Continuity of a linear functional $T$: $ker{T}$ is closed
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I was solving this exercise:
let $T: C^{0}([0,b]) rightarrow C^{0}([0,b])$ s.t. $u mapsto int_{0}^{t}e^{t-x}u(x)dx$.
Prove $T$ is continuous.
Of course $T$ is linear. It's continuous since this functional is bounded (and I have no problem in showing it)
Then, from theory I know that $ker{T}$ must be a closed set: I'd like to show it.
Let $u(x) in C^{0}([0,b])$ s.t. $T(u)=0$.
This amounts to
$int_{0}^{t} e^{t-x} u(x)dx=0$, which implies $u(x)=0, forall x in C^{0}([0,b])$, since $t$ is finite and $e^{t-x} ne0$
Then $ker{T}={u: u(x)=0}$. Intuitively, this set is closed since it contains all its accumulation points. Is this a correct way to say it's closed?
Thanks
functional-analysis
asked Dec 6 '18 at 19:42
VoBVoB
673213
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add a comment |
$begingroup$
I was solving this exercise:
let $T: C^{0}([0,b]) rightarrow C^{0}([0,b])$ s.t. $u mapsto int_{0}^{t}e^{t-x}u(x)dx$.
Prove $T$ is continuous.
Of course $T$ is linear. It's continuous since this functional is bounded (and I have no problem in showing it)
Then, from theory I know that $ker{T}$ must be a closed set: I'd like to show it.
Let $u(x) in C^{0}([0,b])$ s.t. $T(u)=0$.
This amounts to
$int_{0}^{t} e^{t-x} u(x)dx=0$, which implies $u(x)=0, forall x in C^{0}([0,b])$, since $t$ is finite and $e^{t-x} ne0$
Then $ker{T}={u: u(x)=0}$. Intuitively, this set is closed since it contains all its accumulation points. Is this a correct way to say it's closed?
Thanks
functional-analysis
asked Dec 6 '18 at 19:42
VoBVoB
673213
$endgroup$
add a comment |
$begingroup$
I was solving this exercise:
let $T: C^{0}([0,b]) rightarrow C^{0}([0,b])$ s.t. $u mapsto int_{0}^{t}e^{t-x}u(x)dx$.
Prove $T$ is continuous.
Of course $T$ is linear. It's continuous since this functional is bounded (and I have no problem in showing it)
Then, from theory I know that $ker{T}$ must be a closed set: I'd like to show it.
Let $u(x) in C^{0}([0,b])$ s.t. $T(u)=0$.
This amounts to
$int_{0}^{t} e^{t-x} u(x)dx=0$, which implies $u(x)=0, forall x in C^{0}([0,b])$, since $t$ is finite and $e^{t-x} ne0$
Then $ker{T}={u: u(x)=0}$. Intuitively, this set is closed since it contains all its accumulation points. Is this a correct way to say it's closed?
Thanks
functional-analysis
asked Dec 6 '18 at 19:42
VoBVoB
673213
$endgroup$
I was solving this exercise:
let $T: C^{0}([0,b]) rightarrow C^{0}([0,b])$ s.t. $u mapsto int_{0}^{t}e^{t-x}u(x)dx$.
Prove $T$ is continuous.
Of course $T$ is linear. It's continuous since this functional is bounded (and I have no problem in showing it)
Then, from theory I know that $ker{T}$ must be a closed set: I'd like to show it.
Let $u(x) in C^{0}([0,b])$ s.t. $T(u)=0$.
This amounts to
$int_{0}^{t} e^{t-x} u(x)dx=0$, which implies $u(x)=0, forall x in C^{0}([0,b])$, since $t$ is finite and $e^{t-x} ne0$
Then $ker{T}={u: u(x)=0}$. Intuitively, this set is closed since it contains all its accumulation points. Is this a correct way to say it's closed?
Thanks
functional-analysis
functional-analysis
asked Dec 6 '18 at 19:42
VoBVoB
673213
asked Dec 6 '18 at 19:42
VoBVoB
673213
asked Dec 6 '18 at 19:42
VoBVoB
673213
asked Dec 6 '18 at 19:42
VoBVoB
673213
asked Dec 6 '18 at 19:42
VoBVoB
673213
673213
add a comment |
add a comment |
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That set is closed because $T$ is continuous, ${0}$ is closed, and $ker T=T^{-1}bigl({0}bigr)$
answered Dec 6 '18 at 19:58
José Carlos SantosJosé Carlos Santos
155k22124227
$endgroup$
$begingroup$
That was my first guess, then I thought thay the elements of the kernel are functions, and this confused me a little
$endgroup$
– VoB
Dec 6 '18 at 20:13
add a comment |
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1 Answer
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$begingroup$
That set is closed because $T$ is continuous, ${0}$ is closed, and $ker T=T^{-1}bigl({0}bigr)$
answered Dec 6 '18 at 19:58
José Carlos SantosJosé Carlos Santos
155k22124227
$endgroup$
$begingroup$
That was my first guess, then I thought thay the elements of the kernel are functions, and this confused me a little
$endgroup$
– VoB
Dec 6 '18 at 20:13
add a comment |
$begingroup$
That set is closed because $T$ is continuous, ${0}$ is closed, and $ker T=T^{-1}bigl({0}bigr)$
answered Dec 6 '18 at 19:58
José Carlos SantosJosé Carlos Santos
155k22124227
$endgroup$
$begingroup$
That was my first guess, then I thought thay the elements of the kernel are functions, and this confused me a little
$endgroup$
– VoB
Dec 6 '18 at 20:13
add a comment |
$begingroup$
That set is closed because $T$ is continuous, ${0}$ is closed, and $ker T=T^{-1}bigl({0}bigr)$
answered Dec 6 '18 at 19:58
José Carlos SantosJosé Carlos Santos
155k22124227
$endgroup$
That set is closed because $T$ is continuous, ${0}$ is closed, and $ker T=T^{-1}bigl({0}bigr)$
answered Dec 6 '18 at 19:58
José Carlos SantosJosé Carlos Santos
155k22124227
edited Dec 6 '18 at 21:12
answered Dec 6 '18 at 19:58
José Carlos SantosJosé Carlos Santos
155k22124227
answered Dec 6 '18 at 19:58
José Carlos SantosJosé Carlos Santos
155k22124227
answered Dec 6 '18 at 19:58
José Carlos SantosJosé Carlos Santos
155k22124227
155k22124227
$begingroup$
That was my first guess, then I thought thay the elements of the kernel are functions, and this confused me a little
$endgroup$
– VoB
Dec 6 '18 at 20:13
add a comment |
$begingroup$
That was my first guess, then I thought thay the elements of the kernel are functions, and this confused me a little
$endgroup$
– VoB
Dec 6 '18 at 20:13
$begingroup$
That was my first guess, then I thought thay the elements of the kernel are functions, and this confused me a little
$endgroup$
– VoB
Dec 6 '18 at 20:13
$begingroup$
That was my first guess, then I thought thay the elements of the kernel are functions, and this confused me a little
$endgroup$
– VoB
Dec 6 '18 at 20:13
add a comment |
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