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Is my argument true : prove that $int_0^infty fracflambda +gsimfrac1lambda int_0^infty f.$
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Let $fgeq 0$ continuous and integrable over $[0,infty [$ and $ggeq 0$ continuous and bounded $geq 0$ over $[0,infty [$. Let $lambda >0$. Set $$varphi(lambda )=int_0^infty fracflambda +g.$$
Prove that $$varphi(lambda )sim_lambda to infty frac1lambda int_0^infty f.$$
Can I do as following : Since $g$ is bounded,
$$g(x)=mathcal O(1),$$
and thus $$lambda +g(x)=lambda +mathcal O(1)=lambda left(1+mathcal Oleft(frac1lambda right)right)=lambda (1+o(1)).$$
Therefore $$fracflambda +g=fracflambda left(1+o(1)right)=fracflambda left(1+o(1)right),$$
and thus $$int_0^infty fracflambda +g=frac1lambda int_0^infty f+oleft(frac1lambda int_0^infty fright),$$
and thus $$varphi(lambda )sim_lambda to infty frac1lambda int_0^infty f.$$
Is it working ?
Added
We have when $lambda to infty $ that $$fracflambda+g=fracflambda + oleft(fracflambda right).$$
I set $R(lambda,x )=oleft(fracf(x)lambda right)$
Let $varepsilon>0$. There is $M>0$ s.t. $$|R(lambda,x )|leq varepsilonfracflambda ,$$
when $lambda geq M$. Then $$int_0^infty |R(x,lambda )|leq fracvarepsilonlambda int_0^infty f(x),dx,$$
and thus $$int_0^infty R(x,lambda ),dx = oleft(frac1lambda int_0^infty f(x),dxright).$$
real-analysis integration proof-verification asymptotics
edited Jul 18 at 19:28
Michael Hardy
204k23186462
asked Jul 18 at 12:33
Peter
348112
 |Â
show 23 more comments
up vote
0
down vote
favorite
Let $fgeq 0$ continuous and integrable over $[0,infty [$ and $ggeq 0$ continuous and bounded $geq 0$ over $[0,infty [$. Let $lambda >0$. Set $$varphi(lambda )=int_0^infty fracflambda +g.$$
Prove that $$varphi(lambda )sim_lambda to infty frac1lambda int_0^infty f.$$
Can I do as following : Since $g$ is bounded,
$$g(x)=mathcal O(1),$$
and thus $$lambda +g(x)=lambda +mathcal O(1)=lambda left(1+mathcal Oleft(frac1lambda right)right)=lambda (1+o(1)).$$
Therefore $$fracflambda +g=fracflambda left(1+o(1)right)=fracflambda left(1+o(1)right),$$
and thus $$int_0^infty fracflambda +g=frac1lambda int_0^infty f+oleft(frac1lambda int_0^infty fright),$$
and thus $$varphi(lambda )sim_lambda to infty frac1lambda int_0^infty f.$$
Is it working ?
Added
We have when $lambda to infty $ that $$fracflambda+g=fracflambda + oleft(fracflambda right).$$
I set $R(lambda,x )=oleft(fracf(x)lambda right)$
Let $varepsilon>0$. There is $M>0$ s.t. $$|R(lambda,x )|leq varepsilonfracflambda ,$$
when $lambda geq M$. Then $$int_0^infty |R(x,lambda )|leq fracvarepsilonlambda int_0^infty f(x),dx,$$
and thus $$int_0^infty R(x,lambda ),dx = oleft(frac1lambda int_0^infty f(x),dxright).$$
real-analysis integration proof-verification asymptotics
edited Jul 18 at 19:28
Michael Hardy
204k23186462
asked Jul 18 at 12:33
Peter
348112
The argument is obviously wrong, for example when you replace a big-O by a little-o, and much too vague, for example when you integrate a little-o on an unbounded interval. For a proof, simply use $$lambdaleqslantlambda+gleqslantlambda+C$$ for some given finite $C$ and integrate.
– Did
Jul 18 at 12:39
1
As already mentioned, integrating some little-o term on some unbounded interval (or even a nonuniform little-o term on a bounded interval) can lead to surprises. Exercise: Find some regular functions $h_n$ such that $h_n(x)to0$ when $ntoinfty$, for every $x$ in $[0,1]$ (that is, $h_n(x)=o(1)$, right?), and $int_0^1h_n(x)dx$ does not converge to $0$.
– Did
Jul 18 at 12:50
1
As I said already, either you know the reasons why this holds (that $g$ is bounded) and then you could as well use this condition from the start, or you do not, and then such strings of characters can lead you into chaos. IOW, if you want a clean proof, drop the os and Os and work with inequalities valid on (0,oo)...
– Did
Jul 18 at 15:47
1
Q1) Too many posts, I do not know. Q2) For each fixed $x$, yes. Q3) Yes there is a problem, one of uniformity, as already explained.
– Did
Jul 18 at 18:42
1
I find your claim in the final edit (the one that starts with "There is M>0 s.t...." not properly justified. The main criticism with the type of proof you seem hellbent on making is that any complete proof (i.e. which contains all necessary justifications) of that form will necessarily contain the simpler proofs you've been recommended. It will also, then, contain a lot of unnecessary parts and be significantly less clear.
– Antonio Vargas
Jul 18 at 23:10
 |Â
show 23 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $fgeq 0$ continuous and integrable over $[0,infty [$ and $ggeq 0$ continuous and bounded $geq 0$ over $[0,infty [$. Let $lambda >0$. Set $$varphi(lambda )=int_0^infty fracflambda +g.$$
Prove that $$varphi(lambda )sim_lambda to infty frac1lambda int_0^infty f.$$
Can I do as following : Since $g$ is bounded,
$$g(x)=mathcal O(1),$$
and thus $$lambda +g(x)=lambda +mathcal O(1)=lambda left(1+mathcal Oleft(frac1lambda right)right)=lambda (1+o(1)).$$
Therefore $$fracflambda +g=fracflambda left(1+o(1)right)=fracflambda left(1+o(1)right),$$
and thus $$int_0^infty fracflambda +g=frac1lambda int_0^infty f+oleft(frac1lambda int_0^infty fright),$$
and thus $$varphi(lambda )sim_lambda to infty frac1lambda int_0^infty f.$$
Is it working ?
Added
We have when $lambda to infty $ that $$fracflambda+g=fracflambda + oleft(fracflambda right).$$
I set $R(lambda,x )=oleft(fracf(x)lambda right)$
Let $varepsilon>0$. There is $M>0$ s.t. $$|R(lambda,x )|leq varepsilonfracflambda ,$$
when $lambda geq M$. Then $$int_0^infty |R(x,lambda )|leq fracvarepsilonlambda int_0^infty f(x),dx,$$
and thus $$int_0^infty R(x,lambda ),dx = oleft(frac1lambda int_0^infty f(x),dxright).$$
real-analysis integration proof-verification asymptotics
edited Jul 18 at 19:28
Michael Hardy
204k23186462
asked Jul 18 at 12:33
Peter
348112
Let $fgeq 0$ continuous and integrable over $[0,infty [$ and $ggeq 0$ continuous and bounded $geq 0$ over $[0,infty [$. Let $lambda >0$. Set $$varphi(lambda )=int_0^infty fracflambda +g.$$
Prove that $$varphi(lambda )sim_lambda to infty frac1lambda int_0^infty f.$$
Can I do as following : Since $g$ is bounded,
$$g(x)=mathcal O(1),$$
and thus $$lambda +g(x)=lambda +mathcal O(1)=lambda left(1+mathcal Oleft(frac1lambda right)right)=lambda (1+o(1)).$$
Therefore $$fracflambda +g=fracflambda left(1+o(1)right)=fracflambda left(1+o(1)right),$$
and thus $$int_0^infty fracflambda +g=frac1lambda int_0^infty f+oleft(frac1lambda int_0^infty fright),$$
and thus $$varphi(lambda )sim_lambda to infty frac1lambda int_0^infty f.$$
Is it working ?
Added
We have when $lambda to infty $ that $$fracflambda+g=fracflambda + oleft(fracflambda right).$$
I set $R(lambda,x )=oleft(fracf(x)lambda right)$
Let $varepsilon>0$. There is $M>0$ s.t. $$|R(lambda,x )|leq varepsilonfracflambda ,$$
when $lambda geq M$. Then $$int_0^infty |R(x,lambda )|leq fracvarepsilonlambda int_0^infty f(x),dx,$$
and thus $$int_0^infty R(x,lambda ),dx = oleft(frac1lambda int_0^infty f(x),dxright).$$
real-analysis integration proof-verification asymptotics
edited Jul 18 at 19:28
Michael Hardy
204k23186462
asked Jul 18 at 12:33
Peter
348112
edited Jul 18 at 19:28
Michael Hardy
204k23186462
edited Jul 18 at 19:28
Michael Hardy
204k23186462
edited Jul 18 at 19:28
Michael Hardy
204k23186462
204k23186462
asked Jul 18 at 12:33
Peter
348112
asked Jul 18 at 12:33
Peter
348112
asked Jul 18 at 12:33
Peter
348112
348112
The argument is obviously wrong, for example when you replace a big-O by a little-o, and much too vague, for example when you integrate a little-o on an unbounded interval. For a proof, simply use $$lambdaleqslantlambda+gleqslantlambda+C$$ for some given finite $C$ and integrate.
– Did
Jul 18 at 12:39
1
As already mentioned, integrating some little-o term on some unbounded interval (or even a nonuniform little-o term on a bounded interval) can lead to surprises. Exercise: Find some regular functions $h_n$ such that $h_n(x)to0$ when $ntoinfty$, for every $x$ in $[0,1]$ (that is, $h_n(x)=o(1)$, right?), and $int_0^1h_n(x)dx$ does not converge to $0$.
– Did
Jul 18 at 12:50
1
As I said already, either you know the reasons why this holds (that $g$ is bounded) and then you could as well use this condition from the start, or you do not, and then such strings of characters can lead you into chaos. IOW, if you want a clean proof, drop the os and Os and work with inequalities valid on (0,oo)...
– Did
Jul 18 at 15:47
1
Q1) Too many posts, I do not know. Q2) For each fixed $x$, yes. Q3) Yes there is a problem, one of uniformity, as already explained.
– Did
Jul 18 at 18:42
1
I find your claim in the final edit (the one that starts with "There is M>0 s.t...." not properly justified. The main criticism with the type of proof you seem hellbent on making is that any complete proof (i.e. which contains all necessary justifications) of that form will necessarily contain the simpler proofs you've been recommended. It will also, then, contain a lot of unnecessary parts and be significantly less clear.
– Antonio Vargas
Jul 18 at 23:10
 |Â
show 23 more comments
The argument is obviously wrong, for example when you replace a big-O by a little-o, and much too vague, for example when you integrate a little-o on an unbounded interval. For a proof, simply use $$lambdaleqslantlambda+gleqslantlambda+C$$ for some given finite $C$ and integrate.
– Did
Jul 18 at 12:39
1
As already mentioned, integrating some little-o term on some unbounded interval (or even a nonuniform little-o term on a bounded interval) can lead to surprises. Exercise: Find some regular functions $h_n$ such that $h_n(x)to0$ when $ntoinfty$, for every $x$ in $[0,1]$ (that is, $h_n(x)=o(1)$, right?), and $int_0^1h_n(x)dx$ does not converge to $0$.
– Did
Jul 18 at 12:50
1
As I said already, either you know the reasons why this holds (that $g$ is bounded) and then you could as well use this condition from the start, or you do not, and then such strings of characters can lead you into chaos. IOW, if you want a clean proof, drop the os and Os and work with inequalities valid on (0,oo)...
– Did
Jul 18 at 15:47
1
Q1) Too many posts, I do not know. Q2) For each fixed $x$, yes. Q3) Yes there is a problem, one of uniformity, as already explained.
– Did
Jul 18 at 18:42
1
I find your claim in the final edit (the one that starts with "There is M>0 s.t...." not properly justified. The main criticism with the type of proof you seem hellbent on making is that any complete proof (i.e. which contains all necessary justifications) of that form will necessarily contain the simpler proofs you've been recommended. It will also, then, contain a lot of unnecessary parts and be significantly less clear.
– Antonio Vargas
Jul 18 at 23:10
The argument is obviously wrong, for example when you replace a big-O by a little-o, and much too vague, for example when you integrate a little-o on an unbounded interval. For a proof, simply use $$lambdaleqslantlambda+gleqslantlambda+C$$ for some given finite $C$ and integrate.
– Did
Jul 18 at 12:39
The argument is obviously wrong, for example when you replace a big-O by a little-o, and much too vague, for example when you integrate a little-o on an unbounded interval. For a proof, simply use $$lambdaleqslantlambda+gleqslantlambda+C$$ for some given finite $C$ and integrate.
– Did
Jul 18 at 12:39
1
1
As already mentioned, integrating some little-o term on some unbounded interval (or even a nonuniform little-o term on a bounded interval) can lead to surprises. Exercise: Find some regular functions $h_n$ such that $h_n(x)to0$ when $ntoinfty$, for every $x$ in $[0,1]$ (that is, $h_n(x)=o(1)$, right?), and $int_0^1h_n(x)dx$ does not converge to $0$.
– Did
Jul 18 at 12:50
As already mentioned, integrating some little-o term on some unbounded interval (or even a nonuniform little-o term on a bounded interval) can lead to surprises. Exercise: Find some regular functions $h_n$ such that $h_n(x)to0$ when $ntoinfty$, for every $x$ in $[0,1]$ (that is, $h_n(x)=o(1)$, right?), and $int_0^1h_n(x)dx$ does not converge to $0$.
– Did
Jul 18 at 12:50
1
1
As I said already, either you know the reasons why this holds (that $g$ is bounded) and then you could as well use this condition from the start, or you do not, and then such strings of characters can lead you into chaos. IOW, if you want a clean proof, drop the os and Os and work with inequalities valid on (0,oo)...
– Did
Jul 18 at 15:47
As I said already, either you know the reasons why this holds (that $g$ is bounded) and then you could as well use this condition from the start, or you do not, and then such strings of characters can lead you into chaos. IOW, if you want a clean proof, drop the os and Os and work with inequalities valid on (0,oo)...
– Did
Jul 18 at 15:47
1
1
Q1) Too many posts, I do not know. Q2) For each fixed $x$, yes. Q3) Yes there is a problem, one of uniformity, as already explained.
– Did
Jul 18 at 18:42
Q1) Too many posts, I do not know. Q2) For each fixed $x$, yes. Q3) Yes there is a problem, one of uniformity, as already explained.
– Did
Jul 18 at 18:42
1
1
I find your claim in the final edit (the one that starts with "There is M>0 s.t...." not properly justified. The main criticism with the type of proof you seem hellbent on making is that any complete proof (i.e. which contains all necessary justifications) of that form will necessarily contain the simpler proofs you've been recommended. It will also, then, contain a lot of unnecessary parts and be significantly less clear.
– Antonio Vargas
Jul 18 at 23:10
I find your claim in the final edit (the one that starts with "There is M>0 s.t...." not properly justified. The main criticism with the type of proof you seem hellbent on making is that any complete proof (i.e. which contains all necessary justifications) of that form will necessarily contain the simpler proofs you've been recommended. It will also, then, contain a lot of unnecessary parts and be significantly less clear.
– Antonio Vargas
Jul 18 at 23:10
 |Â
show 23 more comments
1 Answer
1
active
oldest
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up vote
2
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accepted
There exists some $M$ such that, for every $xgeq 0$, $0leq g(x)leq M$.
For any $x$, $displaystyle left|frac1lambda +g(x)-frac1lambda right|= fracg(x)lambda(lambda+g(x))leq frac g(x)lambda^2leq frac Mlambda^2$
Thus $$
left|int_0^infty fracflambda +g-fracflambda right|leq int_0^inftyfleft|frac1lambda +g-frac1lambda right|leq fracMlambda^2int_0^inftyf$$
Hence $$int_0^infty fracflambda +g=int_0^inftyfracflambda + Oleft( frac1lambda^2right)=int_0^inftyfracflambda + oleft( frac1lambdaright)$$
edited Jul 18 at 18:45
Did
242k23208443
answered Jul 18 at 13:11
Gabriel Romon
17k43081
at your last equality, is it correct that $$int_0^infty fracflambda +oleft(frac1lambda right)=int_0^infty fracflambda +oleft(frac1lambda int_0^infty fright)$$ ?
– Peter
Jul 19 at 8:37
1
@Peter Yes. In general, if $cneq 0$ is a constant and $g(x)$ is a function, $o(cg(x)) = o(g(x))$. This is very easy to prove if you go back to the definition of little-o.
– Gabriel Romon
Jul 19 at 9:02
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
There exists some $M$ such that, for every $xgeq 0$, $0leq g(x)leq M$.
For any $x$, $displaystyle left|frac1lambda +g(x)-frac1lambda right|= fracg(x)lambda(lambda+g(x))leq frac g(x)lambda^2leq frac Mlambda^2$
Thus $$
left|int_0^infty fracflambda +g-fracflambda right|leq int_0^inftyfleft|frac1lambda +g-frac1lambda right|leq fracMlambda^2int_0^inftyf$$
Hence $$int_0^infty fracflambda +g=int_0^inftyfracflambda + Oleft( frac1lambda^2right)=int_0^inftyfracflambda + oleft( frac1lambdaright)$$
edited Jul 18 at 18:45
Did
242k23208443
answered Jul 18 at 13:11
Gabriel Romon
17k43081
at your last equality, is it correct that $$int_0^infty fracflambda +oleft(frac1lambda right)=int_0^infty fracflambda +oleft(frac1lambda int_0^infty fright)$$ ?
– Peter
Jul 19 at 8:37
1
@Peter Yes. In general, if $cneq 0$ is a constant and $g(x)$ is a function, $o(cg(x)) = o(g(x))$. This is very easy to prove if you go back to the definition of little-o.
– Gabriel Romon
Jul 19 at 9:02
add a comment |Â
up vote
2
down vote
accepted
There exists some $M$ such that, for every $xgeq 0$, $0leq g(x)leq M$.
For any $x$, $displaystyle left|frac1lambda +g(x)-frac1lambda right|= fracg(x)lambda(lambda+g(x))leq frac g(x)lambda^2leq frac Mlambda^2$
Thus $$
left|int_0^infty fracflambda +g-fracflambda right|leq int_0^inftyfleft|frac1lambda +g-frac1lambda right|leq fracMlambda^2int_0^inftyf$$
Hence $$int_0^infty fracflambda +g=int_0^inftyfracflambda + Oleft( frac1lambda^2right)=int_0^inftyfracflambda + oleft( frac1lambdaright)$$
edited Jul 18 at 18:45
Did
242k23208443
answered Jul 18 at 13:11
Gabriel Romon
17k43081
at your last equality, is it correct that $$int_0^infty fracflambda +oleft(frac1lambda right)=int_0^infty fracflambda +oleft(frac1lambda int_0^infty fright)$$ ?
– Peter
Jul 19 at 8:37
1
@Peter Yes. In general, if $cneq 0$ is a constant and $g(x)$ is a function, $o(cg(x)) = o(g(x))$. This is very easy to prove if you go back to the definition of little-o.
– Gabriel Romon
Jul 19 at 9:02
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
There exists some $M$ such that, for every $xgeq 0$, $0leq g(x)leq M$.
For any $x$, $displaystyle left|frac1lambda +g(x)-frac1lambda right|= fracg(x)lambda(lambda+g(x))leq frac g(x)lambda^2leq frac Mlambda^2$
Thus $$
left|int_0^infty fracflambda +g-fracflambda right|leq int_0^inftyfleft|frac1lambda +g-frac1lambda right|leq fracMlambda^2int_0^inftyf$$
Hence $$int_0^infty fracflambda +g=int_0^inftyfracflambda + Oleft( frac1lambda^2right)=int_0^inftyfracflambda + oleft( frac1lambdaright)$$
edited Jul 18 at 18:45
Did
242k23208443
answered Jul 18 at 13:11
Gabriel Romon
17k43081
There exists some $M$ such that, for every $xgeq 0$, $0leq g(x)leq M$.
For any $x$, $displaystyle left|frac1lambda +g(x)-frac1lambda right|= fracg(x)lambda(lambda+g(x))leq frac g(x)lambda^2leq frac Mlambda^2$
Thus $$
left|int_0^infty fracflambda +g-fracflambda right|leq int_0^inftyfleft|frac1lambda +g-frac1lambda right|leq fracMlambda^2int_0^inftyf$$
Hence $$int_0^infty fracflambda +g=int_0^inftyfracflambda + Oleft( frac1lambda^2right)=int_0^inftyfracflambda + oleft( frac1lambdaright)$$
edited Jul 18 at 18:45
Did
242k23208443
answered Jul 18 at 13:11
Gabriel Romon
17k43081
edited Jul 18 at 18:45
Did
242k23208443
edited Jul 18 at 18:45
Did
242k23208443
edited Jul 18 at 18:45
Did
242k23208443
242k23208443
answered Jul 18 at 13:11
Gabriel Romon
17k43081
answered Jul 18 at 13:11
Gabriel Romon
17k43081
answered Jul 18 at 13:11
Gabriel Romon
17k43081
17k43081
at your last equality, is it correct that $$int_0^infty fracflambda +oleft(frac1lambda right)=int_0^infty fracflambda +oleft(frac1lambda int_0^infty fright)$$ ?
– Peter
Jul 19 at 8:37
1
@Peter Yes. In general, if $cneq 0$ is a constant and $g(x)$ is a function, $o(cg(x)) = o(g(x))$. This is very easy to prove if you go back to the definition of little-o.
– Gabriel Romon
Jul 19 at 9:02
add a comment |Â
at your last equality, is it correct that $$int_0^infty fracflambda +oleft(frac1lambda right)=int_0^infty fracflambda +oleft(frac1lambda int_0^infty fright)$$ ?
– Peter
Jul 19 at 8:37
1
@Peter Yes. In general, if $cneq 0$ is a constant and $g(x)$ is a function, $o(cg(x)) = o(g(x))$. This is very easy to prove if you go back to the definition of little-o.
– Gabriel Romon
Jul 19 at 9:02
at your last equality, is it correct that $$int_0^infty fracflambda +oleft(frac1lambda right)=int_0^infty fracflambda +oleft(frac1lambda int_0^infty fright)$$ ?
– Peter
Jul 19 at 8:37
at your last equality, is it correct that $$int_0^infty fracflambda +oleft(frac1lambda right)=int_0^infty fracflambda +oleft(frac1lambda int_0^infty fright)$$ ?
– Peter
Jul 19 at 8:37
1
1
@Peter Yes. In general, if $cneq 0$ is a constant and $g(x)$ is a function, $o(cg(x)) = o(g(x))$. This is very easy to prove if you go back to the definition of little-o.
– Gabriel Romon
Jul 19 at 9:02
@Peter Yes. In general, if $cneq 0$ is a constant and $g(x)$ is a function, $o(cg(x)) = o(g(x))$. This is very easy to prove if you go back to the definition of little-o.
– Gabriel Romon
Jul 19 at 9:02
add a comment |Â
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The argument is obviously wrong, for example when you replace a big-O by a little-o, and much too vague, for example when you integrate a little-o on an unbounded interval. For a proof, simply use $$lambdaleqslantlambda+gleqslantlambda+C$$ for some given finite $C$ and integrate.
– Did
Jul 18 at 12:39
1
As already mentioned, integrating some little-o term on some unbounded interval (or even a nonuniform little-o term on a bounded interval) can lead to surprises. Exercise: Find some regular functions $h_n$ such that $h_n(x)to0$ when $ntoinfty$, for every $x$ in $[0,1]$ (that is, $h_n(x)=o(1)$, right?), and $int_0^1h_n(x)dx$ does not converge to $0$.
– Did
Jul 18 at 12:50
1
As I said already, either you know the reasons why this holds (that $g$ is bounded) and then you could as well use this condition from the start, or you do not, and then such strings of characters can lead you into chaos. IOW, if you want a clean proof, drop the os and Os and work with inequalities valid on (0,oo)...
– Did
Jul 18 at 15:47
1
Q1) Too many posts, I do not know. Q2) For each fixed $x$, yes. Q3) Yes there is a problem, one of uniformity, as already explained.
– Did
Jul 18 at 18:42
1
I find your claim in the final edit (the one that starts with "There is M>0 s.t...." not properly justified. The main criticism with the type of proof you seem hellbent on making is that any complete proof (i.e. which contains all necessary justifications) of that form will necessarily contain the simpler proofs you've been recommended. It will also, then, contain a lot of unnecessary parts and be significantly less clear.
– Antonio Vargas
Jul 18 at 23:10