Functional analysis completeness proof
$begingroup$
Heres a proof for $(C^{0}[0,1],left | . right |_{infty })$ being complete.
In red: Why does $f_k$ being a cauchy sequence imply that it converges? I thought the implication only holds the other way around (i.e. all convergent sequences are cauchy).
In green: Why are they showing that $f_k$ converges to $f$ uniformly. In the usual proofs ive seen of completeness they let a sequence of a sequence (i.e $(f_k)_p$ converge to a function in the vector space wrt to the norm but here $f_k$ isnt a sequence of a sequence so im not sure why they've done that.
functional-analysis
asked Dec 14 '18 at 17:25
NoteBookNoteBook
1197
$endgroup$
add a comment |
$begingroup$
Heres a proof for $(C^{0}[0,1],left | . right |_{infty })$ being complete.
In red: Why does $f_k$ being a cauchy sequence imply that it converges? I thought the implication only holds the other way around (i.e. all convergent sequences are cauchy).
In green: Why are they showing that $f_k$ converges to $f$ uniformly. In the usual proofs ive seen of completeness they let a sequence of a sequence (i.e $(f_k)_p$ converge to a function in the vector space wrt to the norm but here $f_k$ isnt a sequence of a sequence so im not sure why they've done that.
functional-analysis
asked Dec 14 '18 at 17:25
NoteBookNoteBook
1197
$endgroup$
add a comment |
$begingroup$
Heres a proof for $(C^{0}[0,1],left | . right |_{infty })$ being complete.
In red: Why does $f_k$ being a cauchy sequence imply that it converges? I thought the implication only holds the other way around (i.e. all convergent sequences are cauchy).
In green: Why are they showing that $f_k$ converges to $f$ uniformly. In the usual proofs ive seen of completeness they let a sequence of a sequence (i.e $(f_k)_p$ converge to a function in the vector space wrt to the norm but here $f_k$ isnt a sequence of a sequence so im not sure why they've done that.
functional-analysis
asked Dec 14 '18 at 17:25
NoteBookNoteBook
1197
$endgroup$
Heres a proof for $(C^{0}[0,1],left | . right |_{infty })$ being complete.
In red: Why does $f_k$ being a cauchy sequence imply that it converges? I thought the implication only holds the other way around (i.e. all convergent sequences are cauchy).
In green: Why are they showing that $f_k$ converges to $f$ uniformly. In the usual proofs ive seen of completeness they let a sequence of a sequence (i.e $(f_k)_p$ converge to a function in the vector space wrt to the norm but here $f_k$ isnt a sequence of a sequence so im not sure why they've done that.
functional-analysis
functional-analysis
asked Dec 14 '18 at 17:25
NoteBookNoteBook
1197
asked Dec 14 '18 at 17:25
NoteBookNoteBook
1197
asked Dec 14 '18 at 17:25
NoteBookNoteBook
1197
asked Dec 14 '18 at 17:25
NoteBookNoteBook
1197
asked Dec 14 '18 at 17:25
NoteBookNoteBook
1197
1197
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
In red: We know $mathbb{R}$ (or $mathbb{C}$) is complete, so every Cuachy sequence of scalars converges to some scalar.
In green: We are showing that $(C[0,1],|cdot|_infty)$ is complete. By definition this means that every Cauchy sequence must converge to some function under the norm $|cdot|_infty$. This is also called uniform convergence.
answered Dec 14 '18 at 17:30
SmileyCraftSmileyCraft
3,591517
$endgroup$
$begingroup$
Maybe im missing the obvious here but how do we know they are talking about scalars, the vector space is the space of continuous functions so wouldnt $f_k$ be a cont. function? Also where did you get the R and C part from as its not mentioned anywhere in the proof. Im new to functional analysis so still need help with the basic
$endgroup$
– NoteBook
Dec 14 '18 at 17:45
$begingroup$
In red: We are looking at ${f_k(x)}$ which is a sequence of scalars. That $mathbb{R}$ is complete is almost literally an axiom, and that $mathbb{C}$ is complete follows from the completeness of $mathbb{R}$ and the definition of $mathbb{C}$.
$endgroup$
– SmileyCraft
Dec 14 '18 at 17:47
$begingroup$
i get that R and C are complete but how did you deduce that $f_k$ are scalars as it can be a sequence of any continuous functions correct? For example $f_1$ = x where k =1 and $f_2$ = 2x and so on as these are continuous functions for x in [0,1]... that may not be the best example as f is taken to be cauchy but hopefully you get my point
$endgroup$
– NoteBook
Dec 14 '18 at 19:05
1
$begingroup$
We fix some $x$. I quote "so $f_k(x)$ converges for each fixed $xin[0,1]$." Thus $f_k(x)$ does not represent the function $f_k$, but rather the function $f_k$ evaluated at the $x$ that we have fixed.
$endgroup$
– SmileyCraft
Dec 14 '18 at 19:22
add a comment |
$begingroup$
In red: because it s puntual convergente and being Cauchy in $mathbb R$ is equivalente to being convergent.
In green: because convergente in $|cdot|_infty$ is exactly The uniform convergence.
answered Dec 14 '18 at 17:33
Tito EliatronTito Eliatron
1,601622
$endgroup$
$begingroup$
If I understand your first point properly, you're saying that because $f_n$ is pointwise convergent, you have that $f_n(x)$ is convergent in $mathbb{R}$, and hence Cauchy. But, if that's the case, there is no such assumption that $f_n(x)$ pointwise converges. The Cauchiness is proven from the uniform Cauchiness, and convergence is deduced from completeness.
$endgroup$
– Theo Bendit
Dec 14 '18 at 18:22
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
$begingroup$
In red: We know $mathbb{R}$ (or $mathbb{C}$) is complete, so every Cuachy sequence of scalars converges to some scalar.
In green: We are showing that $(C[0,1],|cdot|_infty)$ is complete. By definition this means that every Cauchy sequence must converge to some function under the norm $|cdot|_infty$. This is also called uniform convergence.
answered Dec 14 '18 at 17:30
SmileyCraftSmileyCraft
3,591517
$endgroup$
$begingroup$
Maybe im missing the obvious here but how do we know they are talking about scalars, the vector space is the space of continuous functions so wouldnt $f_k$ be a cont. function? Also where did you get the R and C part from as its not mentioned anywhere in the proof. Im new to functional analysis so still need help with the basic
$endgroup$
– NoteBook
Dec 14 '18 at 17:45
$begingroup$
In red: We are looking at ${f_k(x)}$ which is a sequence of scalars. That $mathbb{R}$ is complete is almost literally an axiom, and that $mathbb{C}$ is complete follows from the completeness of $mathbb{R}$ and the definition of $mathbb{C}$.
$endgroup$
– SmileyCraft
Dec 14 '18 at 17:47
$begingroup$
i get that R and C are complete but how did you deduce that $f_k$ are scalars as it can be a sequence of any continuous functions correct? For example $f_1$ = x where k =1 and $f_2$ = 2x and so on as these are continuous functions for x in [0,1]... that may not be the best example as f is taken to be cauchy but hopefully you get my point
$endgroup$
– NoteBook
Dec 14 '18 at 19:05
1
$begingroup$
We fix some $x$. I quote "so $f_k(x)$ converges for each fixed $xin[0,1]$." Thus $f_k(x)$ does not represent the function $f_k$, but rather the function $f_k$ evaluated at the $x$ that we have fixed.
$endgroup$
– SmileyCraft
Dec 14 '18 at 19:22
add a comment |
$begingroup$
In red: We know $mathbb{R}$ (or $mathbb{C}$) is complete, so every Cuachy sequence of scalars converges to some scalar.
In green: We are showing that $(C[0,1],|cdot|_infty)$ is complete. By definition this means that every Cauchy sequence must converge to some function under the norm $|cdot|_infty$. This is also called uniform convergence.
answered Dec 14 '18 at 17:30
SmileyCraftSmileyCraft
3,591517
$endgroup$
$begingroup$
Maybe im missing the obvious here but how do we know they are talking about scalars, the vector space is the space of continuous functions so wouldnt $f_k$ be a cont. function? Also where did you get the R and C part from as its not mentioned anywhere in the proof. Im new to functional analysis so still need help with the basic
$endgroup$
– NoteBook
Dec 14 '18 at 17:45
$begingroup$
In red: We are looking at ${f_k(x)}$ which is a sequence of scalars. That $mathbb{R}$ is complete is almost literally an axiom, and that $mathbb{C}$ is complete follows from the completeness of $mathbb{R}$ and the definition of $mathbb{C}$.
$endgroup$
– SmileyCraft
Dec 14 '18 at 17:47
$begingroup$
i get that R and C are complete but how did you deduce that $f_k$ are scalars as it can be a sequence of any continuous functions correct? For example $f_1$ = x where k =1 and $f_2$ = 2x and so on as these are continuous functions for x in [0,1]... that may not be the best example as f is taken to be cauchy but hopefully you get my point
$endgroup$
– NoteBook
Dec 14 '18 at 19:05
1
$begingroup$
We fix some $x$. I quote "so $f_k(x)$ converges for each fixed $xin[0,1]$." Thus $f_k(x)$ does not represent the function $f_k$, but rather the function $f_k$ evaluated at the $x$ that we have fixed.
$endgroup$
– SmileyCraft
Dec 14 '18 at 19:22
add a comment |
$begingroup$
In red: We know $mathbb{R}$ (or $mathbb{C}$) is complete, so every Cuachy sequence of scalars converges to some scalar.
In green: We are showing that $(C[0,1],|cdot|_infty)$ is complete. By definition this means that every Cauchy sequence must converge to some function under the norm $|cdot|_infty$. This is also called uniform convergence.
answered Dec 14 '18 at 17:30
SmileyCraftSmileyCraft
3,591517
$endgroup$
In red: We know $mathbb{R}$ (or $mathbb{C}$) is complete, so every Cuachy sequence of scalars converges to some scalar.
In green: We are showing that $(C[0,1],|cdot|_infty)$ is complete. By definition this means that every Cauchy sequence must converge to some function under the norm $|cdot|_infty$. This is also called uniform convergence.
answered Dec 14 '18 at 17:30
SmileyCraftSmileyCraft
3,591517
answered Dec 14 '18 at 17:30
SmileyCraftSmileyCraft
3,591517
answered Dec 14 '18 at 17:30
SmileyCraftSmileyCraft
3,591517
answered Dec 14 '18 at 17:30
SmileyCraftSmileyCraft
3,591517
3,591517
$begingroup$
Maybe im missing the obvious here but how do we know they are talking about scalars, the vector space is the space of continuous functions so wouldnt $f_k$ be a cont. function? Also where did you get the R and C part from as its not mentioned anywhere in the proof. Im new to functional analysis so still need help with the basic
$endgroup$
– NoteBook
Dec 14 '18 at 17:45
$begingroup$
In red: We are looking at ${f_k(x)}$ which is a sequence of scalars. That $mathbb{R}$ is complete is almost literally an axiom, and that $mathbb{C}$ is complete follows from the completeness of $mathbb{R}$ and the definition of $mathbb{C}$.
$endgroup$
– SmileyCraft
Dec 14 '18 at 17:47
$begingroup$
i get that R and C are complete but how did you deduce that $f_k$ are scalars as it can be a sequence of any continuous functions correct? For example $f_1$ = x where k =1 and $f_2$ = 2x and so on as these are continuous functions for x in [0,1]... that may not be the best example as f is taken to be cauchy but hopefully you get my point
$endgroup$
– NoteBook
Dec 14 '18 at 19:05
1
$begingroup$
We fix some $x$. I quote "so $f_k(x)$ converges for each fixed $xin[0,1]$." Thus $f_k(x)$ does not represent the function $f_k$, but rather the function $f_k$ evaluated at the $x$ that we have fixed.
$endgroup$
– SmileyCraft
Dec 14 '18 at 19:22
add a comment |
$begingroup$
Maybe im missing the obvious here but how do we know they are talking about scalars, the vector space is the space of continuous functions so wouldnt $f_k$ be a cont. function? Also where did you get the R and C part from as its not mentioned anywhere in the proof. Im new to functional analysis so still need help with the basic
$endgroup$
– NoteBook
Dec 14 '18 at 17:45
$begingroup$
In red: We are looking at ${f_k(x)}$ which is a sequence of scalars. That $mathbb{R}$ is complete is almost literally an axiom, and that $mathbb{C}$ is complete follows from the completeness of $mathbb{R}$ and the definition of $mathbb{C}$.
$endgroup$
– SmileyCraft
Dec 14 '18 at 17:47
$begingroup$
i get that R and C are complete but how did you deduce that $f_k$ are scalars as it can be a sequence of any continuous functions correct? For example $f_1$ = x where k =1 and $f_2$ = 2x and so on as these are continuous functions for x in [0,1]... that may not be the best example as f is taken to be cauchy but hopefully you get my point
$endgroup$
– NoteBook
Dec 14 '18 at 19:05
1
$begingroup$
We fix some $x$. I quote "so $f_k(x)$ converges for each fixed $xin[0,1]$." Thus $f_k(x)$ does not represent the function $f_k$, but rather the function $f_k$ evaluated at the $x$ that we have fixed.
$endgroup$
– SmileyCraft
Dec 14 '18 at 19:22
$begingroup$
Maybe im missing the obvious here but how do we know they are talking about scalars, the vector space is the space of continuous functions so wouldnt $f_k$ be a cont. function? Also where did you get the R and C part from as its not mentioned anywhere in the proof. Im new to functional analysis so still need help with the basic
$endgroup$
– NoteBook
Dec 14 '18 at 17:45
$begingroup$
Maybe im missing the obvious here but how do we know they are talking about scalars, the vector space is the space of continuous functions so wouldnt $f_k$ be a cont. function? Also where did you get the R and C part from as its not mentioned anywhere in the proof. Im new to functional analysis so still need help with the basic
$endgroup$
– NoteBook
Dec 14 '18 at 17:45
$begingroup$
In red: We are looking at ${f_k(x)}$ which is a sequence of scalars. That $mathbb{R}$ is complete is almost literally an axiom, and that $mathbb{C}$ is complete follows from the completeness of $mathbb{R}$ and the definition of $mathbb{C}$.
$endgroup$
– SmileyCraft
Dec 14 '18 at 17:47
$begingroup$
In red: We are looking at ${f_k(x)}$ which is a sequence of scalars. That $mathbb{R}$ is complete is almost literally an axiom, and that $mathbb{C}$ is complete follows from the completeness of $mathbb{R}$ and the definition of $mathbb{C}$.
$endgroup$
– SmileyCraft
Dec 14 '18 at 17:47
$begingroup$
i get that R and C are complete but how did you deduce that $f_k$ are scalars as it can be a sequence of any continuous functions correct? For example $f_1$ = x where k =1 and $f_2$ = 2x and so on as these are continuous functions for x in [0,1]... that may not be the best example as f is taken to be cauchy but hopefully you get my point
$endgroup$
– NoteBook
Dec 14 '18 at 19:05
$begingroup$
i get that R and C are complete but how did you deduce that $f_k$ are scalars as it can be a sequence of any continuous functions correct? For example $f_1$ = x where k =1 and $f_2$ = 2x and so on as these are continuous functions for x in [0,1]... that may not be the best example as f is taken to be cauchy but hopefully you get my point
$endgroup$
– NoteBook
Dec 14 '18 at 19:05
1
1
$begingroup$
We fix some $x$. I quote "so $f_k(x)$ converges for each fixed $xin[0,1]$." Thus $f_k(x)$ does not represent the function $f_k$, but rather the function $f_k$ evaluated at the $x$ that we have fixed.
$endgroup$
– SmileyCraft
Dec 14 '18 at 19:22
$begingroup$
We fix some $x$. I quote "so $f_k(x)$ converges for each fixed $xin[0,1]$." Thus $f_k(x)$ does not represent the function $f_k$, but rather the function $f_k$ evaluated at the $x$ that we have fixed.
$endgroup$
– SmileyCraft
Dec 14 '18 at 19:22
add a comment |
$begingroup$
In red: because it s puntual convergente and being Cauchy in $mathbb R$ is equivalente to being convergent.
In green: because convergente in $|cdot|_infty$ is exactly The uniform convergence.
answered Dec 14 '18 at 17:33
Tito EliatronTito Eliatron
1,601622
$endgroup$
$begingroup$
If I understand your first point properly, you're saying that because $f_n$ is pointwise convergent, you have that $f_n(x)$ is convergent in $mathbb{R}$, and hence Cauchy. But, if that's the case, there is no such assumption that $f_n(x)$ pointwise converges. The Cauchiness is proven from the uniform Cauchiness, and convergence is deduced from completeness.
$endgroup$
– Theo Bendit
Dec 14 '18 at 18:22
add a comment |
$begingroup$
In red: because it s puntual convergente and being Cauchy in $mathbb R$ is equivalente to being convergent.
In green: because convergente in $|cdot|_infty$ is exactly The uniform convergence.
answered Dec 14 '18 at 17:33
Tito EliatronTito Eliatron
1,601622
$endgroup$
$begingroup$
If I understand your first point properly, you're saying that because $f_n$ is pointwise convergent, you have that $f_n(x)$ is convergent in $mathbb{R}$, and hence Cauchy. But, if that's the case, there is no such assumption that $f_n(x)$ pointwise converges. The Cauchiness is proven from the uniform Cauchiness, and convergence is deduced from completeness.
$endgroup$
– Theo Bendit
Dec 14 '18 at 18:22
add a comment |
$begingroup$
In red: because it s puntual convergente and being Cauchy in $mathbb R$ is equivalente to being convergent.
In green: because convergente in $|cdot|_infty$ is exactly The uniform convergence.
answered Dec 14 '18 at 17:33
Tito EliatronTito Eliatron
1,601622
$endgroup$
In red: because it s puntual convergente and being Cauchy in $mathbb R$ is equivalente to being convergent.
In green: because convergente in $|cdot|_infty$ is exactly The uniform convergence.
answered Dec 14 '18 at 17:33
Tito EliatronTito Eliatron
1,601622
answered Dec 14 '18 at 17:33
Tito EliatronTito Eliatron
1,601622
answered Dec 14 '18 at 17:33
Tito EliatronTito Eliatron
1,601622
answered Dec 14 '18 at 17:33
Tito EliatronTito Eliatron
1,601622
1,601622
$begingroup$
If I understand your first point properly, you're saying that because $f_n$ is pointwise convergent, you have that $f_n(x)$ is convergent in $mathbb{R}$, and hence Cauchy. But, if that's the case, there is no such assumption that $f_n(x)$ pointwise converges. The Cauchiness is proven from the uniform Cauchiness, and convergence is deduced from completeness.
$endgroup$
– Theo Bendit
Dec 14 '18 at 18:22
add a comment |
$begingroup$
If I understand your first point properly, you're saying that because $f_n$ is pointwise convergent, you have that $f_n(x)$ is convergent in $mathbb{R}$, and hence Cauchy. But, if that's the case, there is no such assumption that $f_n(x)$ pointwise converges. The Cauchiness is proven from the uniform Cauchiness, and convergence is deduced from completeness.
$endgroup$
– Theo Bendit
Dec 14 '18 at 18:22
$begingroup$
If I understand your first point properly, you're saying that because $f_n$ is pointwise convergent, you have that $f_n(x)$ is convergent in $mathbb{R}$, and hence Cauchy. But, if that's the case, there is no such assumption that $f_n(x)$ pointwise converges. The Cauchiness is proven from the uniform Cauchiness, and convergence is deduced from completeness.
$endgroup$
– Theo Bendit
Dec 14 '18 at 18:22
$begingroup$
If I understand your first point properly, you're saying that because $f_n$ is pointwise convergent, you have that $f_n(x)$ is convergent in $mathbb{R}$, and hence Cauchy. But, if that's the case, there is no such assumption that $f_n(x)$ pointwise converges. The Cauchiness is proven from the uniform Cauchiness, and convergence is deduced from completeness.
$endgroup$
– Theo Bendit
Dec 14 '18 at 18:22
add a comment |
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