Norm of Volterra Operator with $L^1$












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?










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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
















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?










share|cite|improve this question






















  • 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














1












1








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?










share|cite|improve this question













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






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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


















  • 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










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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 $Vf_n$.






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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 $Vf_n$.






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      0














      $|V|leq 1$ 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 $Vf_n$.






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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 $Vf_n$.






        share|cite|improve this answer












        $|V|leq 1$ 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 $Vf_n$.







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        answered Nov 27 '18 at 23:34









        Kavi Rama MurthyKavi Rama Murthy

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