Closed operator in sobolev spaces
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So suppose we look at the operator $S:H_0^1(0,1) rightarrow L^2(0,1), umapsto u' $,
where $H_0^1(0,1)$ denotes the closure of the infinitely differentiable functions compactly supported in $(0,1)$ in $H^1(0,1)$ and $u'$ the weak derivative of $u$. Is this operator closed? I remember that one can show that the operator $A:C^1[0,1] rightarrow C[0,1] $ is closed by using a theorem for a sequence of differentiable functions, where under certain conditions the limit of the sequence was differentiable too.
Is there a similiar theorem for functions in $H_0^1(0,1)$?
functional-analysis analysis sobolev-spaces
asked Jan 23 at 16:34
KatakuriKatakuri
286
$endgroup$
add a comment |
$begingroup$
So suppose we look at the operator $S:H_0^1(0,1) rightarrow L^2(0,1), umapsto u' $,
where $H_0^1(0,1)$ denotes the closure of the infinitely differentiable functions compactly supported in $(0,1)$ in $H^1(0,1)$ and $u'$ the weak derivative of $u$. Is this operator closed? I remember that one can show that the operator $A:C^1[0,1] rightarrow C[0,1] $ is closed by using a theorem for a sequence of differentiable functions, where under certain conditions the limit of the sequence was differentiable too.
Is there a similiar theorem for functions in $H_0^1(0,1)$?
functional-analysis analysis sobolev-spaces
asked Jan 23 at 16:34
KatakuriKatakuri
286
$endgroup$
1
$begingroup$
Well your operator $S$ in this case is even bounded if I'm not mistaken.
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– The Brainlet Exterminator
Jan 23 at 17:36
$begingroup$
The operator $S$ is also closed if consider as operator from $D(S)subset L^2$ to $L^2$. The proof follows directly from the definition of closedness, weak derivatives.
$endgroup$
– daw
Jan 24 at 7:30
add a comment |
$begingroup$
So suppose we look at the operator $S:H_0^1(0,1) rightarrow L^2(0,1), umapsto u' $,
where $H_0^1(0,1)$ denotes the closure of the infinitely differentiable functions compactly supported in $(0,1)$ in $H^1(0,1)$ and $u'$ the weak derivative of $u$. Is this operator closed? I remember that one can show that the operator $A:C^1[0,1] rightarrow C[0,1] $ is closed by using a theorem for a sequence of differentiable functions, where under certain conditions the limit of the sequence was differentiable too.
Is there a similiar theorem for functions in $H_0^1(0,1)$?
functional-analysis analysis sobolev-spaces
asked Jan 23 at 16:34
KatakuriKatakuri
286
$endgroup$
So suppose we look at the operator $S:H_0^1(0,1) rightarrow L^2(0,1), umapsto u' $,
where $H_0^1(0,1)$ denotes the closure of the infinitely differentiable functions compactly supported in $(0,1)$ in $H^1(0,1)$ and $u'$ the weak derivative of $u$. Is this operator closed? I remember that one can show that the operator $A:C^1[0,1] rightarrow C[0,1] $ is closed by using a theorem for a sequence of differentiable functions, where under certain conditions the limit of the sequence was differentiable too.
Is there a similiar theorem for functions in $H_0^1(0,1)$?
functional-analysis analysis sobolev-spaces
functional-analysis analysis sobolev-spaces
asked Jan 23 at 16:34
KatakuriKatakuri
286
asked Jan 23 at 16:34
KatakuriKatakuri
286
asked Jan 23 at 16:34
KatakuriKatakuri
286
asked Jan 23 at 16:34
KatakuriKatakuri
286
asked Jan 23 at 16:34
KatakuriKatakuri
286
286
1
$begingroup$
Well your operator $S$ in this case is even bounded if I'm not mistaken.
$endgroup$
– The Brainlet Exterminator
Jan 23 at 17:36
$begingroup$
The operator $S$ is also closed if consider as operator from $D(S)subset L^2$ to $L^2$. The proof follows directly from the definition of closedness, weak derivatives.
$endgroup$
– daw
Jan 24 at 7:30
add a comment |
1
$begingroup$
Well your operator $S$ in this case is even bounded if I'm not mistaken.
$endgroup$
– The Brainlet Exterminator
Jan 23 at 17:36
$begingroup$
The operator $S$ is also closed if consider as operator from $D(S)subset L^2$ to $L^2$. The proof follows directly from the definition of closedness, weak derivatives.
$endgroup$
– daw
Jan 24 at 7:30
1
1
$begingroup$
Well your operator $S$ in this case is even bounded if I'm not mistaken.
$endgroup$
– The Brainlet Exterminator
Jan 23 at 17:36
$begingroup$
Well your operator $S$ in this case is even bounded if I'm not mistaken.
$endgroup$
– The Brainlet Exterminator
Jan 23 at 17:36
$begingroup$
The operator $S$ is also closed if consider as operator from $D(S)subset L^2$ to $L^2$. The proof follows directly from the definition of closedness, weak derivatives.
$endgroup$
– daw
Jan 24 at 7:30
$begingroup$
The operator $S$ is also closed if consider as operator from $D(S)subset L^2$ to $L^2$. The proof follows directly from the definition of closedness, weak derivatives.
$endgroup$
– daw
Jan 24 at 7:30
add a comment |
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$begingroup$
Well your operator $S$ in this case is even bounded if I'm not mistaken.
$endgroup$
– The Brainlet Exterminator
Jan 23 at 17:36
$begingroup$
The operator $S$ is also closed if consider as operator from $D(S)subset L^2$ to $L^2$. The proof follows directly from the definition of closedness, weak derivatives.
$endgroup$
– daw
Jan 24 at 7:30