Proving $Phi$ is differentiable and find $D Phi$
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Let $E = { f : [0,1] to mathbb{R} : f(0)=0 , f $ is continuously differential$ }$
With $||f|| = sup_{t in [0,1]} |f'(t)|$ and Let $Phi(f)(t) = int limits_{0}^{t} f(u)du$ with $||f|| = sup_{t in [0,1]} |f(t)|$,
prove that $Phi$ is differentiable and compute $DPhi$ ?
I need hint make sense of the question, like the limits is over real numbers or function that there norm tends to zero.
real-analysis multivariable-calculus
asked Nov 17 at 13:11
Ahmad
2,4831625
add a comment |
up vote
0
down vote
favorite
Let $E = { f : [0,1] to mathbb{R} : f(0)=0 , f $ is continuously differential$ }$
With $||f|| = sup_{t in [0,1]} |f'(t)|$ and Let $Phi(f)(t) = int limits_{0}^{t} f(u)du$ with $||f|| = sup_{t in [0,1]} |f(t)|$,
prove that $Phi$ is differentiable and compute $DPhi$ ?
I need hint make sense of the question, like the limits is over real numbers or function that there norm tends to zero.
real-analysis multivariable-calculus
asked Nov 17 at 13:11
Ahmad
2,4831625
$Phi(f+h)(t)=ldots$
– Fakemistake
Nov 17 at 13:18
$Phi(f+h)(t)$ = $Phi(f)(t) + Phi(h)(t)$ , but how does this helps ? @Fakemistake
– Ahmad
Nov 17 at 13:19
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $E = { f : [0,1] to mathbb{R} : f(0)=0 , f $ is continuously differential$ }$
With $||f|| = sup_{t in [0,1]} |f'(t)|$ and Let $Phi(f)(t) = int limits_{0}^{t} f(u)du$ with $||f|| = sup_{t in [0,1]} |f(t)|$,
prove that $Phi$ is differentiable and compute $DPhi$ ?
I need hint make sense of the question, like the limits is over real numbers or function that there norm tends to zero.
real-analysis multivariable-calculus
asked Nov 17 at 13:11
Ahmad
2,4831625
Let $E = { f : [0,1] to mathbb{R} : f(0)=0 , f $ is continuously differential$ }$
With $||f|| = sup_{t in [0,1]} |f'(t)|$ and Let $Phi(f)(t) = int limits_{0}^{t} f(u)du$ with $||f|| = sup_{t in [0,1]} |f(t)|$,
prove that $Phi$ is differentiable and compute $DPhi$ ?
I need hint make sense of the question, like the limits is over real numbers or function that there norm tends to zero.
real-analysis multivariable-calculus
real-analysis multivariable-calculus
asked Nov 17 at 13:11
Ahmad
2,4831625
asked Nov 17 at 13:11
Ahmad
2,4831625
asked Nov 17 at 13:11
Ahmad
2,4831625
asked Nov 17 at 13:11
Ahmad
2,4831625
asked Nov 17 at 13:11
Ahmad
2,4831625
2,4831625
$Phi(f+h)(t)=ldots$
– Fakemistake
Nov 17 at 13:18
$Phi(f+h)(t)$ = $Phi(f)(t) + Phi(h)(t)$ , but how does this helps ? @Fakemistake
– Ahmad
Nov 17 at 13:19
add a comment |
$Phi(f+h)(t)=ldots$
– Fakemistake
Nov 17 at 13:18
$Phi(f+h)(t)$ = $Phi(f)(t) + Phi(h)(t)$ , but how does this helps ? @Fakemistake
– Ahmad
Nov 17 at 13:19
$Phi(f+h)(t)=ldots$
– Fakemistake
Nov 17 at 13:18
$Phi(f+h)(t)=ldots$
– Fakemistake
Nov 17 at 13:18
$Phi(f+h)(t)$ = $Phi(f)(t) + Phi(h)(t)$ , but how does this helps ? @Fakemistake
– Ahmad
Nov 17 at 13:19
$Phi(f+h)(t)$ = $Phi(f)(t) + Phi(h)(t)$ , but how does this helps ? @Fakemistake
– Ahmad
Nov 17 at 13:19
add a comment |
1 Answer
1
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up vote
2
down vote
Let $a,binmathbb{R}$, $Phi(af+bg)=aPhi(f)+bPhi(g)$. This implies that $Phi$ il linear and equal to its derivative.
answered Nov 17 at 13:49
Tsemo Aristide
54.6k11444
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Let $a,binmathbb{R}$, $Phi(af+bg)=aPhi(f)+bPhi(g)$. This implies that $Phi$ il linear and equal to its derivative.
answered Nov 17 at 13:49
Tsemo Aristide
54.6k11444
add a comment |
up vote
2
down vote
Let $a,binmathbb{R}$, $Phi(af+bg)=aPhi(f)+bPhi(g)$. This implies that $Phi$ il linear and equal to its derivative.
answered Nov 17 at 13:49
Tsemo Aristide
54.6k11444
add a comment |
up vote
2
down vote
up vote
2
down vote
Let $a,binmathbb{R}$, $Phi(af+bg)=aPhi(f)+bPhi(g)$. This implies that $Phi$ il linear and equal to its derivative.
answered Nov 17 at 13:49
Tsemo Aristide
54.6k11444
Let $a,binmathbb{R}$, $Phi(af+bg)=aPhi(f)+bPhi(g)$. This implies that $Phi$ il linear and equal to its derivative.
answered Nov 17 at 13:49
Tsemo Aristide
54.6k11444
answered Nov 17 at 13:49
Tsemo Aristide
54.6k11444
answered Nov 17 at 13:49
Tsemo Aristide
54.6k11444
answered Nov 17 at 13:49
Tsemo Aristide
54.6k11444
54.6k11444
add a comment |
add a comment |
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$Phi(f+h)(t)=ldots$
– Fakemistake
Nov 17 at 13:18
$Phi(f+h)(t)$ = $Phi(f)(t) + Phi(h)(t)$ , but how does this helps ? @Fakemistake
– Ahmad
Nov 17 at 13:19