Mixing
Injectivity of an integral operator
$begingroup$
we have the following integral operator:
$T:C^0[0,1] rightarrow C^0[0,1]$, with $Tu(x)=e^x int_{0}^{x}e^{-t}u(t)dt$.
This operator is not injective since I found $Tu(x)=0$ for $u(t)=e^t(2t-x)$. So $ker(T) ne langle {vec 0} rangle$, thus it's not injective.
Am I wrong?
functional-analysis operator-theory
asked Jan 25 at 10:59
VoBVoB
738513
$endgroup$
add a comment |
$begingroup$
we have the following integral operator:
$T:C^0[0,1] rightarrow C^0[0,1]$, with $Tu(x)=e^x int_{0}^{x}e^{-t}u(t)dt$.
This operator is not injective since I found $Tu(x)=0$ for $u(t)=e^t(2t-x)$. So $ker(T) ne langle {vec 0} rangle$, thus it's not injective.
Am I wrong?
functional-analysis operator-theory
asked Jan 25 at 10:59
VoBVoB
738513
$endgroup$
$begingroup$
It's wrong because it is meaningless : your function $u$ must depend either on $x$ or on $t$, not both...
$endgroup$
– Jean Marie
Jan 25 at 11:10
$begingroup$
On a different level, had you worked in $C^1$, i.e., considered $T:u in C^1 to v in C^1$ (with the same definition), the operator is a bijection because it has an inverse which is $T^{-1}: v to u = v' - v$ (I discovered it using Laplace Transform)
$endgroup$
– Jean Marie
Jan 25 at 11:34
add a comment |
$begingroup$
we have the following integral operator:
$T:C^0[0,1] rightarrow C^0[0,1]$, with $Tu(x)=e^x int_{0}^{x}e^{-t}u(t)dt$.
This operator is not injective since I found $Tu(x)=0$ for $u(t)=e^t(2t-x)$. So $ker(T) ne langle {vec 0} rangle$, thus it's not injective.
Am I wrong?
functional-analysis operator-theory
asked Jan 25 at 10:59
VoBVoB
738513
$endgroup$
we have the following integral operator:
$T:C^0[0,1] rightarrow C^0[0,1]$, with $Tu(x)=e^x int_{0}^{x}e^{-t}u(t)dt$.
This operator is not injective since I found $Tu(x)=0$ for $u(t)=e^t(2t-x)$. So $ker(T) ne langle {vec 0} rangle$, thus it's not injective.
Am I wrong?
functional-analysis operator-theory
functional-analysis operator-theory
asked Jan 25 at 10:59
VoBVoB
738513
asked Jan 25 at 10:59
VoBVoB
738513
asked Jan 25 at 10:59
VoBVoB
738513
asked Jan 25 at 10:59
VoBVoB
738513
asked Jan 25 at 10:59
VoBVoB
738513
738513
$begingroup$
It's wrong because it is meaningless : your function $u$ must depend either on $x$ or on $t$, not both...
$endgroup$
– Jean Marie
Jan 25 at 11:10
$begingroup$
On a different level, had you worked in $C^1$, i.e., considered $T:u in C^1 to v in C^1$ (with the same definition), the operator is a bijection because it has an inverse which is $T^{-1}: v to u = v' - v$ (I discovered it using Laplace Transform)
$endgroup$
– Jean Marie
Jan 25 at 11:34
add a comment |
$begingroup$
It's wrong because it is meaningless : your function $u$ must depend either on $x$ or on $t$, not both...
$endgroup$
– Jean Marie
Jan 25 at 11:10
$begingroup$
On a different level, had you worked in $C^1$, i.e., considered $T:u in C^1 to v in C^1$ (with the same definition), the operator is a bijection because it has an inverse which is $T^{-1}: v to u = v' - v$ (I discovered it using Laplace Transform)
$endgroup$
– Jean Marie
Jan 25 at 11:34
$begingroup$
It's wrong because it is meaningless : your function $u$ must depend either on $x$ or on $t$, not both...
$endgroup$
– Jean Marie
Jan 25 at 11:10
$begingroup$
It's wrong because it is meaningless : your function $u$ must depend either on $x$ or on $t$, not both...
$endgroup$
– Jean Marie
Jan 25 at 11:10
$begingroup$
On a different level, had you worked in $C^1$, i.e., considered $T:u in C^1 to v in C^1$ (with the same definition), the operator is a bijection because it has an inverse which is $T^{-1}: v to u = v' - v$ (I discovered it using Laplace Transform)
$endgroup$
– Jean Marie
Jan 25 at 11:34
$begingroup$
On a different level, had you worked in $C^1$, i.e., considered $T:u in C^1 to v in C^1$ (with the same definition), the operator is a bijection because it has an inverse which is $T^{-1}: v to u = v' - v$ (I discovered it using Laplace Transform)
$endgroup$
– Jean Marie
Jan 25 at 11:34
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You are wrong ! Your function $u$ contains the parameter $x$. $u$ has to be independent of $x$.
$T$ is injective. To this end let $u in C^0[0,1]$ and $Tu=0$.
This means that
$e^x int_{0}^{x}e^{-t}u(t)dt=0$ for all $x in [0,1].$
Hence $int_{0}^{x}e^{-t}u(t)dt=0$ for all $x in [0,1]$ and therefore $frac{d}{dx}int_{0}^{x}e^{-t}u(t)dt=e^{-x}u(x)=0$ for all $x in [0,1]$.
Conclusion: $u(x)=0$ for all $x in [0,1]$.
answered Jan 25 at 11:09
FredFred
48.5k11849
$endgroup$
$begingroup$
Thanks. The issue was that the extremal of the integral was coincident with the parameter.
$endgroup$
– VoB
Jan 25 at 11:59
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You are wrong ! Your function $u$ contains the parameter $x$. $u$ has to be independent of $x$.
$T$ is injective. To this end let $u in C^0[0,1]$ and $Tu=0$.
This means that
$e^x int_{0}^{x}e^{-t}u(t)dt=0$ for all $x in [0,1].$
Hence $int_{0}^{x}e^{-t}u(t)dt=0$ for all $x in [0,1]$ and therefore $frac{d}{dx}int_{0}^{x}e^{-t}u(t)dt=e^{-x}u(x)=0$ for all $x in [0,1]$.
Conclusion: $u(x)=0$ for all $x in [0,1]$.
answered Jan 25 at 11:09
FredFred
48.5k11849
$endgroup$
$begingroup$
Thanks. The issue was that the extremal of the integral was coincident with the parameter.
$endgroup$
– VoB
Jan 25 at 11:59
add a comment |
$begingroup$
You are wrong ! Your function $u$ contains the parameter $x$. $u$ has to be independent of $x$.
$T$ is injective. To this end let $u in C^0[0,1]$ and $Tu=0$.
This means that
$e^x int_{0}^{x}e^{-t}u(t)dt=0$ for all $x in [0,1].$
Hence $int_{0}^{x}e^{-t}u(t)dt=0$ for all $x in [0,1]$ and therefore $frac{d}{dx}int_{0}^{x}e^{-t}u(t)dt=e^{-x}u(x)=0$ for all $x in [0,1]$.
Conclusion: $u(x)=0$ for all $x in [0,1]$.
answered Jan 25 at 11:09
FredFred
48.5k11849
$endgroup$
$begingroup$
Thanks. The issue was that the extremal of the integral was coincident with the parameter.
$endgroup$
– VoB
Jan 25 at 11:59
add a comment |
$begingroup$
You are wrong ! Your function $u$ contains the parameter $x$. $u$ has to be independent of $x$.
$T$ is injective. To this end let $u in C^0[0,1]$ and $Tu=0$.
This means that
$e^x int_{0}^{x}e^{-t}u(t)dt=0$ for all $x in [0,1].$
Hence $int_{0}^{x}e^{-t}u(t)dt=0$ for all $x in [0,1]$ and therefore $frac{d}{dx}int_{0}^{x}e^{-t}u(t)dt=e^{-x}u(x)=0$ for all $x in [0,1]$.
Conclusion: $u(x)=0$ for all $x in [0,1]$.
answered Jan 25 at 11:09
FredFred
48.5k11849
$endgroup$
You are wrong ! Your function $u$ contains the parameter $x$. $u$ has to be independent of $x$.
$T$ is injective. To this end let $u in C^0[0,1]$ and $Tu=0$.
This means that
$e^x int_{0}^{x}e^{-t}u(t)dt=0$ for all $x in [0,1].$
Hence $int_{0}^{x}e^{-t}u(t)dt=0$ for all $x in [0,1]$ and therefore $frac{d}{dx}int_{0}^{x}e^{-t}u(t)dt=e^{-x}u(x)=0$ for all $x in [0,1]$.
Conclusion: $u(x)=0$ for all $x in [0,1]$.
answered Jan 25 at 11:09
FredFred
48.5k11849
answered Jan 25 at 11:09
FredFred
48.5k11849
answered Jan 25 at 11:09
FredFred
48.5k11849
answered Jan 25 at 11:09
FredFred
48.5k11849
48.5k11849
$begingroup$
Thanks. The issue was that the extremal of the integral was coincident with the parameter.
$endgroup$
– VoB
Jan 25 at 11:59
add a comment |
$begingroup$
Thanks. The issue was that the extremal of the integral was coincident with the parameter.
$endgroup$
– VoB
Jan 25 at 11:59
$begingroup$
Thanks. The issue was that the extremal of the integral was coincident with the parameter.
$endgroup$
– VoB
Jan 25 at 11:59
$begingroup$
Thanks. The issue was that the extremal of the integral was coincident with the parameter.
$endgroup$
– VoB
Jan 25 at 11:59
add a comment |
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$begingroup$
It's wrong because it is meaningless : your function $u$ must depend either on $x$ or on $t$, not both...
$endgroup$
– Jean Marie
Jan 25 at 11:10
$begingroup$
On a different level, had you worked in $C^1$, i.e., considered $T:u in C^1 to v in C^1$ (with the same definition), the operator is a bijection because it has an inverse which is $T^{-1}: v to u = v' - v$ (I discovered it using Laplace Transform)
$endgroup$
– Jean Marie
Jan 25 at 11:34