Inverse function that does not have a inverse function given by an integralConvergence of Integral Implies Uniform convergence of Equicontinuous FamilyHow to prove that this function is continuous at zero?Integral Estimate Using a Function and its InverseThe function is continuous but not uniformly continuous at $[0,1) cup (1,2]$.Show that $f$ is uniformly continuous if sequence function $f$ is uniformly convergentContinuous function that has limit at infinity is uniformly continuous (another viewpoint)Radial function and integralDoes the step function defined via a function $f$, converge (in $L^1$) to the function $f$?Finding an infimum and showing it is never obtained for a function on a normed vector space.Prove that $f * phi _n$ uniformly converges to $f$.
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Inverse function that does not have a inverse function given by an integral
Convergence of Integral Implies Uniform convergence of Equicontinuous FamilyHow to prove that this function is continuous at zero?Integral Estimate Using a Function and its InverseThe function is continuous but not uniformly continuous at $[0,1) cup (1,2]$.Show that $f$ is uniformly continuous if sequence function $f$ is uniformly convergentContinuous function that has limit at infinity is uniformly continuous (another viewpoint)Radial function and integralDoes the step function defined via a function $f$, converge (in $L^1$) to the function $f$?Finding an infimum and showing it is never obtained for a function on a normed vector space.Prove that $f * phi _n$ uniformly converges to $f$.
$begingroup$
I have the following exercise:
1) Consider $X = left( (C[0,1], mathbbR), d_infty right) $ and
$Y = left( (C^1[0,1], mathbbR), d_infty right) $ and the operator
$Phi:X rightarrow Y$ given by
$$
Phi(f)(x) = int_0^xf(s);ds, ;;forall x in [0,1].
$$
Show that
(a) $Phi$ is uniformly continuous;
(b) $Phi$ has an inverse $Phi^-1$;
(c) $Phi^-1$ is not continuous;
(d) Consider now $X$ and
$Z = left( (C^1[0,1], mathbbR), d_1, infty right) $, where $d_1, infty (f,g)= d_infty(f,g) + d_infty(f',g')$, and the same operator $Phi : X rightarrow Z$.
With this metric $d_1, infty$, do we take $Phi^-1$ continuous?
So, I know which the Fundamental Theorem of Calculus assures me that $Phi$ belongs to Y. This implies that $Phi$ is well defined.
The proof of item (a) I had that
$$
| Phi(f)(x)-Phi(f)(y)|=
Big|int_0^xf(s);ds - int_0^yf(s);dsBig| =\
Big|int_0^xf(s);ds - int_0^xf(s);ds - int_x^yf(s);dsBig|=
Big|int_x^yf(s);dsBig| leq
int_x^y|f(s)|;ds.
$$
How $f: [0,1] rightarrow mathbbR$, $f in C[0,1]$, we know that $f$ is bounded, so there is $k in mathbbR_+$ that $|f(x)| leq k, forall x in [0,1]$.
Hence
$$
| Phi(f)(x)-Phi(f)(y)| leq int_x^y|f(s)|;ds leq int_x^yk;ds = kint_x^y;ds = k|y-x|.
$$
This implies that $Phi$ is Lipschitz and hence $Phi$ is uniformly continuous.
(b) I know by FTC which for given $fin X$, we have $Phi$ such that
$$ Phi'(f)(x)=
dfracddx Phi(f)(x) =
dfracddx int_0^xf(s);ds =
f(x), forall x in [0,1].$$
That is, the inverse of $Phi$ is $Phi^-1:Y rightarrow X$ given by $Phi^-1(Phi(f)) = (Phi(f))'=f$.
But I don't know how to ensure that $Phi^-1$ is a bijection function
But, I can't figure out to solve (b) (conclusion), (c) and (d).
analysis
edited Mar 29 at 20:55
Martín-Blas Pérez Pinilla
35.1k42971
asked Mar 29 at 16:45
Thiago AlexandreThiago Alexandre
1347
$endgroup$
add a comment |
$begingroup$
I have the following exercise:
1) Consider $X = left( (C[0,1], mathbbR), d_infty right) $ and
$Y = left( (C^1[0,1], mathbbR), d_infty right) $ and the operator
$Phi:X rightarrow Y$ given by
$$
Phi(f)(x) = int_0^xf(s);ds, ;;forall x in [0,1].
$$
Show that
(a) $Phi$ is uniformly continuous;
(b) $Phi$ has an inverse $Phi^-1$;
(c) $Phi^-1$ is not continuous;
(d) Consider now $X$ and
$Z = left( (C^1[0,1], mathbbR), d_1, infty right) $, where $d_1, infty (f,g)= d_infty(f,g) + d_infty(f',g')$, and the same operator $Phi : X rightarrow Z$.
With this metric $d_1, infty$, do we take $Phi^-1$ continuous?
So, I know which the Fundamental Theorem of Calculus assures me that $Phi$ belongs to Y. This implies that $Phi$ is well defined.
The proof of item (a) I had that
$$
| Phi(f)(x)-Phi(f)(y)|=
Big|int_0^xf(s);ds - int_0^yf(s);dsBig| =\
Big|int_0^xf(s);ds - int_0^xf(s);ds - int_x^yf(s);dsBig|=
Big|int_x^yf(s);dsBig| leq
int_x^y|f(s)|;ds.
$$
How $f: [0,1] rightarrow mathbbR$, $f in C[0,1]$, we know that $f$ is bounded, so there is $k in mathbbR_+$ that $|f(x)| leq k, forall x in [0,1]$.
Hence
$$
| Phi(f)(x)-Phi(f)(y)| leq int_x^y|f(s)|;ds leq int_x^yk;ds = kint_x^y;ds = k|y-x|.
$$
This implies that $Phi$ is Lipschitz and hence $Phi$ is uniformly continuous.
(b) I know by FTC which for given $fin X$, we have $Phi$ such that
$$ Phi'(f)(x)=
dfracddx Phi(f)(x) =
dfracddx int_0^xf(s);ds =
f(x), forall x in [0,1].$$
That is, the inverse of $Phi$ is $Phi^-1:Y rightarrow X$ given by $Phi^-1(Phi(f)) = (Phi(f))'=f$.
But I don't know how to ensure that $Phi^-1$ is a bijection function
But, I can't figure out to solve (b) (conclusion), (c) and (d).
analysis
edited Mar 29 at 20:55
Martín-Blas Pérez Pinilla
35.1k42971
asked Mar 29 at 16:45
Thiago AlexandreThiago Alexandre
1347
$endgroup$
$begingroup$
For (b), (c), (d), what have you tried and where are you stuck?
$endgroup$
– Saad
Mar 29 at 16:47
$begingroup$
@Saad for (b) I know by FTC which for given $f in X$, we have $Phi$ such that $Phi' (x) = f(x), forall x in [0,1]$. So I need to show that $Phi^-1$ given by $Phi^-1(Phi(f)) = f(x) forall x$. I think that is correct by FTC. But I don't have sure and I don't know how to ensure that $Phi$ is a bijection function.
$endgroup$
– Thiago Alexandre
Mar 29 at 17:00
add a comment |
$begingroup$
I have the following exercise:
1) Consider $X = left( (C[0,1], mathbbR), d_infty right) $ and
$Y = left( (C^1[0,1], mathbbR), d_infty right) $ and the operator
$Phi:X rightarrow Y$ given by
$$
Phi(f)(x) = int_0^xf(s);ds, ;;forall x in [0,1].
$$
Show that
(a) $Phi$ is uniformly continuous;
(b) $Phi$ has an inverse $Phi^-1$;
(c) $Phi^-1$ is not continuous;
(d) Consider now $X$ and
$Z = left( (C^1[0,1], mathbbR), d_1, infty right) $, where $d_1, infty (f,g)= d_infty(f,g) + d_infty(f',g')$, and the same operator $Phi : X rightarrow Z$.
With this metric $d_1, infty$, do we take $Phi^-1$ continuous?
So, I know which the Fundamental Theorem of Calculus assures me that $Phi$ belongs to Y. This implies that $Phi$ is well defined.
The proof of item (a) I had that
$$
| Phi(f)(x)-Phi(f)(y)|=
Big|int_0^xf(s);ds - int_0^yf(s);dsBig| =\
Big|int_0^xf(s);ds - int_0^xf(s);ds - int_x^yf(s);dsBig|=
Big|int_x^yf(s);dsBig| leq
int_x^y|f(s)|;ds.
$$
How $f: [0,1] rightarrow mathbbR$, $f in C[0,1]$, we know that $f$ is bounded, so there is $k in mathbbR_+$ that $|f(x)| leq k, forall x in [0,1]$.
Hence
$$
| Phi(f)(x)-Phi(f)(y)| leq int_x^y|f(s)|;ds leq int_x^yk;ds = kint_x^y;ds = k|y-x|.
$$
This implies that $Phi$ is Lipschitz and hence $Phi$ is uniformly continuous.
(b) I know by FTC which for given $fin X$, we have $Phi$ such that
$$ Phi'(f)(x)=
dfracddx Phi(f)(x) =
dfracddx int_0^xf(s);ds =
f(x), forall x in [0,1].$$
That is, the inverse of $Phi$ is $Phi^-1:Y rightarrow X$ given by $Phi^-1(Phi(f)) = (Phi(f))'=f$.
But I don't know how to ensure that $Phi^-1$ is a bijection function
But, I can't figure out to solve (b) (conclusion), (c) and (d).
analysis
edited Mar 29 at 20:55
Martín-Blas Pérez Pinilla
35.1k42971
asked Mar 29 at 16:45
Thiago AlexandreThiago Alexandre
1347
$endgroup$
I have the following exercise:
1) Consider $X = left( (C[0,1], mathbbR), d_infty right) $ and
$Y = left( (C^1[0,1], mathbbR), d_infty right) $ and the operator
$Phi:X rightarrow Y$ given by
$$
Phi(f)(x) = int_0^xf(s);ds, ;;forall x in [0,1].
$$
Show that
(a) $Phi$ is uniformly continuous;
(b) $Phi$ has an inverse $Phi^-1$;
(c) $Phi^-1$ is not continuous;
(d) Consider now $X$ and
$Z = left( (C^1[0,1], mathbbR), d_1, infty right) $, where $d_1, infty (f,g)= d_infty(f,g) + d_infty(f',g')$, and the same operator $Phi : X rightarrow Z$.
With this metric $d_1, infty$, do we take $Phi^-1$ continuous?
So, I know which the Fundamental Theorem of Calculus assures me that $Phi$ belongs to Y. This implies that $Phi$ is well defined.
The proof of item (a) I had that
$$
| Phi(f)(x)-Phi(f)(y)|=
Big|int_0^xf(s);ds - int_0^yf(s);dsBig| =\
Big|int_0^xf(s);ds - int_0^xf(s);ds - int_x^yf(s);dsBig|=
Big|int_x^yf(s);dsBig| leq
int_x^y|f(s)|;ds.
$$
How $f: [0,1] rightarrow mathbbR$, $f in C[0,1]$, we know that $f$ is bounded, so there is $k in mathbbR_+$ that $|f(x)| leq k, forall x in [0,1]$.
Hence
$$
| Phi(f)(x)-Phi(f)(y)| leq int_x^y|f(s)|;ds leq int_x^yk;ds = kint_x^y;ds = k|y-x|.
$$
This implies that $Phi$ is Lipschitz and hence $Phi$ is uniformly continuous.
(b) I know by FTC which for given $fin X$, we have $Phi$ such that
$$ Phi'(f)(x)=
dfracddx Phi(f)(x) =
dfracddx int_0^xf(s);ds =
f(x), forall x in [0,1].$$
That is, the inverse of $Phi$ is $Phi^-1:Y rightarrow X$ given by $Phi^-1(Phi(f)) = (Phi(f))'=f$.
But I don't know how to ensure that $Phi^-1$ is a bijection function
But, I can't figure out to solve (b) (conclusion), (c) and (d).
analysis
analysis
edited Mar 29 at 20:55
Martín-Blas Pérez Pinilla
35.1k42971
asked Mar 29 at 16:45
Thiago AlexandreThiago Alexandre
1347
edited Mar 29 at 20:55
Martín-Blas Pérez Pinilla
35.1k42971
asked Mar 29 at 16:45
Thiago AlexandreThiago Alexandre
1347
edited Mar 29 at 20:55
Martín-Blas Pérez Pinilla
35.1k42971
edited Mar 29 at 20:55
Martín-Blas Pérez Pinilla
35.1k42971
edited Mar 29 at 20:55
Martín-Blas Pérez Pinilla
35.1k42971
35.1k42971
asked Mar 29 at 16:45
Thiago AlexandreThiago Alexandre
1347
asked Mar 29 at 16:45
Thiago AlexandreThiago Alexandre
1347
asked Mar 29 at 16:45
Thiago AlexandreThiago Alexandre
1347
1347
$begingroup$
For (b), (c), (d), what have you tried and where are you stuck?
$endgroup$
– Saad
Mar 29 at 16:47
$begingroup$
@Saad for (b) I know by FTC which for given $f in X$, we have $Phi$ such that $Phi' (x) = f(x), forall x in [0,1]$. So I need to show that $Phi^-1$ given by $Phi^-1(Phi(f)) = f(x) forall x$. I think that is correct by FTC. But I don't have sure and I don't know how to ensure that $Phi$ is a bijection function.
$endgroup$
– Thiago Alexandre
Mar 29 at 17:00
add a comment |
$begingroup$
For (b), (c), (d), what have you tried and where are you stuck?
$endgroup$
– Saad
Mar 29 at 16:47
$begingroup$
@Saad for (b) I know by FTC which for given $f in X$, we have $Phi$ such that $Phi' (x) = f(x), forall x in [0,1]$. So I need to show that $Phi^-1$ given by $Phi^-1(Phi(f)) = f(x) forall x$. I think that is correct by FTC. But I don't have sure and I don't know how to ensure that $Phi$ is a bijection function.
$endgroup$
– Thiago Alexandre
Mar 29 at 17:00
$begingroup$
For (b), (c), (d), what have you tried and where are you stuck?
$endgroup$
– Saad
Mar 29 at 16:47
$begingroup$
For (b), (c), (d), what have you tried and where are you stuck?
$endgroup$
– Saad
Mar 29 at 16:47
$begingroup$
@Saad for (b) I know by FTC which for given $f in X$, we have $Phi$ such that $Phi' (x) = f(x), forall x in [0,1]$. So I need to show that $Phi^-1$ given by $Phi^-1(Phi(f)) = f(x) forall x$. I think that is correct by FTC. But I don't have sure and I don't know how to ensure that $Phi$ is a bijection function.
$endgroup$
– Thiago Alexandre
Mar 29 at 17:00
$begingroup$
@Saad for (b) I know by FTC which for given $f in X$, we have $Phi$ such that $Phi' (x) = f(x), forall x in [0,1]$. So I need to show that $Phi^-1$ given by $Phi^-1(Phi(f)) = f(x) forall x$. I think that is correct by FTC. But I don't have sure and I don't know how to ensure that $Phi$ is a bijection function.
$endgroup$
– Thiago Alexandre
Mar 29 at 17:00
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
(b) $Phi$ is injective but not surjective. For all $f$ we have $Phi(f)(0) = 0$. Obviously, $Phi(X)$ is strictly smaller than $Y$. But is true that (I will use another name) $Psi(f) = f'$ verifies $forall fin X: Psi(Phi((f)) = f$ and that $Psi$ is defined in the whole $Y$ (is an extension of $Phi^-1$).
(c) Take a sequence $f_nin Y$ s.t. $f_nto fin Y$ uniformly but $f_n'notto f'$ uniformly.
(d) I will use again the name $Psi$:
$$
d_infty(Psi(f),Psi(g)) = d_infty(f',g')le d_1,infty(cdots).
$$
answered Mar 29 at 20:53
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35.1k42971
$endgroup$
$begingroup$
Thanks for your help me.I will focus first on item b. After that I will to (c) and (d). Well, You said for me that 1) $Phi$ is injective 2) $Phi$ is not surjective. 3) $forall f in X$, we have $Phi(f)(0) = 0$. 4) $Phi(X) subsetneqq Y$. 5) we have an extension $Psi $ of $Phi^-1$ that $Psi (Phi (f)) = f$. I can understand 3) and 5). The assertive 3) is obvious. And 5) I know that is a FTC but I can't get all.
$endgroup$
– Thiago Alexandre
Mar 29 at 23:23
$begingroup$
I got it the item (b). You said for me defined $Psi: Y rightarrow X$ given by $Psi (f) = f'$. So with this we obtain $Phi circ Psi = Id_Y$ and $Psi circ Phi = Id_X$. The both we get by FTC. Now, I don't know how sequence $f_n in Y$ we need to take for show that $Psi = Phi^-1$ is not continuous.
$endgroup$
– Thiago Alexandre
Mar 30 at 2:19
1
$begingroup$
@ThiagoAlexandre, see mathcounterexamples.net/….
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 30 at 10:40
1
$begingroup$
@ThiagoAlexandre, $Phi circPsi = Id_Y$ is actually false (again, because $Phi(f)(0) = 0$).
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 30 at 10:43
$begingroup$
Thanks. I got it.
$endgroup$
– Thiago Alexandre
Mar 30 at 17:48
add a comment |
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1 Answer
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1 Answer
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$begingroup$
(b) $Phi$ is injective but not surjective. For all $f$ we have $Phi(f)(0) = 0$. Obviously, $Phi(X)$ is strictly smaller than $Y$. But is true that (I will use another name) $Psi(f) = f'$ verifies $forall fin X: Psi(Phi((f)) = f$ and that $Psi$ is defined in the whole $Y$ (is an extension of $Phi^-1$).
(c) Take a sequence $f_nin Y$ s.t. $f_nto fin Y$ uniformly but $f_n'notto f'$ uniformly.
(d) I will use again the name $Psi$:
$$
d_infty(Psi(f),Psi(g)) = d_infty(f',g')le d_1,infty(cdots).
$$
answered Mar 29 at 20:53
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35.1k42971
$endgroup$
$begingroup$
Thanks for your help me.I will focus first on item b. After that I will to (c) and (d). Well, You said for me that 1) $Phi$ is injective 2) $Phi$ is not surjective. 3) $forall f in X$, we have $Phi(f)(0) = 0$. 4) $Phi(X) subsetneqq Y$. 5) we have an extension $Psi $ of $Phi^-1$ that $Psi (Phi (f)) = f$. I can understand 3) and 5). The assertive 3) is obvious. And 5) I know that is a FTC but I can't get all.
$endgroup$
– Thiago Alexandre
Mar 29 at 23:23
$begingroup$
I got it the item (b). You said for me defined $Psi: Y rightarrow X$ given by $Psi (f) = f'$. So with this we obtain $Phi circ Psi = Id_Y$ and $Psi circ Phi = Id_X$. The both we get by FTC. Now, I don't know how sequence $f_n in Y$ we need to take for show that $Psi = Phi^-1$ is not continuous.
$endgroup$
– Thiago Alexandre
Mar 30 at 2:19
1
$begingroup$
@ThiagoAlexandre, see mathcounterexamples.net/….
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 30 at 10:40
1
$begingroup$
@ThiagoAlexandre, $Phi circPsi = Id_Y$ is actually false (again, because $Phi(f)(0) = 0$).
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 30 at 10:43
$begingroup$
Thanks. I got it.
$endgroup$
– Thiago Alexandre
Mar 30 at 17:48
add a comment |
$begingroup$
(b) $Phi$ is injective but not surjective. For all $f$ we have $Phi(f)(0) = 0$. Obviously, $Phi(X)$ is strictly smaller than $Y$. But is true that (I will use another name) $Psi(f) = f'$ verifies $forall fin X: Psi(Phi((f)) = f$ and that $Psi$ is defined in the whole $Y$ (is an extension of $Phi^-1$).
(c) Take a sequence $f_nin Y$ s.t. $f_nto fin Y$ uniformly but $f_n'notto f'$ uniformly.
(d) I will use again the name $Psi$:
$$
d_infty(Psi(f),Psi(g)) = d_infty(f',g')le d_1,infty(cdots).
$$
answered Mar 29 at 20:53
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35.1k42971
$endgroup$
$begingroup$
Thanks for your help me.I will focus first on item b. After that I will to (c) and (d). Well, You said for me that 1) $Phi$ is injective 2) $Phi$ is not surjective. 3) $forall f in X$, we have $Phi(f)(0) = 0$. 4) $Phi(X) subsetneqq Y$. 5) we have an extension $Psi $ of $Phi^-1$ that $Psi (Phi (f)) = f$. I can understand 3) and 5). The assertive 3) is obvious. And 5) I know that is a FTC but I can't get all.
$endgroup$
– Thiago Alexandre
Mar 29 at 23:23
$begingroup$
I got it the item (b). You said for me defined $Psi: Y rightarrow X$ given by $Psi (f) = f'$. So with this we obtain $Phi circ Psi = Id_Y$ and $Psi circ Phi = Id_X$. The both we get by FTC. Now, I don't know how sequence $f_n in Y$ we need to take for show that $Psi = Phi^-1$ is not continuous.
$endgroup$
– Thiago Alexandre
Mar 30 at 2:19
1
$begingroup$
@ThiagoAlexandre, see mathcounterexamples.net/….
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 30 at 10:40
1
$begingroup$
@ThiagoAlexandre, $Phi circPsi = Id_Y$ is actually false (again, because $Phi(f)(0) = 0$).
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 30 at 10:43
$begingroup$
Thanks. I got it.
$endgroup$
– Thiago Alexandre
Mar 30 at 17:48
add a comment |
$begingroup$
(b) $Phi$ is injective but not surjective. For all $f$ we have $Phi(f)(0) = 0$. Obviously, $Phi(X)$ is strictly smaller than $Y$. But is true that (I will use another name) $Psi(f) = f'$ verifies $forall fin X: Psi(Phi((f)) = f$ and that $Psi$ is defined in the whole $Y$ (is an extension of $Phi^-1$).
(c) Take a sequence $f_nin Y$ s.t. $f_nto fin Y$ uniformly but $f_n'notto f'$ uniformly.
(d) I will use again the name $Psi$:
$$
d_infty(Psi(f),Psi(g)) = d_infty(f',g')le d_1,infty(cdots).
$$
answered Mar 29 at 20:53
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35.1k42971
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(b) $Phi$ is injective but not surjective. For all $f$ we have $Phi(f)(0) = 0$. Obviously, $Phi(X)$ is strictly smaller than $Y$. But is true that (I will use another name) $Psi(f) = f'$ verifies $forall fin X: Psi(Phi((f)) = f$ and that $Psi$ is defined in the whole $Y$ (is an extension of $Phi^-1$).
(c) Take a sequence $f_nin Y$ s.t. $f_nto fin Y$ uniformly but $f_n'notto f'$ uniformly.
(d) I will use again the name $Psi$:
$$
d_infty(Psi(f),Psi(g)) = d_infty(f',g')le d_1,infty(cdots).
$$
answered Mar 29 at 20:53
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35.1k42971
answered Mar 29 at 20:53
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35.1k42971
answered Mar 29 at 20:53
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35.1k42971
answered Mar 29 at 20:53
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35.1k42971
35.1k42971
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Thanks for your help me.I will focus first on item b. After that I will to (c) and (d). Well, You said for me that 1) $Phi$ is injective 2) $Phi$ is not surjective. 3) $forall f in X$, we have $Phi(f)(0) = 0$. 4) $Phi(X) subsetneqq Y$. 5) we have an extension $Psi $ of $Phi^-1$ that $Psi (Phi (f)) = f$. I can understand 3) and 5). The assertive 3) is obvious. And 5) I know that is a FTC but I can't get all.
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– Thiago Alexandre
Mar 29 at 23:23
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I got it the item (b). You said for me defined $Psi: Y rightarrow X$ given by $Psi (f) = f'$. So with this we obtain $Phi circ Psi = Id_Y$ and $Psi circ Phi = Id_X$. The both we get by FTC. Now, I don't know how sequence $f_n in Y$ we need to take for show that $Psi = Phi^-1$ is not continuous.
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– Thiago Alexandre
Mar 30 at 2:19
1
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@ThiagoAlexandre, see mathcounterexamples.net/….
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– Martín-Blas Pérez Pinilla
Mar 30 at 10:40
1
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@ThiagoAlexandre, $Phi circPsi = Id_Y$ is actually false (again, because $Phi(f)(0) = 0$).
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– Martín-Blas Pérez Pinilla
Mar 30 at 10:43
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Thanks. I got it.
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– Thiago Alexandre
Mar 30 at 17:48
add a comment |
$begingroup$
Thanks for your help me.I will focus first on item b. After that I will to (c) and (d). Well, You said for me that 1) $Phi$ is injective 2) $Phi$ is not surjective. 3) $forall f in X$, we have $Phi(f)(0) = 0$. 4) $Phi(X) subsetneqq Y$. 5) we have an extension $Psi $ of $Phi^-1$ that $Psi (Phi (f)) = f$. I can understand 3) and 5). The assertive 3) is obvious. And 5) I know that is a FTC but I can't get all.
$endgroup$
– Thiago Alexandre
Mar 29 at 23:23
$begingroup$
I got it the item (b). You said for me defined $Psi: Y rightarrow X$ given by $Psi (f) = f'$. So with this we obtain $Phi circ Psi = Id_Y$ and $Psi circ Phi = Id_X$. The both we get by FTC. Now, I don't know how sequence $f_n in Y$ we need to take for show that $Psi = Phi^-1$ is not continuous.
$endgroup$
– Thiago Alexandre
Mar 30 at 2:19
1
$begingroup$
@ThiagoAlexandre, see mathcounterexamples.net/….
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 30 at 10:40
1
$begingroup$
@ThiagoAlexandre, $Phi circPsi = Id_Y$ is actually false (again, because $Phi(f)(0) = 0$).
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 30 at 10:43
$begingroup$
Thanks. I got it.
$endgroup$
– Thiago Alexandre
Mar 30 at 17:48
$begingroup$
Thanks for your help me.I will focus first on item b. After that I will to (c) and (d). Well, You said for me that 1) $Phi$ is injective 2) $Phi$ is not surjective. 3) $forall f in X$, we have $Phi(f)(0) = 0$. 4) $Phi(X) subsetneqq Y$. 5) we have an extension $Psi $ of $Phi^-1$ that $Psi (Phi (f)) = f$. I can understand 3) and 5). The assertive 3) is obvious. And 5) I know that is a FTC but I can't get all.
$endgroup$
– Thiago Alexandre
Mar 29 at 23:23
$begingroup$
Thanks for your help me.I will focus first on item b. After that I will to (c) and (d). Well, You said for me that 1) $Phi$ is injective 2) $Phi$ is not surjective. 3) $forall f in X$, we have $Phi(f)(0) = 0$. 4) $Phi(X) subsetneqq Y$. 5) we have an extension $Psi $ of $Phi^-1$ that $Psi (Phi (f)) = f$. I can understand 3) and 5). The assertive 3) is obvious. And 5) I know that is a FTC but I can't get all.
$endgroup$
– Thiago Alexandre
Mar 29 at 23:23
$begingroup$
I got it the item (b). You said for me defined $Psi: Y rightarrow X$ given by $Psi (f) = f'$. So with this we obtain $Phi circ Psi = Id_Y$ and $Psi circ Phi = Id_X$. The both we get by FTC. Now, I don't know how sequence $f_n in Y$ we need to take for show that $Psi = Phi^-1$ is not continuous.
$endgroup$
– Thiago Alexandre
Mar 30 at 2:19
$begingroup$
I got it the item (b). You said for me defined $Psi: Y rightarrow X$ given by $Psi (f) = f'$. So with this we obtain $Phi circ Psi = Id_Y$ and $Psi circ Phi = Id_X$. The both we get by FTC. Now, I don't know how sequence $f_n in Y$ we need to take for show that $Psi = Phi^-1$ is not continuous.
$endgroup$
– Thiago Alexandre
Mar 30 at 2:19
1
1
$begingroup$
@ThiagoAlexandre, see mathcounterexamples.net/….
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 30 at 10:40
$begingroup$
@ThiagoAlexandre, see mathcounterexamples.net/….
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 30 at 10:40
1
1
$begingroup$
@ThiagoAlexandre, $Phi circPsi = Id_Y$ is actually false (again, because $Phi(f)(0) = 0$).
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 30 at 10:43
$begingroup$
@ThiagoAlexandre, $Phi circPsi = Id_Y$ is actually false (again, because $Phi(f)(0) = 0$).
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 30 at 10:43
$begingroup$
Thanks. I got it.
$endgroup$
– Thiago Alexandre
Mar 30 at 17:48
$begingroup$
Thanks. I got it.
$endgroup$
– Thiago Alexandre
Mar 30 at 17:48
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$begingroup$
For (b), (c), (d), what have you tried and where are you stuck?
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– Saad
Mar 29 at 16:47
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@Saad for (b) I know by FTC which for given $f in X$, we have $Phi$ such that $Phi' (x) = f(x), forall x in [0,1]$. So I need to show that $Phi^-1$ given by $Phi^-1(Phi(f)) = f(x) forall x$. I think that is correct by FTC. But I don't have sure and I don't know how to ensure that $Phi$ is a bijection function.
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– Thiago Alexandre
Mar 29 at 17:00