$operatorname{Ker}(T)$, $T(f) = f'g + fg'$
$begingroup$
Let $g in C^{1}[0,1]$
Let $T:C^{1}[0,1] to C[0,1]$ linear transformation such that $T(f) = (fg)'$
I have to calculate $operatorname{Ker}(T)$
I know if $f in operatorname{Ker}(T)$ then $(fg)' = 0$ so $fg = c$ in $[0,1]$
If $c not = 0$, then $g(x) not = 0$ $forall x in [0,1]$
Thus $f = frac{c}{g}$
But if there exists $x in [0,1]$ such that $g(x) = 0$ I don't know how to proceed
Can anybody help me?
linear-transformations
edited Dec 2 '18 at 0:46
Bernard
119k639112
asked Dec 2 '18 at 0:30
ZAFZAF
4357
$endgroup$
add a comment |
$begingroup$
Let $g in C^{1}[0,1]$
Let $T:C^{1}[0,1] to C[0,1]$ linear transformation such that $T(f) = (fg)'$
I have to calculate $operatorname{Ker}(T)$
I know if $f in operatorname{Ker}(T)$ then $(fg)' = 0$ so $fg = c$ in $[0,1]$
If $c not = 0$, then $g(x) not = 0$ $forall x in [0,1]$
Thus $f = frac{c}{g}$
But if there exists $x in [0,1]$ such that $g(x) = 0$ I don't know how to proceed
Can anybody help me?
linear-transformations
edited Dec 2 '18 at 0:46
Bernard
119k639112
asked Dec 2 '18 at 0:30
ZAFZAF
4357
$endgroup$
$begingroup$
If $g(x)=0$ for some $x,$ then $c = 0.$ IS this of help?
$endgroup$
– Will M.
Dec 2 '18 at 0:34
$begingroup$
No really, I thought about that but I still do not deduce any expression for f
$endgroup$
– ZAF
Dec 2 '18 at 0:35
add a comment |
$begingroup$
Let $g in C^{1}[0,1]$
Let $T:C^{1}[0,1] to C[0,1]$ linear transformation such that $T(f) = (fg)'$
I have to calculate $operatorname{Ker}(T)$
I know if $f in operatorname{Ker}(T)$ then $(fg)' = 0$ so $fg = c$ in $[0,1]$
If $c not = 0$, then $g(x) not = 0$ $forall x in [0,1]$
Thus $f = frac{c}{g}$
But if there exists $x in [0,1]$ such that $g(x) = 0$ I don't know how to proceed
Can anybody help me?
linear-transformations
edited Dec 2 '18 at 0:46
Bernard
119k639112
asked Dec 2 '18 at 0:30
ZAFZAF
4357
$endgroup$
Let $g in C^{1}[0,1]$
Let $T:C^{1}[0,1] to C[0,1]$ linear transformation such that $T(f) = (fg)'$
I have to calculate $operatorname{Ker}(T)$
I know if $f in operatorname{Ker}(T)$ then $(fg)' = 0$ so $fg = c$ in $[0,1]$
If $c not = 0$, then $g(x) not = 0$ $forall x in [0,1]$
Thus $f = frac{c}{g}$
But if there exists $x in [0,1]$ such that $g(x) = 0$ I don't know how to proceed
Can anybody help me?
linear-transformations
linear-transformations
edited Dec 2 '18 at 0:46
Bernard
119k639112
asked Dec 2 '18 at 0:30
ZAFZAF
4357
edited Dec 2 '18 at 0:46
Bernard
119k639112
asked Dec 2 '18 at 0:30
ZAFZAF
4357
edited Dec 2 '18 at 0:46
Bernard
119k639112
edited Dec 2 '18 at 0:46
Bernard
119k639112
edited Dec 2 '18 at 0:46
Bernard
119k639112
119k639112
asked Dec 2 '18 at 0:30
ZAFZAF
4357
asked Dec 2 '18 at 0:30
ZAFZAF
4357
asked Dec 2 '18 at 0:30
ZAFZAF
4357
4357
$begingroup$
If $g(x)=0$ for some $x,$ then $c = 0.$ IS this of help?
$endgroup$
– Will M.
Dec 2 '18 at 0:34
$begingroup$
No really, I thought about that but I still do not deduce any expression for f
$endgroup$
– ZAF
Dec 2 '18 at 0:35
add a comment |
$begingroup$
If $g(x)=0$ for some $x,$ then $c = 0.$ IS this of help?
$endgroup$
– Will M.
Dec 2 '18 at 0:34
$begingroup$
No really, I thought about that but I still do not deduce any expression for f
$endgroup$
– ZAF
Dec 2 '18 at 0:35
$begingroup$
If $g(x)=0$ for some $x,$ then $c = 0.$ IS this of help?
$endgroup$
– Will M.
Dec 2 '18 at 0:34
$begingroup$
If $g(x)=0$ for some $x,$ then $c = 0.$ IS this of help?
$endgroup$
– Will M.
Dec 2 '18 at 0:34
$begingroup$
No really, I thought about that but I still do not deduce any expression for f
$endgroup$
– ZAF
Dec 2 '18 at 0:35
$begingroup$
No really, I thought about that but I still do not deduce any expression for f
$endgroup$
– ZAF
Dec 2 '18 at 0:35
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If $c=0$ the answer $ker (T)={f:f(x)=0$ on the open set ${y:g(y) neq 0} }$.
answered Dec 2 '18 at 0:39
Kavi Rama MurthyKavi Rama Murthy
52.8k32055
$endgroup$
add a comment |
$begingroup$
Noting as you did that
$$
fg=c
$$
for some constant $c$, we can deduce that if $g$ vanishes anywhere, $c=0$, and $f$ then is forced to vanish whenever $g$ doesn't, indeed
$$
ker(T)={ fin C^1[0,1]:;f(x)=0 ;forall x;text{s.t}; g(x)ne 0}
$$
If $g$ does not vanish, we have the simpler expression $f=frac{c}{g}$ for a constant $c$.
answered Dec 2 '18 at 0:41
qbertqbert
22.1k32460
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $c=0$ the answer $ker (T)={f:f(x)=0$ on the open set ${y:g(y) neq 0} }$.
answered Dec 2 '18 at 0:39
Kavi Rama MurthyKavi Rama Murthy
52.8k32055
$endgroup$
add a comment |
$begingroup$
If $c=0$ the answer $ker (T)={f:f(x)=0$ on the open set ${y:g(y) neq 0} }$.
answered Dec 2 '18 at 0:39
Kavi Rama MurthyKavi Rama Murthy
52.8k32055
$endgroup$
add a comment |
$begingroup$
If $c=0$ the answer $ker (T)={f:f(x)=0$ on the open set ${y:g(y) neq 0} }$.
answered Dec 2 '18 at 0:39
Kavi Rama MurthyKavi Rama Murthy
52.8k32055
$endgroup$
If $c=0$ the answer $ker (T)={f:f(x)=0$ on the open set ${y:g(y) neq 0} }$.
answered Dec 2 '18 at 0:39
Kavi Rama MurthyKavi Rama Murthy
52.8k32055
answered Dec 2 '18 at 0:39
Kavi Rama MurthyKavi Rama Murthy
52.8k32055
answered Dec 2 '18 at 0:39
Kavi Rama MurthyKavi Rama Murthy
52.8k32055
answered Dec 2 '18 at 0:39
Kavi Rama MurthyKavi Rama Murthy
52.8k32055
52.8k32055
add a comment |
add a comment |
$begingroup$
Noting as you did that
$$
fg=c
$$
for some constant $c$, we can deduce that if $g$ vanishes anywhere, $c=0$, and $f$ then is forced to vanish whenever $g$ doesn't, indeed
$$
ker(T)={ fin C^1[0,1]:;f(x)=0 ;forall x;text{s.t}; g(x)ne 0}
$$
If $g$ does not vanish, we have the simpler expression $f=frac{c}{g}$ for a constant $c$.
answered Dec 2 '18 at 0:41
qbertqbert
22.1k32460
$endgroup$
add a comment |
$begingroup$
Noting as you did that
$$
fg=c
$$
for some constant $c$, we can deduce that if $g$ vanishes anywhere, $c=0$, and $f$ then is forced to vanish whenever $g$ doesn't, indeed
$$
ker(T)={ fin C^1[0,1]:;f(x)=0 ;forall x;text{s.t}; g(x)ne 0}
$$
If $g$ does not vanish, we have the simpler expression $f=frac{c}{g}$ for a constant $c$.
answered Dec 2 '18 at 0:41
qbertqbert
22.1k32460
$endgroup$
add a comment |
$begingroup$
Noting as you did that
$$
fg=c
$$
for some constant $c$, we can deduce that if $g$ vanishes anywhere, $c=0$, and $f$ then is forced to vanish whenever $g$ doesn't, indeed
$$
ker(T)={ fin C^1[0,1]:;f(x)=0 ;forall x;text{s.t}; g(x)ne 0}
$$
If $g$ does not vanish, we have the simpler expression $f=frac{c}{g}$ for a constant $c$.
answered Dec 2 '18 at 0:41
qbertqbert
22.1k32460
$endgroup$
Noting as you did that
$$
fg=c
$$
for some constant $c$, we can deduce that if $g$ vanishes anywhere, $c=0$, and $f$ then is forced to vanish whenever $g$ doesn't, indeed
$$
ker(T)={ fin C^1[0,1]:;f(x)=0 ;forall x;text{s.t}; g(x)ne 0}
$$
If $g$ does not vanish, we have the simpler expression $f=frac{c}{g}$ for a constant $c$.
answered Dec 2 '18 at 0:41
qbertqbert
22.1k32460
answered Dec 2 '18 at 0:41
qbertqbert
22.1k32460
answered Dec 2 '18 at 0:41
qbertqbert
22.1k32460
answered Dec 2 '18 at 0:41
qbertqbert
22.1k32460
22.1k32460
add a comment |
add a comment |
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$begingroup$
If $g(x)=0$ for some $x,$ then $c = 0.$ IS this of help?
$endgroup$
– Will M.
Dec 2 '18 at 0:34
$begingroup$
No really, I thought about that but I still do not deduce any expression for f
$endgroup$
– ZAF
Dec 2 '18 at 0:35