Norm of Volterra Operator with $L^1$
I am trying to find the Operator norm for the following operator
$$(Vf)(t):(C[0,1],L^1)to (C[0,1],L^1)$$
$$fmapsto int_0^t f(s)ds$$
What I have done
$$|Vf(t)|=sup_{|f|_{L^1}=1}{|int_0^t f(s) |ds}_{L^1}=sup_{|{f}|_{L^1}=1}int_0^1|{int_0^t f(s)| ~ds}dtleq sup_{|{f}|_{L^1}=1}int_0^1int_0^t |{f(s)}|~dsdt .$$
Tonelli:
$$=sup_{|{f}|_{L^1}=1}int_0^tint_0^1 |f(s)|~ dt ds =sup_{|{f}|_{L^1}=1}int_0^t |f(s)|~ds leq int_0^1|f(s)|~ds =|{f}|_{L^1}=1 $$
this doesnt seem all too correct. How do I continue?
functional-analysis
asked Nov 27 '18 at 22:10
orangeorange
615215
add a comment |
I am trying to find the Operator norm for the following operator
$$(Vf)(t):(C[0,1],L^1)to (C[0,1],L^1)$$
$$fmapsto int_0^t f(s)ds$$
What I have done
$$|Vf(t)|=sup_{|f|_{L^1}=1}{|int_0^t f(s) |ds}_{L^1}=sup_{|{f}|_{L^1}=1}int_0^1|{int_0^t f(s)| ~ds}dtleq sup_{|{f}|_{L^1}=1}int_0^1int_0^t |{f(s)}|~dsdt .$$
Tonelli:
$$=sup_{|{f}|_{L^1}=1}int_0^tint_0^1 |f(s)|~ dt ds =sup_{|{f}|_{L^1}=1}int_0^t |f(s)|~ds leq int_0^1|f(s)|~ds =|{f}|_{L^1}=1 $$
this doesnt seem all too correct. How do I continue?
functional-analysis
asked Nov 27 '18 at 22:10
orangeorange
615215
Is $V$ a map on $C[0,1]$ or $L^1$? Your notation is very confusing.
– copper.hat
Nov 28 '18 at 0:03
If your notation means that V is acting on the space $C[0,1]$ equipped with the $L^1$ norm, then this question is a duplicate of math.stackexchange.com/questions/2997663
– Angelo Lucia
Nov 28 '18 at 1:17
add a comment |
I am trying to find the Operator norm for the following operator
$$(Vf)(t):(C[0,1],L^1)to (C[0,1],L^1)$$
$$fmapsto int_0^t f(s)ds$$
What I have done
$$|Vf(t)|=sup_{|f|_{L^1}=1}{|int_0^t f(s) |ds}_{L^1}=sup_{|{f}|_{L^1}=1}int_0^1|{int_0^t f(s)| ~ds}dtleq sup_{|{f}|_{L^1}=1}int_0^1int_0^t |{f(s)}|~dsdt .$$
Tonelli:
$$=sup_{|{f}|_{L^1}=1}int_0^tint_0^1 |f(s)|~ dt ds =sup_{|{f}|_{L^1}=1}int_0^t |f(s)|~ds leq int_0^1|f(s)|~ds =|{f}|_{L^1}=1 $$
this doesnt seem all too correct. How do I continue?
functional-analysis
asked Nov 27 '18 at 22:10
orangeorange
615215
I am trying to find the Operator norm for the following operator
$$(Vf)(t):(C[0,1],L^1)to (C[0,1],L^1)$$
$$fmapsto int_0^t f(s)ds$$
What I have done
$$|Vf(t)|=sup_{|f|_{L^1}=1}{|int_0^t f(s) |ds}_{L^1}=sup_{|{f}|_{L^1}=1}int_0^1|{int_0^t f(s)| ~ds}dtleq sup_{|{f}|_{L^1}=1}int_0^1int_0^t |{f(s)}|~dsdt .$$
Tonelli:
$$=sup_{|{f}|_{L^1}=1}int_0^tint_0^1 |f(s)|~ dt ds =sup_{|{f}|_{L^1}=1}int_0^t |f(s)|~ds leq int_0^1|f(s)|~ds =|{f}|_{L^1}=1 $$
this doesnt seem all too correct. How do I continue?
functional-analysis
functional-analysis
asked Nov 27 '18 at 22:10
orangeorange
615215
asked Nov 27 '18 at 22:10
orangeorange
615215
asked Nov 27 '18 at 22:10
orangeorange
615215
asked Nov 27 '18 at 22:10
orangeorange
615215
asked Nov 27 '18 at 22:10
orangeorange
615215
615215
Is $V$ a map on $C[0,1]$ or $L^1$? Your notation is very confusing.
– copper.hat
Nov 28 '18 at 0:03
If your notation means that V is acting on the space $C[0,1]$ equipped with the $L^1$ norm, then this question is a duplicate of math.stackexchange.com/questions/2997663
– Angelo Lucia
Nov 28 '18 at 1:17
add a comment |
Is $V$ a map on $C[0,1]$ or $L^1$? Your notation is very confusing.
– copper.hat
Nov 28 '18 at 0:03
If your notation means that V is acting on the space $C[0,1]$ equipped with the $L^1$ norm, then this question is a duplicate of math.stackexchange.com/questions/2997663
– Angelo Lucia
Nov 28 '18 at 1:17
Is $V$ a map on $C[0,1]$ or $L^1$? Your notation is very confusing.
– copper.hat
Nov 28 '18 at 0:03
Is $V$ a map on $C[0,1]$ or $L^1$? Your notation is very confusing.
– copper.hat
Nov 28 '18 at 0:03
If your notation means that V is acting on the space $C[0,1]$ equipped with the $L^1$ norm, then this question is a duplicate of math.stackexchange.com/questions/2997663
– Angelo Lucia
Nov 28 '18 at 1:17
If your notation means that V is acting on the space $C[0,1]$ equipped with the $L^1$ norm, then this question is a duplicate of math.stackexchange.com/questions/2997663
– Angelo Lucia
Nov 28 '18 at 1:17
add a comment |
1 Answer
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$|V|leq 1$ is correct. To get $|V|geq 1$ define $f_n(s)=nI_{(0,frac 1 n)}$. (For each $s$, the function $f_n(s)$ is a constant function in $L^{1}$). I leave it to you to compute $Vf_n$.
answered Nov 27 '18 at 23:34
Kavi Rama MurthyKavi Rama Murthy
52.2k32055
add a comment |
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1 Answer
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active
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$|V|leq 1$ is correct. To get $|V|geq 1$ define $f_n(s)=nI_{(0,frac 1 n)}$. (For each $s$, the function $f_n(s)$ is a constant function in $L^{1}$). I leave it to you to compute $Vf_n$.
answered Nov 27 '18 at 23:34
Kavi Rama MurthyKavi Rama Murthy
52.2k32055
add a comment |
$|V|leq 1$ is correct. To get $|V|geq 1$ define $f_n(s)=nI_{(0,frac 1 n)}$. (For each $s$, the function $f_n(s)$ is a constant function in $L^{1}$). I leave it to you to compute $Vf_n$.
answered Nov 27 '18 at 23:34
Kavi Rama MurthyKavi Rama Murthy
52.2k32055
add a comment |
$|V|leq 1$ is correct. To get $|V|geq 1$ define $f_n(s)=nI_{(0,frac 1 n)}$. (For each $s$, the function $f_n(s)$ is a constant function in $L^{1}$). I leave it to you to compute $Vf_n$.
answered Nov 27 '18 at 23:34
Kavi Rama MurthyKavi Rama Murthy
52.2k32055
$|V|leq 1$ is correct. To get $|V|geq 1$ define $f_n(s)=nI_{(0,frac 1 n)}$. (For each $s$, the function $f_n(s)$ is a constant function in $L^{1}$). I leave it to you to compute $Vf_n$.
answered Nov 27 '18 at 23:34
Kavi Rama MurthyKavi Rama Murthy
52.2k32055
answered Nov 27 '18 at 23:34
Kavi Rama MurthyKavi Rama Murthy
52.2k32055
answered Nov 27 '18 at 23:34
Kavi Rama MurthyKavi Rama Murthy
52.2k32055
answered Nov 27 '18 at 23:34
Kavi Rama MurthyKavi Rama Murthy
52.2k32055
52.2k32055
add a comment |
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Is $V$ a map on $C[0,1]$ or $L^1$? Your notation is very confusing.
– copper.hat
Nov 28 '18 at 0:03
If your notation means that V is acting on the space $C[0,1]$ equipped with the $L^1$ norm, then this question is a duplicate of math.stackexchange.com/questions/2997663
– Angelo Lucia
Nov 28 '18 at 1:17