrt ke^{-k^2x^2}$. This is a standard example. Verify that $f_k(x)in L^2(mathbb{R})$ but $|f_k'(x)|_{L^2}^2toinfty$ as $ntoinfty$.
answered Nov 15 at 16:21
Hirak
2,54911336
add a comment |
up vote
1
down vote
One way is to take a sequence ${f_k}$ of functions such that $|f_k|$ is some fixed constant but $|f_k'|$ becomes large as $ntoinfty$. You may take $f_k(x)=sqrt ke^{-k^2x^2}$. This is a standard example. Verify that $f_k(x)in L^2(mathbb{R})$ but $|f_k'(x)|_{L^2}^2toinfty$ as $ntoinfty$.
answered Nov 15 at 16:21
Hirak
2,54911336
add a comment |
up vote
1
down vote
up vote
1
down vote
One way is to take a sequence ${f_k}$ of functions such that $|f_k|$ is some fixed constant but $|f_k'|$ becomes large as $ntoinfty$. You may take $f_k(x)=sqrt ke^{-k^2x^2}$. This is a standard example. Verify that $f_k(x)in L^2(mathbb{R})$ but $|f_k'(x)|_{L^2}^2toinfty$ as $ntoinfty$.
answered Nov 15 at 16:21
Hirak
2,54911336
One way is to take a sequence ${f_k}$ of functions such that $|f_k|$ is some fixed constant but $|f_k'|$ becomes large as $ntoinfty$. You may take $f_k(x)=sqrt ke^{-k^2x^2}$. This is a standard example. Verify that $f_k(x)in L^2(mathbb{R})$ but $|f_k'(x)|_{L^2}^2toinfty$ as $ntoinfty$.
answered Nov 15 at 16:21
Hirak
2,54911336
answered Nov 15 at 16:21
Hirak
2,54911336
answered Nov 15 at 16:21
Hirak
2,54911336
answered Nov 15 at 16:21
Hirak
2,54911336
2,54911336
add a comment |
add a comment |
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