Mixing
$frac{d}{dx}:C^1([0,1])to C([0,1])$ as a closed, unbounded operator
$begingroup$
First recall that a (potentially unbounded) operator $T:D(T)subseteq Xto Y$ is closed whenever $(x_n)_{n=1}^inftysubset D(T)$ convergent to $xin X_0$ and $(Tx_n)_{n=1}^infty$ convergent in $X_1$ together imply that $xin D(T)$ and $Tx=lim_{ntoinfty}Tx_n$.
On the Wikipedia page for unbounded operators, differentiation $frac{d}{dx}:C^1([0,1])to C([0,1])$ is given as an example of a closed, unbounded operator. In this example, $X,Y$ are both taken to be $C([a,b])$ and $D(T)=C^1([a,b])$. As our unbounded operator we have $T=frac{d}{dx}$.
Whilst I have no trouble seeing that in this example $T$ is unbounded, I am not so convinced that $T$ is closed. In particular, I know that $C^1([0,1])$ is dense in $C([0,1])$, but not closed; so it isn't so obvious that a convergent sequence $f_nin C^1([0,1])$ should have limit $fin C^1([0,1])$. How, exactly, should one proceed in trying to show this?
functional-analysis analysis operator-theory relations unbounded-operators
asked Jan 21 at 10:37
Jeremy Jeffrey JamesJeremy Jeffrey James
999715
$endgroup$
add a comment |
$begingroup$
First recall that a (potentially unbounded) operator $T:D(T)subseteq Xto Y$ is closed whenever $(x_n)_{n=1}^inftysubset D(T)$ convergent to $xin X_0$ and $(Tx_n)_{n=1}^infty$ convergent in $X_1$ together imply that $xin D(T)$ and $Tx=lim_{ntoinfty}Tx_n$.
On the Wikipedia page for unbounded operators, differentiation $frac{d}{dx}:C^1([0,1])to C([0,1])$ is given as an example of a closed, unbounded operator. In this example, $X,Y$ are both taken to be $C([a,b])$ and $D(T)=C^1([a,b])$. As our unbounded operator we have $T=frac{d}{dx}$.
Whilst I have no trouble seeing that in this example $T$ is unbounded, I am not so convinced that $T$ is closed. In particular, I know that $C^1([0,1])$ is dense in $C([0,1])$, but not closed; so it isn't so obvious that a convergent sequence $f_nin C^1([0,1])$ should have limit $fin C^1([0,1])$. How, exactly, should one proceed in trying to show this?
functional-analysis analysis operator-theory relations unbounded-operators
asked Jan 21 at 10:37
Jeremy Jeffrey JamesJeremy Jeffrey James
999715
$endgroup$
1
$begingroup$
I added a source in the Wikipedia page which gives the details. It is Example 4.13-4 .
$endgroup$
– Pedro
Jan 21 at 10:54
$begingroup$
Just to check, that's the Kreyszig reference?
$endgroup$
– Jeremy Jeffrey James
Jan 21 at 10:56
$begingroup$
Yes, it is Kreyszig.
$endgroup$
– Pedro
Jan 21 at 10:59
$begingroup$
Thanks for that, Pedro, I'll be sure to check it out.
$endgroup$
– Jeremy Jeffrey James
Jan 21 at 10:59
add a comment |
$begingroup$
First recall that a (potentially unbounded) operator $T:D(T)subseteq Xto Y$ is closed whenever $(x_n)_{n=1}^inftysubset D(T)$ convergent to $xin X_0$ and $(Tx_n)_{n=1}^infty$ convergent in $X_1$ together imply that $xin D(T)$ and $Tx=lim_{ntoinfty}Tx_n$.
On the Wikipedia page for unbounded operators, differentiation $frac{d}{dx}:C^1([0,1])to C([0,1])$ is given as an example of a closed, unbounded operator. In this example, $X,Y$ are both taken to be $C([a,b])$ and $D(T)=C^1([a,b])$. As our unbounded operator we have $T=frac{d}{dx}$.
Whilst I have no trouble seeing that in this example $T$ is unbounded, I am not so convinced that $T$ is closed. In particular, I know that $C^1([0,1])$ is dense in $C([0,1])$, but not closed; so it isn't so obvious that a convergent sequence $f_nin C^1([0,1])$ should have limit $fin C^1([0,1])$. How, exactly, should one proceed in trying to show this?
functional-analysis analysis operator-theory relations unbounded-operators
asked Jan 21 at 10:37
Jeremy Jeffrey JamesJeremy Jeffrey James
999715
$endgroup$
First recall that a (potentially unbounded) operator $T:D(T)subseteq Xto Y$ is closed whenever $(x_n)_{n=1}^inftysubset D(T)$ convergent to $xin X_0$ and $(Tx_n)_{n=1}^infty$ convergent in $X_1$ together imply that $xin D(T)$ and $Tx=lim_{ntoinfty}Tx_n$.
On the Wikipedia page for unbounded operators, differentiation $frac{d}{dx}:C^1([0,1])to C([0,1])$ is given as an example of a closed, unbounded operator. In this example, $X,Y$ are both taken to be $C([a,b])$ and $D(T)=C^1([a,b])$. As our unbounded operator we have $T=frac{d}{dx}$.
Whilst I have no trouble seeing that in this example $T$ is unbounded, I am not so convinced that $T$ is closed. In particular, I know that $C^1([0,1])$ is dense in $C([0,1])$, but not closed; so it isn't so obvious that a convergent sequence $f_nin C^1([0,1])$ should have limit $fin C^1([0,1])$. How, exactly, should one proceed in trying to show this?
functional-analysis analysis operator-theory relations unbounded-operators
functional-analysis analysis operator-theory relations unbounded-operators
asked Jan 21 at 10:37
Jeremy Jeffrey JamesJeremy Jeffrey James
999715
asked Jan 21 at 10:37
Jeremy Jeffrey JamesJeremy Jeffrey James
999715
asked Jan 21 at 10:37
Jeremy Jeffrey JamesJeremy Jeffrey James
999715
asked Jan 21 at 10:37
Jeremy Jeffrey JamesJeremy Jeffrey James
999715
asked Jan 21 at 10:37
Jeremy Jeffrey JamesJeremy Jeffrey James
999715
999715
1
$begingroup$
I added a source in the Wikipedia page which gives the details. It is Example 4.13-4 .
$endgroup$
– Pedro
Jan 21 at 10:54
$begingroup$
Just to check, that's the Kreyszig reference?
$endgroup$
– Jeremy Jeffrey James
Jan 21 at 10:56
$begingroup$
Yes, it is Kreyszig.
$endgroup$
– Pedro
Jan 21 at 10:59
$begingroup$
Thanks for that, Pedro, I'll be sure to check it out.
$endgroup$
– Jeremy Jeffrey James
Jan 21 at 10:59
add a comment |
1
$begingroup$
I added a source in the Wikipedia page which gives the details. It is Example 4.13-4 .
$endgroup$
– Pedro
Jan 21 at 10:54
$begingroup$
Just to check, that's the Kreyszig reference?
$endgroup$
– Jeremy Jeffrey James
Jan 21 at 10:56
$begingroup$
Yes, it is Kreyszig.
$endgroup$
– Pedro
Jan 21 at 10:59
$begingroup$
Thanks for that, Pedro, I'll be sure to check it out.
$endgroup$
– Jeremy Jeffrey James
Jan 21 at 10:59
1
1
$begingroup$
I added a source in the Wikipedia page which gives the details. It is Example 4.13-4 .
$endgroup$
– Pedro
Jan 21 at 10:54
$begingroup$
I added a source in the Wikipedia page which gives the details. It is Example 4.13-4 .
$endgroup$
– Pedro
Jan 21 at 10:54
$begingroup$
Just to check, that's the Kreyszig reference?
$endgroup$
– Jeremy Jeffrey James
Jan 21 at 10:56
$begingroup$
Just to check, that's the Kreyszig reference?
$endgroup$
– Jeremy Jeffrey James
Jan 21 at 10:56
$begingroup$
Yes, it is Kreyszig.
$endgroup$
– Pedro
Jan 21 at 10:59
$begingroup$
Yes, it is Kreyszig.
$endgroup$
– Pedro
Jan 21 at 10:59
$begingroup$
Thanks for that, Pedro, I'll be sure to check it out.
$endgroup$
– Jeremy Jeffrey James
Jan 21 at 10:59
$begingroup$
Thanks for that, Pedro, I'll be sure to check it out.
$endgroup$
– Jeremy Jeffrey James
Jan 21 at 10:59
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
To check that $frac{d}{dx}$ is closed, you should check that if $f_n in C^1([0,1])$ and $f_n to f$, $f_n' to g$ in $C([0,1])$, then $f in C^1([0,1])$ and $f' = g$.
By the fundamental theorem of calculus, we can write $f_n(x) = f_n(0) + int_0^x f_n'(s) ds$. In particular, by sending $n to infty$,
$$f(x) = f(0) + int_0^x g(s) ds$$
so $f in C^1([0,1])$ and $f' = g$.
answered Jan 21 at 10:42
Rhys SteeleRhys Steele
7,0251829
$endgroup$
$begingroup$
I knew that it had to involve the integral somehow - this clearly identifies how; a silly oversight on my part.
$endgroup$
– Jeremy Jeffrey James
Jan 21 at 10:45
add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
To check that $frac{d}{dx}$ is closed, you should check that if $f_n in C^1([0,1])$ and $f_n to f$, $f_n' to g$ in $C([0,1])$, then $f in C^1([0,1])$ and $f' = g$.
By the fundamental theorem of calculus, we can write $f_n(x) = f_n(0) + int_0^x f_n'(s) ds$. In particular, by sending $n to infty$,
$$f(x) = f(0) + int_0^x g(s) ds$$
so $f in C^1([0,1])$ and $f' = g$.
answered Jan 21 at 10:42
Rhys SteeleRhys Steele
7,0251829
$endgroup$
$begingroup$
I knew that it had to involve the integral somehow - this clearly identifies how; a silly oversight on my part.
$endgroup$
– Jeremy Jeffrey James
Jan 21 at 10:45
add a comment |
$begingroup$
To check that $frac{d}{dx}$ is closed, you should check that if $f_n in C^1([0,1])$ and $f_n to f$, $f_n' to g$ in $C([0,1])$, then $f in C^1([0,1])$ and $f' = g$.
By the fundamental theorem of calculus, we can write $f_n(x) = f_n(0) + int_0^x f_n'(s) ds$. In particular, by sending $n to infty$,
$$f(x) = f(0) + int_0^x g(s) ds$$
so $f in C^1([0,1])$ and $f' = g$.
answered Jan 21 at 10:42
Rhys SteeleRhys Steele
7,0251829
$endgroup$
$begingroup$
I knew that it had to involve the integral somehow - this clearly identifies how; a silly oversight on my part.
$endgroup$
– Jeremy Jeffrey James
Jan 21 at 10:45
add a comment |
$begingroup$
To check that $frac{d}{dx}$ is closed, you should check that if $f_n in C^1([0,1])$ and $f_n to f$, $f_n' to g$ in $C([0,1])$, then $f in C^1([0,1])$ and $f' = g$.
By the fundamental theorem of calculus, we can write $f_n(x) = f_n(0) + int_0^x f_n'(s) ds$. In particular, by sending $n to infty$,
$$f(x) = f(0) + int_0^x g(s) ds$$
so $f in C^1([0,1])$ and $f' = g$.
answered Jan 21 at 10:42
Rhys SteeleRhys Steele
7,0251829
$endgroup$
To check that $frac{d}{dx}$ is closed, you should check that if $f_n in C^1([0,1])$ and $f_n to f$, $f_n' to g$ in $C([0,1])$, then $f in C^1([0,1])$ and $f' = g$.
By the fundamental theorem of calculus, we can write $f_n(x) = f_n(0) + int_0^x f_n'(s) ds$. In particular, by sending $n to infty$,
$$f(x) = f(0) + int_0^x g(s) ds$$
so $f in C^1([0,1])$ and $f' = g$.
answered Jan 21 at 10:42
Rhys SteeleRhys Steele
7,0251829
answered Jan 21 at 10:42
Rhys SteeleRhys Steele
7,0251829
answered Jan 21 at 10:42
Rhys SteeleRhys Steele
7,0251829
answered Jan 21 at 10:42
Rhys SteeleRhys Steele
7,0251829
7,0251829
$begingroup$
I knew that it had to involve the integral somehow - this clearly identifies how; a silly oversight on my part.
$endgroup$
– Jeremy Jeffrey James
Jan 21 at 10:45
add a comment |
$begingroup$
I knew that it had to involve the integral somehow - this clearly identifies how; a silly oversight on my part.
$endgroup$
– Jeremy Jeffrey James
Jan 21 at 10:45
$begingroup$
I knew that it had to involve the integral somehow - this clearly identifies how; a silly oversight on my part.
$endgroup$
– Jeremy Jeffrey James
Jan 21 at 10:45
$begingroup$
I knew that it had to involve the integral somehow - this clearly identifies how; a silly oversight on my part.
$endgroup$
– Jeremy Jeffrey James
Jan 21 at 10:45
add a comment |
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$begingroup$
I added a source in the Wikipedia page which gives the details. It is Example 4.13-4 .
$endgroup$
– Pedro
Jan 21 at 10:54
$begingroup$
Just to check, that's the Kreyszig reference?
$endgroup$
– Jeremy Jeffrey James
Jan 21 at 10:56
$begingroup$
Yes, it is Kreyszig.
$endgroup$
– Pedro
Jan 21 at 10:59
$begingroup$
Thanks for that, Pedro, I'll be sure to check it out.
$endgroup$
– Jeremy Jeffrey James
Jan 21 at 10:59