Define $Tf(x) = int_{0}^{x} f(t)dt$ Then, choose the correct option
$begingroup$
$(C[0, 1], ||.||_{infty})$ denotes the set of all real-valued continuous functions on $[0, 1]$ with $||f||_{infty}$:=
$sup{|f(t)|, t ∈ [0, 1]}$. For each $x in [0, 1],$ define
$Tf(x) = int_{0}^{x} f(t)dt$
Then, choose the correct option
$1.$ $T$ is injective but not surjective.
$2.$ $T$ is surjective but not injective.
$3.$ $T$ is bijective.
$4.$ $T$ is neither injective nor surjective.
My attempt : i thinks T is bijective take $f(x)= x$ option $3)$ will true
Is its True ?
Any hints/solution will be appreciated
real-analysis normed-spaces
asked Jan 7 at 10:54
jasminejasmine
1,980420
$endgroup$
add a comment |
$begingroup$
$(C[0, 1], ||.||_{infty})$ denotes the set of all real-valued continuous functions on $[0, 1]$ with $||f||_{infty}$:=
$sup{|f(t)|, t ∈ [0, 1]}$. For each $x in [0, 1],$ define
$Tf(x) = int_{0}^{x} f(t)dt$
Then, choose the correct option
$1.$ $T$ is injective but not surjective.
$2.$ $T$ is surjective but not injective.
$3.$ $T$ is bijective.
$4.$ $T$ is neither injective nor surjective.
My attempt : i thinks T is bijective take $f(x)= x$ option $3)$ will true
Is its True ?
Any hints/solution will be appreciated
real-analysis normed-spaces
asked Jan 7 at 10:54
jasminejasmine
1,980420
$endgroup$
$begingroup$
What is $T$? Is $T$ a function?
$endgroup$
– YiFan
Jan 7 at 12:16
1
$begingroup$
If $Tf=Tg$ then the function $h(x)=(Tf)(x)-(Tg)(x) =0$ for all $xin (0,1),$ so for all $xin (0,1)$ we have $f(x)-g(x)=(Tf)'(x)-(Tg)'(x)=h'(x)=0.$
$endgroup$
– DanielWainfleet
Jan 7 at 12:41
2
$begingroup$
@YiFan. $T$ is a function from $C[0,1]$ into $C[0,1]$ such that $T(f)(x)=int_0^xf(t) dt$ for all $fin C[0,1]$ and all $xin [0,1].$ It is common in this topic to write $Tf$ for $T(f)$ and to write $Tf(x)$ for $T(f)(x)$.
$endgroup$
– DanielWainfleet
Jan 7 at 12:52
add a comment |
$begingroup$
$(C[0, 1], ||.||_{infty})$ denotes the set of all real-valued continuous functions on $[0, 1]$ with $||f||_{infty}$:=
$sup{|f(t)|, t ∈ [0, 1]}$. For each $x in [0, 1],$ define
$Tf(x) = int_{0}^{x} f(t)dt$
Then, choose the correct option
$1.$ $T$ is injective but not surjective.
$2.$ $T$ is surjective but not injective.
$3.$ $T$ is bijective.
$4.$ $T$ is neither injective nor surjective.
My attempt : i thinks T is bijective take $f(x)= x$ option $3)$ will true
Is its True ?
Any hints/solution will be appreciated
real-analysis normed-spaces
asked Jan 7 at 10:54
jasminejasmine
1,980420
$endgroup$
$(C[0, 1], ||.||_{infty})$ denotes the set of all real-valued continuous functions on $[0, 1]$ with $||f||_{infty}$:=
$sup{|f(t)|, t ∈ [0, 1]}$. For each $x in [0, 1],$ define
$Tf(x) = int_{0}^{x} f(t)dt$
Then, choose the correct option
$1.$ $T$ is injective but not surjective.
$2.$ $T$ is surjective but not injective.
$3.$ $T$ is bijective.
$4.$ $T$ is neither injective nor surjective.
My attempt : i thinks T is bijective take $f(x)= x$ option $3)$ will true
Is its True ?
Any hints/solution will be appreciated
real-analysis normed-spaces
real-analysis normed-spaces
asked Jan 7 at 10:54
jasminejasmine
1,980420
asked Jan 7 at 10:54
jasminejasmine
1,980420
asked Jan 7 at 10:54
jasminejasmine
1,980420
asked Jan 7 at 10:54
jasminejasmine
1,980420
asked Jan 7 at 10:54
jasminejasmine
1,980420
1,980420
$begingroup$
What is $T$? Is $T$ a function?
$endgroup$
– YiFan
Jan 7 at 12:16
1
$begingroup$
If $Tf=Tg$ then the function $h(x)=(Tf)(x)-(Tg)(x) =0$ for all $xin (0,1),$ so for all $xin (0,1)$ we have $f(x)-g(x)=(Tf)'(x)-(Tg)'(x)=h'(x)=0.$
$endgroup$
– DanielWainfleet
Jan 7 at 12:41
2
$begingroup$
@YiFan. $T$ is a function from $C[0,1]$ into $C[0,1]$ such that $T(f)(x)=int_0^xf(t) dt$ for all $fin C[0,1]$ and all $xin [0,1].$ It is common in this topic to write $Tf$ for $T(f)$ and to write $Tf(x)$ for $T(f)(x)$.
$endgroup$
– DanielWainfleet
Jan 7 at 12:52
add a comment |
$begingroup$
What is $T$? Is $T$ a function?
$endgroup$
– YiFan
Jan 7 at 12:16
1
$begingroup$
If $Tf=Tg$ then the function $h(x)=(Tf)(x)-(Tg)(x) =0$ for all $xin (0,1),$ so for all $xin (0,1)$ we have $f(x)-g(x)=(Tf)'(x)-(Tg)'(x)=h'(x)=0.$
$endgroup$
– DanielWainfleet
Jan 7 at 12:41
2
$begingroup$
@YiFan. $T$ is a function from $C[0,1]$ into $C[0,1]$ such that $T(f)(x)=int_0^xf(t) dt$ for all $fin C[0,1]$ and all $xin [0,1].$ It is common in this topic to write $Tf$ for $T(f)$ and to write $Tf(x)$ for $T(f)(x)$.
$endgroup$
– DanielWainfleet
Jan 7 at 12:52
$begingroup$
What is $T$? Is $T$ a function?
$endgroup$
– YiFan
Jan 7 at 12:16
$begingroup$
What is $T$? Is $T$ a function?
$endgroup$
– YiFan
Jan 7 at 12:16
1
1
$begingroup$
If $Tf=Tg$ then the function $h(x)=(Tf)(x)-(Tg)(x) =0$ for all $xin (0,1),$ so for all $xin (0,1)$ we have $f(x)-g(x)=(Tf)'(x)-(Tg)'(x)=h'(x)=0.$
$endgroup$
– DanielWainfleet
Jan 7 at 12:41
$begingroup$
If $Tf=Tg$ then the function $h(x)=(Tf)(x)-(Tg)(x) =0$ for all $xin (0,1),$ so for all $xin (0,1)$ we have $f(x)-g(x)=(Tf)'(x)-(Tg)'(x)=h'(x)=0.$
$endgroup$
– DanielWainfleet
Jan 7 at 12:41
2
2
$begingroup$
@YiFan. $T$ is a function from $C[0,1]$ into $C[0,1]$ such that $T(f)(x)=int_0^xf(t) dt$ for all $fin C[0,1]$ and all $xin [0,1].$ It is common in this topic to write $Tf$ for $T(f)$ and to write $Tf(x)$ for $T(f)(x)$.
$endgroup$
– DanielWainfleet
Jan 7 at 12:52
$begingroup$
@YiFan. $T$ is a function from $C[0,1]$ into $C[0,1]$ such that $T(f)(x)=int_0^xf(t) dt$ for all $fin C[0,1]$ and all $xin [0,1].$ It is common in this topic to write $Tf$ for $T(f)$ and to write $Tf(x)$ for $T(f)(x)$.
$endgroup$
– DanielWainfleet
Jan 7 at 12:52
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If $g$ is a continuous function then it follows by the fundamental theorem of calculus that $int_{0}^{x}g(t)dt$ is differentiable. Consider how this affects surjectivity.
A linear map, $T$ is injective if and only if $ker T = {0}$. What can you conclude about $f$ if for every $x$
$$int_{0}^{x}f(t)dt = 0$$
Hint apply fundamental theorem of calculus
answered Jan 7 at 11:02
Olof RubinOlof Rubin
878317
$endgroup$
$begingroup$
..@Olof then which option is correct ?
$endgroup$
– jasmine
Jan 7 at 11:10
$begingroup$
What happens when you apply FTC to the integral equation above?
$endgroup$
– Olof Rubin
Jan 7 at 11:11
$begingroup$
..@Olof $f(x)- f(0)= 0 $ , $f(x) = f(0)$ that mean option $2$ is correct ?? is its correct
$endgroup$
– jasmine
Jan 7 at 11:14
$begingroup$
I was thinking more along the lines of differentiating both sides to obtain that $f(x) = 0$ and therefore the only element in the kernel is the zero function which implies injectivity. How do you come to the conclusion that $T$ is surjective?
$endgroup$
– Olof Rubin
Jan 7 at 11:17
1
$begingroup$
$Tf(x)$ is differentiable. If $T$ were surjective then that would mean that every continuous function $g$ could be written as $g = Tf$ but then this also implies that every continuous function is differentiable! Is that true? if not then $T$ is definitely not surjective. $T$ is injective by the reasoning i supplied above
$endgroup$
– Olof Rubin
Jan 7 at 11:26
|
show 4 more comments
$begingroup$
If $T$ maps from $C[0,1]$ to $C[0,1]$, $T$ is not surjective. For every $f$ we have $Tf(0)=0$, but not every function $g$ in $C[0,1]$ has $g(0)=0$. Hence not every $gin C[0,1]$ is of the form $g=Tf$ for some $fin C[0,1]$.
answered Jan 7 at 11:03
Student7Student7
1839
$endgroup$
$begingroup$
what is the correct option
$endgroup$
– jasmine
Jan 7 at 11:24
$begingroup$
The correct option is 1.
$endgroup$
– Student7
Jan 7 at 11:29
1
$begingroup$
The correct option is 1., since by my answer above $T$ is not surjective and additionally $T$ is injective as Olof Rubin answered: (more explicitely:) Since we have $(Tf)'(x)=f(x)$ (*) by the fundamental theorem of calculus, it holds that if $Tf=0$, i. e. $Tf$ is the constant zero function, then $(Tf)'=0$ and hence $f=0$ by (*). So $T$ is injective, because it is linear and the kernel is ${0}$.
$endgroup$
– Student7
Jan 7 at 12:40
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 $g$ is a continuous function then it follows by the fundamental theorem of calculus that $int_{0}^{x}g(t)dt$ is differentiable. Consider how this affects surjectivity.
A linear map, $T$ is injective if and only if $ker T = {0}$. What can you conclude about $f$ if for every $x$
$$int_{0}^{x}f(t)dt = 0$$
Hint apply fundamental theorem of calculus
answered Jan 7 at 11:02
Olof RubinOlof Rubin
878317
$endgroup$
$begingroup$
..@Olof then which option is correct ?
$endgroup$
– jasmine
Jan 7 at 11:10
$begingroup$
What happens when you apply FTC to the integral equation above?
$endgroup$
– Olof Rubin
Jan 7 at 11:11
$begingroup$
..@Olof $f(x)- f(0)= 0 $ , $f(x) = f(0)$ that mean option $2$ is correct ?? is its correct
$endgroup$
– jasmine
Jan 7 at 11:14
$begingroup$
I was thinking more along the lines of differentiating both sides to obtain that $f(x) = 0$ and therefore the only element in the kernel is the zero function which implies injectivity. How do you come to the conclusion that $T$ is surjective?
$endgroup$
– Olof Rubin
Jan 7 at 11:17
1
$begingroup$
$Tf(x)$ is differentiable. If $T$ were surjective then that would mean that every continuous function $g$ could be written as $g = Tf$ but then this also implies that every continuous function is differentiable! Is that true? if not then $T$ is definitely not surjective. $T$ is injective by the reasoning i supplied above
$endgroup$
– Olof Rubin
Jan 7 at 11:26
|
show 4 more comments
$begingroup$
If $g$ is a continuous function then it follows by the fundamental theorem of calculus that $int_{0}^{x}g(t)dt$ is differentiable. Consider how this affects surjectivity.
A linear map, $T$ is injective if and only if $ker T = {0}$. What can you conclude about $f$ if for every $x$
$$int_{0}^{x}f(t)dt = 0$$
Hint apply fundamental theorem of calculus
answered Jan 7 at 11:02
Olof RubinOlof Rubin
878317
$endgroup$
$begingroup$
..@Olof then which option is correct ?
$endgroup$
– jasmine
Jan 7 at 11:10
$begingroup$
What happens when you apply FTC to the integral equation above?
$endgroup$
– Olof Rubin
Jan 7 at 11:11
$begingroup$
..@Olof $f(x)- f(0)= 0 $ , $f(x) = f(0)$ that mean option $2$ is correct ?? is its correct
$endgroup$
– jasmine
Jan 7 at 11:14
$begingroup$
I was thinking more along the lines of differentiating both sides to obtain that $f(x) = 0$ and therefore the only element in the kernel is the zero function which implies injectivity. How do you come to the conclusion that $T$ is surjective?
$endgroup$
– Olof Rubin
Jan 7 at 11:17
1
$begingroup$
$Tf(x)$ is differentiable. If $T$ were surjective then that would mean that every continuous function $g$ could be written as $g = Tf$ but then this also implies that every continuous function is differentiable! Is that true? if not then $T$ is definitely not surjective. $T$ is injective by the reasoning i supplied above
$endgroup$
– Olof Rubin
Jan 7 at 11:26
|
show 4 more comments
$begingroup$
If $g$ is a continuous function then it follows by the fundamental theorem of calculus that $int_{0}^{x}g(t)dt$ is differentiable. Consider how this affects surjectivity.
A linear map, $T$ is injective if and only if $ker T = {0}$. What can you conclude about $f$ if for every $x$
$$int_{0}^{x}f(t)dt = 0$$
Hint apply fundamental theorem of calculus
answered Jan 7 at 11:02
Olof RubinOlof Rubin
878317
$endgroup$
If $g$ is a continuous function then it follows by the fundamental theorem of calculus that $int_{0}^{x}g(t)dt$ is differentiable. Consider how this affects surjectivity.
A linear map, $T$ is injective if and only if $ker T = {0}$. What can you conclude about $f$ if for every $x$
$$int_{0}^{x}f(t)dt = 0$$
Hint apply fundamental theorem of calculus
answered Jan 7 at 11:02
Olof RubinOlof Rubin
878317
answered Jan 7 at 11:02
Olof RubinOlof Rubin
878317
answered Jan 7 at 11:02
Olof RubinOlof Rubin
878317
answered Jan 7 at 11:02
Olof RubinOlof Rubin
878317
878317
$begingroup$
..@Olof then which option is correct ?
$endgroup$
– jasmine
Jan 7 at 11:10
$begingroup$
What happens when you apply FTC to the integral equation above?
$endgroup$
– Olof Rubin
Jan 7 at 11:11
$begingroup$
..@Olof $f(x)- f(0)= 0 $ , $f(x) = f(0)$ that mean option $2$ is correct ?? is its correct
$endgroup$
– jasmine
Jan 7 at 11:14
$begingroup$
I was thinking more along the lines of differentiating both sides to obtain that $f(x) = 0$ and therefore the only element in the kernel is the zero function which implies injectivity. How do you come to the conclusion that $T$ is surjective?
$endgroup$
– Olof Rubin
Jan 7 at 11:17
1
$begingroup$
$Tf(x)$ is differentiable. If $T$ were surjective then that would mean that every continuous function $g$ could be written as $g = Tf$ but then this also implies that every continuous function is differentiable! Is that true? if not then $T$ is definitely not surjective. $T$ is injective by the reasoning i supplied above
$endgroup$
– Olof Rubin
Jan 7 at 11:26
|
show 4 more comments
$begingroup$
..@Olof then which option is correct ?
$endgroup$
– jasmine
Jan 7 at 11:10
$begingroup$
What happens when you apply FTC to the integral equation above?
$endgroup$
– Olof Rubin
Jan 7 at 11:11
$begingroup$
..@Olof $f(x)- f(0)= 0 $ , $f(x) = f(0)$ that mean option $2$ is correct ?? is its correct
$endgroup$
– jasmine
Jan 7 at 11:14
$begingroup$
I was thinking more along the lines of differentiating both sides to obtain that $f(x) = 0$ and therefore the only element in the kernel is the zero function which implies injectivity. How do you come to the conclusion that $T$ is surjective?
$endgroup$
– Olof Rubin
Jan 7 at 11:17
1
$begingroup$
$Tf(x)$ is differentiable. If $T$ were surjective then that would mean that every continuous function $g$ could be written as $g = Tf$ but then this also implies that every continuous function is differentiable! Is that true? if not then $T$ is definitely not surjective. $T$ is injective by the reasoning i supplied above
$endgroup$
– Olof Rubin
Jan 7 at 11:26
$begingroup$
..@Olof then which option is correct ?
$endgroup$
– jasmine
Jan 7 at 11:10
$begingroup$
..@Olof then which option is correct ?
$endgroup$
– jasmine
Jan 7 at 11:10
$begingroup$
What happens when you apply FTC to the integral equation above?
$endgroup$
– Olof Rubin
Jan 7 at 11:11
$begingroup$
What happens when you apply FTC to the integral equation above?
$endgroup$
– Olof Rubin
Jan 7 at 11:11
$begingroup$
..@Olof $f(x)- f(0)= 0 $ , $f(x) = f(0)$ that mean option $2$ is correct ?? is its correct
$endgroup$
– jasmine
Jan 7 at 11:14
$begingroup$
..@Olof $f(x)- f(0)= 0 $ , $f(x) = f(0)$ that mean option $2$ is correct ?? is its correct
$endgroup$
– jasmine
Jan 7 at 11:14
$begingroup$
I was thinking more along the lines of differentiating both sides to obtain that $f(x) = 0$ and therefore the only element in the kernel is the zero function which implies injectivity. How do you come to the conclusion that $T$ is surjective?
$endgroup$
– Olof Rubin
Jan 7 at 11:17
$begingroup$
I was thinking more along the lines of differentiating both sides to obtain that $f(x) = 0$ and therefore the only element in the kernel is the zero function which implies injectivity. How do you come to the conclusion that $T$ is surjective?
$endgroup$
– Olof Rubin
Jan 7 at 11:17
1
1
$begingroup$
$Tf(x)$ is differentiable. If $T$ were surjective then that would mean that every continuous function $g$ could be written as $g = Tf$ but then this also implies that every continuous function is differentiable! Is that true? if not then $T$ is definitely not surjective. $T$ is injective by the reasoning i supplied above
$endgroup$
– Olof Rubin
Jan 7 at 11:26
$begingroup$
$Tf(x)$ is differentiable. If $T$ were surjective then that would mean that every continuous function $g$ could be written as $g = Tf$ but then this also implies that every continuous function is differentiable! Is that true? if not then $T$ is definitely not surjective. $T$ is injective by the reasoning i supplied above
$endgroup$
– Olof Rubin
Jan 7 at 11:26
|
show 4 more comments
$begingroup$
If $T$ maps from $C[0,1]$ to $C[0,1]$, $T$ is not surjective. For every $f$ we have $Tf(0)=0$, but not every function $g$ in $C[0,1]$ has $g(0)=0$. Hence not every $gin C[0,1]$ is of the form $g=Tf$ for some $fin C[0,1]$.
answered Jan 7 at 11:03
Student7Student7
1839
$endgroup$
$begingroup$
what is the correct option
$endgroup$
– jasmine
Jan 7 at 11:24
$begingroup$
The correct option is 1.
$endgroup$
– Student7
Jan 7 at 11:29
1
$begingroup$
The correct option is 1., since by my answer above $T$ is not surjective and additionally $T$ is injective as Olof Rubin answered: (more explicitely:) Since we have $(Tf)'(x)=f(x)$ (*) by the fundamental theorem of calculus, it holds that if $Tf=0$, i. e. $Tf$ is the constant zero function, then $(Tf)'=0$ and hence $f=0$ by (*). So $T$ is injective, because it is linear and the kernel is ${0}$.
$endgroup$
– Student7
Jan 7 at 12:40
add a comment |
$begingroup$
If $T$ maps from $C[0,1]$ to $C[0,1]$, $T$ is not surjective. For every $f$ we have $Tf(0)=0$, but not every function $g$ in $C[0,1]$ has $g(0)=0$. Hence not every $gin C[0,1]$ is of the form $g=Tf$ for some $fin C[0,1]$.
answered Jan 7 at 11:03
Student7Student7
1839
$endgroup$
$begingroup$
what is the correct option
$endgroup$
– jasmine
Jan 7 at 11:24
$begingroup$
The correct option is 1.
$endgroup$
– Student7
Jan 7 at 11:29
1
$begingroup$
The correct option is 1., since by my answer above $T$ is not surjective and additionally $T$ is injective as Olof Rubin answered: (more explicitely:) Since we have $(Tf)'(x)=f(x)$ (*) by the fundamental theorem of calculus, it holds that if $Tf=0$, i. e. $Tf$ is the constant zero function, then $(Tf)'=0$ and hence $f=0$ by (*). So $T$ is injective, because it is linear and the kernel is ${0}$.
$endgroup$
– Student7
Jan 7 at 12:40
add a comment |
$begingroup$
If $T$ maps from $C[0,1]$ to $C[0,1]$, $T$ is not surjective. For every $f$ we have $Tf(0)=0$, but not every function $g$ in $C[0,1]$ has $g(0)=0$. Hence not every $gin C[0,1]$ is of the form $g=Tf$ for some $fin C[0,1]$.
answered Jan 7 at 11:03
Student7Student7
1839
$endgroup$
If $T$ maps from $C[0,1]$ to $C[0,1]$, $T$ is not surjective. For every $f$ we have $Tf(0)=0$, but not every function $g$ in $C[0,1]$ has $g(0)=0$. Hence not every $gin C[0,1]$ is of the form $g=Tf$ for some $fin C[0,1]$.
answered Jan 7 at 11:03
Student7Student7
1839
answered Jan 7 at 11:03
Student7Student7
1839
answered Jan 7 at 11:03
Student7Student7
1839
answered Jan 7 at 11:03
Student7Student7
1839
1839
$begingroup$
what is the correct option
$endgroup$
– jasmine
Jan 7 at 11:24
$begingroup$
The correct option is 1.
$endgroup$
– Student7
Jan 7 at 11:29
1
$begingroup$
The correct option is 1., since by my answer above $T$ is not surjective and additionally $T$ is injective as Olof Rubin answered: (more explicitely:) Since we have $(Tf)'(x)=f(x)$ (*) by the fundamental theorem of calculus, it holds that if $Tf=0$, i. e. $Tf$ is the constant zero function, then $(Tf)'=0$ and hence $f=0$ by (*). So $T$ is injective, because it is linear and the kernel is ${0}$.
$endgroup$
– Student7
Jan 7 at 12:40
add a comment |
$begingroup$
what is the correct option
$endgroup$
– jasmine
Jan 7 at 11:24
$begingroup$
The correct option is 1.
$endgroup$
– Student7
Jan 7 at 11:29
1
$begingroup$
The correct option is 1., since by my answer above $T$ is not surjective and additionally $T$ is injective as Olof Rubin answered: (more explicitely:) Since we have $(Tf)'(x)=f(x)$ (*) by the fundamental theorem of calculus, it holds that if $Tf=0$, i. e. $Tf$ is the constant zero function, then $(Tf)'=0$ and hence $f=0$ by (*). So $T$ is injective, because it is linear and the kernel is ${0}$.
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– Student7
Jan 7 at 12:40
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what is the correct option
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– jasmine
Jan 7 at 11:24
$begingroup$
what is the correct option
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– jasmine
Jan 7 at 11:24
$begingroup$
The correct option is 1.
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– Student7
Jan 7 at 11:29
$begingroup$
The correct option is 1.
$endgroup$
– Student7
Jan 7 at 11:29
1
1
$begingroup$
The correct option is 1., since by my answer above $T$ is not surjective and additionally $T$ is injective as Olof Rubin answered: (more explicitely:) Since we have $(Tf)'(x)=f(x)$ (*) by the fundamental theorem of calculus, it holds that if $Tf=0$, i. e. $Tf$ is the constant zero function, then $(Tf)'=0$ and hence $f=0$ by (*). So $T$ is injective, because it is linear and the kernel is ${0}$.
$endgroup$
– Student7
Jan 7 at 12:40
$begingroup$
The correct option is 1., since by my answer above $T$ is not surjective and additionally $T$ is injective as Olof Rubin answered: (more explicitely:) Since we have $(Tf)'(x)=f(x)$ (*) by the fundamental theorem of calculus, it holds that if $Tf=0$, i. e. $Tf$ is the constant zero function, then $(Tf)'=0$ and hence $f=0$ by (*). So $T$ is injective, because it is linear and the kernel is ${0}$.
$endgroup$
– Student7
Jan 7 at 12:40
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$begingroup$
What is $T$? Is $T$ a function?
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– YiFan
Jan 7 at 12:16
1
$begingroup$
If $Tf=Tg$ then the function $h(x)=(Tf)(x)-(Tg)(x) =0$ for all $xin (0,1),$ so for all $xin (0,1)$ we have $f(x)-g(x)=(Tf)'(x)-(Tg)'(x)=h'(x)=0.$
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– DanielWainfleet
Jan 7 at 12:41
2
$begingroup$
@YiFan. $T$ is a function from $C[0,1]$ into $C[0,1]$ such that $T(f)(x)=int_0^xf(t) dt$ for all $fin C[0,1]$ and all $xin [0,1].$ It is common in this topic to write $Tf$ for $T(f)$ and to write $Tf(x)$ for $T(f)(x)$.
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– DanielWainfleet
Jan 7 at 12:52