Define $Tf(x) = int_{0}^{x} f(t)dt$ Then, choose the correct option












0












$begingroup$


$(C[0, 1], ||.||_{infty})$ 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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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
















0












$begingroup$


$(C[0, 1], ||.||_{infty})$ 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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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














0












0








0





$begingroup$


$(C[0, 1], ||.||_{infty})$ 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







share|cite|improve this question









$endgroup$




$(C[0, 1], ||.||_{infty})$ 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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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


















  • $begingroup$
    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










2 Answers
2






active

oldest

votes


















3












$begingroup$

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






share|cite|improve this answer









$endgroup$













  • $begingroup$
    ..@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





















2












$begingroup$

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 $fin C[0,1]$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

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






share|cite|improve this answer









$endgroup$













  • $begingroup$
    ..@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


















3












$begingroup$

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






share|cite|improve this answer









$endgroup$













  • $begingroup$
    ..@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
















3












3








3





$begingroup$

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






share|cite|improve this answer









$endgroup$



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







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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




















  • $begingroup$
    ..@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













2












$begingroup$

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 $fin C[0,1]$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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


















2












$begingroup$

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 $fin C[0,1]$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















2












2








2





$begingroup$

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 $fin C[0,1]$.






share|cite|improve this answer









$endgroup$



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 $fin C[0,1]$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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




















  • $begingroup$
    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


















$begingroup$
what is the correct option
$endgroup$
– jasmine
Jan 7 at 11:24






$begingroup$
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




$begingroup$
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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