Continous function $r(x)$ on $[0,1]$ such that...












1












$begingroup$


The problem is related to this question (https://math.stackexchange.com/posts/3073378/edit).



Is there any general way to construct a non-zero continous function $r(x)$ on $[0,1]$, independent of parameters $a$ and $b$ such that $$lim_{ntoinfty}sum_{k=0}^{n}frac{1}{n}gleft(frac kn,a,bright)rleft(frac knright)=0$$ where $g(x,a,b)$ is GIVEN continuous (hence bounded) function on $[0,1]$ with parameters $a,b,$ with $0<a<1$ and $binmathbb{R}$.



As an example, we could possibly work with $g(x,a,b)=e^{bx+a}$ though my original $g$ is something else.



Since $g$ is bounded, my intution tells me that there should exist some non-zero $f$ independent of parameters $a,b$ killing that limit.






real-analysis calculus limits analysis summation







share|cite|improve this question











$endgroup$












  • $begingroup$
    What do you mean by independent of $a,b$? Do you want a single function $r(x)$ that works for every possible values of $a,b$?
    $endgroup$
    – BigbearZzz
    Jan 14 at 17:33












  • $begingroup$
    @BigbearZzz, Exactly! A single function which works for every $a,b$.
    $endgroup$
    – ersh
    Jan 14 at 17:54






  • 1




    $begingroup$
    Unfortunately the answer is negative. The conditions can force $r(x)$ to be zero identically.
    $endgroup$
    – BigbearZzz
    Jan 14 at 17:59
















1












$begingroup$


The problem is related to this question (https://math.stackexchange.com/posts/3073378/edit).



Is there any general way to construct a non-zero continous function $r(x)$ on $[0,1]$, independent of parameters $a$ and $b$ such that $$lim_{ntoinfty}sum_{k=0}^{n}frac{1}{n}gleft(frac kn,a,bright)rleft(frac knright)=0$$ where $g(x,a,b)$ is GIVEN continuous (hence bounded) function on $[0,1]$ with parameters $a,b,$ with $0<a<1$ and $binmathbb{R}$.



As an example, we could possibly work with $g(x,a,b)=e^{bx+a}$ though my original $g$ is something else.



Since $g$ is bounded, my intution tells me that there should exist some non-zero $f$ independent of parameters $a,b$ killing that limit.






real-analysis calculus limits analysis summation







share|cite|improve this question











$endgroup$












  • $begingroup$
    What do you mean by independent of $a,b$? Do you want a single function $r(x)$ that works for every possible values of $a,b$?
    $endgroup$
    – BigbearZzz
    Jan 14 at 17:33












  • $begingroup$
    @BigbearZzz, Exactly! A single function which works for every $a,b$.
    $endgroup$
    – ersh
    Jan 14 at 17:54






  • 1




    $begingroup$
    Unfortunately the answer is negative. The conditions can force $r(x)$ to be zero identically.
    $endgroup$
    – BigbearZzz
    Jan 14 at 17:59














1












1








1





$begingroup$


The problem is related to this question (https://math.stackexchange.com/posts/3073378/edit).



Is there any general way to construct a non-zero continous function $r(x)$ on $[0,1]$, independent of parameters $a$ and $b$ such that $$lim_{ntoinfty}sum_{k=0}^{n}frac{1}{n}gleft(frac kn,a,bright)rleft(frac knright)=0$$ where $g(x,a,b)$ is GIVEN continuous (hence bounded) function on $[0,1]$ with parameters $a,b,$ with $0<a<1$ and $binmathbb{R}$.



As an example, we could possibly work with $g(x,a,b)=e^{bx+a}$ though my original $g$ is something else.



Since $g$ is bounded, my intution tells me that there should exist some non-zero $f$ independent of parameters $a,b$ killing that limit.






real-analysis calculus limits analysis summation







share|cite|improve this question











$endgroup$




The problem is related to this question (https://math.stackexchange.com/posts/3073378/edit).



Is there any general way to construct a non-zero continous function $r(x)$ on $[0,1]$, independent of parameters $a$ and $b$ such that $$lim_{ntoinfty}sum_{k=0}^{n}frac{1}{n}gleft(frac kn,a,bright)rleft(frac knright)=0$$ where $g(x,a,b)$ is GIVEN continuous (hence bounded) function on $[0,1]$ with parameters $a,b,$ with $0<a<1$ and $binmathbb{R}$.



As an example, we could possibly work with $g(x,a,b)=e^{bx+a}$ though my original $g$ is something else.



Since $g$ is bounded, my intution tells me that there should exist some non-zero $f$ independent of parameters $a,b$ killing that limit.






real-analysis calculus limits analysis summation




real-analysis calculus limits analysis summation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 14 at 17:56







ersh

















asked Jan 14 at 17:07









ershersh

438113




438113












  • $begingroup$
    What do you mean by independent of $a,b$? Do you want a single function $r(x)$ that works for every possible values of $a,b$?
    $endgroup$
    – BigbearZzz
    Jan 14 at 17:33












  • $begingroup$
    @BigbearZzz, Exactly! A single function which works for every $a,b$.
    $endgroup$
    – ersh
    Jan 14 at 17:54






  • 1




    $begingroup$
    Unfortunately the answer is negative. The conditions can force $r(x)$ to be zero identically.
    $endgroup$
    – BigbearZzz
    Jan 14 at 17:59


















  • $begingroup$
    What do you mean by independent of $a,b$? Do you want a single function $r(x)$ that works for every possible values of $a,b$?
    $endgroup$
    – BigbearZzz
    Jan 14 at 17:33












  • $begingroup$
    @BigbearZzz, Exactly! A single function which works for every $a,b$.
    $endgroup$
    – ersh
    Jan 14 at 17:54






  • 1




    $begingroup$
    Unfortunately the answer is negative. The conditions can force $r(x)$ to be zero identically.
    $endgroup$
    – BigbearZzz
    Jan 14 at 17:59
















$begingroup$
What do you mean by independent of $a,b$? Do you want a single function $r(x)$ that works for every possible values of $a,b$?
$endgroup$
– BigbearZzz
Jan 14 at 17:33






$begingroup$
What do you mean by independent of $a,b$? Do you want a single function $r(x)$ that works for every possible values of $a,b$?
$endgroup$
– BigbearZzz
Jan 14 at 17:33














$begingroup$
@BigbearZzz, Exactly! A single function which works for every $a,b$.
$endgroup$
– ersh
Jan 14 at 17:54




$begingroup$
@BigbearZzz, Exactly! A single function which works for every $a,b$.
$endgroup$
– ersh
Jan 14 at 17:54




1




1




$begingroup$
Unfortunately the answer is negative. The conditions can force $r(x)$ to be zero identically.
$endgroup$
– BigbearZzz
Jan 14 at 17:59




$begingroup$
Unfortunately the answer is negative. The conditions can force $r(x)$ to be zero identically.
$endgroup$
– BigbearZzz
Jan 14 at 17:59










1 Answer
1






active

oldest

votes


















1












$begingroup$

Since we have
$$
lim_{ntoinfty}sum_{k=0}^{n}frac{1}{n}gleft(frac kn,a,bright)rleft(frac knright)=int_0^1g(x,a,b)r(x), dx,
$$
we want the function $r(x)$ to satisfy
$$
int_0^1g(x,a,b)r(x), dx=0
$$
for all $(a,b)in(0,1)timesBbb R$. In general, it is not true that there would exists a nonzero $r(x)$ with the above property.



Indeed, consider the family of bounded continuous functions
$$
mathcal F=left{ frac1{1+b^2}+ sin(x)^{1/a} : (a,b)in(0,1)timesBbb R right},
$$
it is not hard to see that $D={1,sin(x),sin(x)^2,dots}$ is a subset of $mathcal F$. According to the Stone-Weierstrass theorem, the span of $D$ is dense in $C[0,1]$. Since
$$
int_0^1 g(x)r(x), dx=0
$$
for all $gin D$, by passing to the limit (using the density result above) we conclude that
$$
int_0^1 f(x)r(x), dx=0
$$
for all $fin C[0,1]$. It follows that the only $r(x)$ that works is $r(x)equiv 0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Wonderful counterexample!! Thank you! To be specific my $$g=frac{1}{e^{ax+e^{b-x}}+e^{ax}}$$. My different approach to the problem I am solving, suggests me that there should exist a non-zero $f$ for this choice of $g$. Can we think of such $f$ in this case?
    $endgroup$
    – ersh
    Jan 14 at 18:13








  • 1




    $begingroup$
    @ersh Perhaps you should post that as a separate question with that specific formula for $g(x,a,b)$ and link to this question. Tomorrow I might have a look at it since today I'm tired already. Glad I could help.
    $endgroup$
    – BigbearZzz
    Jan 14 at 18:14












  • $begingroup$
    Yeah, Thank you!
    $endgroup$
    – ersh
    Jan 14 at 19:19






  • 1




    $begingroup$
    @ersh So if you're happy with this answer please consider accepting it so the question wouldn't be left in the unanswered queue.
    $endgroup$
    – BigbearZzz
    Jan 14 at 19:23










  • $begingroup$
    Sorry. I forgot. I am quite well satisfied with your example.
    $endgroup$
    – ersh
    Jan 14 at 20:28











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1 Answer
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1 Answer
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1












$begingroup$

Since we have
$$
lim_{ntoinfty}sum_{k=0}^{n}frac{1}{n}gleft(frac kn,a,bright)rleft(frac knright)=int_0^1g(x,a,b)r(x), dx,
$$
we want the function $r(x)$ to satisfy
$$
int_0^1g(x,a,b)r(x), dx=0
$$
for all $(a,b)in(0,1)timesBbb R$. In general, it is not true that there would exists a nonzero $r(x)$ with the above property.



Indeed, consider the family of bounded continuous functions
$$
mathcal F=left{ frac1{1+b^2}+ sin(x)^{1/a} : (a,b)in(0,1)timesBbb R right},
$$
it is not hard to see that $D={1,sin(x),sin(x)^2,dots}$ is a subset of $mathcal F$. According to the Stone-Weierstrass theorem, the span of $D$ is dense in $C[0,1]$. Since
$$
int_0^1 g(x)r(x), dx=0
$$
for all $gin D$, by passing to the limit (using the density result above) we conclude that
$$
int_0^1 f(x)r(x), dx=0
$$
for all $fin C[0,1]$. It follows that the only $r(x)$ that works is $r(x)equiv 0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Wonderful counterexample!! Thank you! To be specific my $$g=frac{1}{e^{ax+e^{b-x}}+e^{ax}}$$. My different approach to the problem I am solving, suggests me that there should exist a non-zero $f$ for this choice of $g$. Can we think of such $f$ in this case?
    $endgroup$
    – ersh
    Jan 14 at 18:13








  • 1




    $begingroup$
    @ersh Perhaps you should post that as a separate question with that specific formula for $g(x,a,b)$ and link to this question. Tomorrow I might have a look at it since today I'm tired already. Glad I could help.
    $endgroup$
    – BigbearZzz
    Jan 14 at 18:14












  • $begingroup$
    Yeah, Thank you!
    $endgroup$
    – ersh
    Jan 14 at 19:19






  • 1




    $begingroup$
    @ersh So if you're happy with this answer please consider accepting it so the question wouldn't be left in the unanswered queue.
    $endgroup$
    – BigbearZzz
    Jan 14 at 19:23










  • $begingroup$
    Sorry. I forgot. I am quite well satisfied with your example.
    $endgroup$
    – ersh
    Jan 14 at 20:28
















1












$begingroup$

Since we have
$$
lim_{ntoinfty}sum_{k=0}^{n}frac{1}{n}gleft(frac kn,a,bright)rleft(frac knright)=int_0^1g(x,a,b)r(x), dx,
$$
we want the function $r(x)$ to satisfy
$$
int_0^1g(x,a,b)r(x), dx=0
$$
for all $(a,b)in(0,1)timesBbb R$. In general, it is not true that there would exists a nonzero $r(x)$ with the above property.



Indeed, consider the family of bounded continuous functions
$$
mathcal F=left{ frac1{1+b^2}+ sin(x)^{1/a} : (a,b)in(0,1)timesBbb R right},
$$
it is not hard to see that $D={1,sin(x),sin(x)^2,dots}$ is a subset of $mathcal F$. According to the Stone-Weierstrass theorem, the span of $D$ is dense in $C[0,1]$. Since
$$
int_0^1 g(x)r(x), dx=0
$$
for all $gin D$, by passing to the limit (using the density result above) we conclude that
$$
int_0^1 f(x)r(x), dx=0
$$
for all $fin C[0,1]$. It follows that the only $r(x)$ that works is $r(x)equiv 0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Wonderful counterexample!! Thank you! To be specific my $$g=frac{1}{e^{ax+e^{b-x}}+e^{ax}}$$. My different approach to the problem I am solving, suggests me that there should exist a non-zero $f$ for this choice of $g$. Can we think of such $f$ in this case?
    $endgroup$
    – ersh
    Jan 14 at 18:13








  • 1




    $begingroup$
    @ersh Perhaps you should post that as a separate question with that specific formula for $g(x,a,b)$ and link to this question. Tomorrow I might have a look at it since today I'm tired already. Glad I could help.
    $endgroup$
    – BigbearZzz
    Jan 14 at 18:14












  • $begingroup$
    Yeah, Thank you!
    $endgroup$
    – ersh
    Jan 14 at 19:19






  • 1




    $begingroup$
    @ersh So if you're happy with this answer please consider accepting it so the question wouldn't be left in the unanswered queue.
    $endgroup$
    – BigbearZzz
    Jan 14 at 19:23










  • $begingroup$
    Sorry. I forgot. I am quite well satisfied with your example.
    $endgroup$
    – ersh
    Jan 14 at 20:28














1












1








1





$begingroup$

Since we have
$$
lim_{ntoinfty}sum_{k=0}^{n}frac{1}{n}gleft(frac kn,a,bright)rleft(frac knright)=int_0^1g(x,a,b)r(x), dx,
$$
we want the function $r(x)$ to satisfy
$$
int_0^1g(x,a,b)r(x), dx=0
$$
for all $(a,b)in(0,1)timesBbb R$. In general, it is not true that there would exists a nonzero $r(x)$ with the above property.



Indeed, consider the family of bounded continuous functions
$$
mathcal F=left{ frac1{1+b^2}+ sin(x)^{1/a} : (a,b)in(0,1)timesBbb R right},
$$
it is not hard to see that $D={1,sin(x),sin(x)^2,dots}$ is a subset of $mathcal F$. According to the Stone-Weierstrass theorem, the span of $D$ is dense in $C[0,1]$. Since
$$
int_0^1 g(x)r(x), dx=0
$$
for all $gin D$, by passing to the limit (using the density result above) we conclude that
$$
int_0^1 f(x)r(x), dx=0
$$
for all $fin C[0,1]$. It follows that the only $r(x)$ that works is $r(x)equiv 0$.






share|cite|improve this answer









$endgroup$



Since we have
$$
lim_{ntoinfty}sum_{k=0}^{n}frac{1}{n}gleft(frac kn,a,bright)rleft(frac knright)=int_0^1g(x,a,b)r(x), dx,
$$
we want the function $r(x)$ to satisfy
$$
int_0^1g(x,a,b)r(x), dx=0
$$
for all $(a,b)in(0,1)timesBbb R$. In general, it is not true that there would exists a nonzero $r(x)$ with the above property.



Indeed, consider the family of bounded continuous functions
$$
mathcal F=left{ frac1{1+b^2}+ sin(x)^{1/a} : (a,b)in(0,1)timesBbb R right},
$$
it is not hard to see that $D={1,sin(x),sin(x)^2,dots}$ is a subset of $mathcal F$. According to the Stone-Weierstrass theorem, the span of $D$ is dense in $C[0,1]$. Since
$$
int_0^1 g(x)r(x), dx=0
$$
for all $gin D$, by passing to the limit (using the density result above) we conclude that
$$
int_0^1 f(x)r(x), dx=0
$$
for all $fin C[0,1]$. It follows that the only $r(x)$ that works is $r(x)equiv 0$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 14 at 17:56









BigbearZzzBigbearZzz

8,94521652




8,94521652












  • $begingroup$
    Wonderful counterexample!! Thank you! To be specific my $$g=frac{1}{e^{ax+e^{b-x}}+e^{ax}}$$. My different approach to the problem I am solving, suggests me that there should exist a non-zero $f$ for this choice of $g$. Can we think of such $f$ in this case?
    $endgroup$
    – ersh
    Jan 14 at 18:13








  • 1




    $begingroup$
    @ersh Perhaps you should post that as a separate question with that specific formula for $g(x,a,b)$ and link to this question. Tomorrow I might have a look at it since today I'm tired already. Glad I could help.
    $endgroup$
    – BigbearZzz
    Jan 14 at 18:14












  • $begingroup$
    Yeah, Thank you!
    $endgroup$
    – ersh
    Jan 14 at 19:19






  • 1




    $begingroup$
    @ersh So if you're happy with this answer please consider accepting it so the question wouldn't be left in the unanswered queue.
    $endgroup$
    – BigbearZzz
    Jan 14 at 19:23










  • $begingroup$
    Sorry. I forgot. I am quite well satisfied with your example.
    $endgroup$
    – ersh
    Jan 14 at 20:28


















  • $begingroup$
    Wonderful counterexample!! Thank you! To be specific my $$g=frac{1}{e^{ax+e^{b-x}}+e^{ax}}$$. My different approach to the problem I am solving, suggests me that there should exist a non-zero $f$ for this choice of $g$. Can we think of such $f$ in this case?
    $endgroup$
    – ersh
    Jan 14 at 18:13








  • 1




    $begingroup$
    @ersh Perhaps you should post that as a separate question with that specific formula for $g(x,a,b)$ and link to this question. Tomorrow I might have a look at it since today I'm tired already. Glad I could help.
    $endgroup$
    – BigbearZzz
    Jan 14 at 18:14












  • $begingroup$
    Yeah, Thank you!
    $endgroup$
    – ersh
    Jan 14 at 19:19






  • 1




    $begingroup$
    @ersh So if you're happy with this answer please consider accepting it so the question wouldn't be left in the unanswered queue.
    $endgroup$
    – BigbearZzz
    Jan 14 at 19:23










  • $begingroup$
    Sorry. I forgot. I am quite well satisfied with your example.
    $endgroup$
    – ersh
    Jan 14 at 20:28
















$begingroup$
Wonderful counterexample!! Thank you! To be specific my $$g=frac{1}{e^{ax+e^{b-x}}+e^{ax}}$$. My different approach to the problem I am solving, suggests me that there should exist a non-zero $f$ for this choice of $g$. Can we think of such $f$ in this case?
$endgroup$
– ersh
Jan 14 at 18:13






$begingroup$
Wonderful counterexample!! Thank you! To be specific my $$g=frac{1}{e^{ax+e^{b-x}}+e^{ax}}$$. My different approach to the problem I am solving, suggests me that there should exist a non-zero $f$ for this choice of $g$. Can we think of such $f$ in this case?
$endgroup$
– ersh
Jan 14 at 18:13






1




1




$begingroup$
@ersh Perhaps you should post that as a separate question with that specific formula for $g(x,a,b)$ and link to this question. Tomorrow I might have a look at it since today I'm tired already. Glad I could help.
$endgroup$
– BigbearZzz
Jan 14 at 18:14






$begingroup$
@ersh Perhaps you should post that as a separate question with that specific formula for $g(x,a,b)$ and link to this question. Tomorrow I might have a look at it since today I'm tired already. Glad I could help.
$endgroup$
– BigbearZzz
Jan 14 at 18:14














$begingroup$
Yeah, Thank you!
$endgroup$
– ersh
Jan 14 at 19:19




$begingroup$
Yeah, Thank you!
$endgroup$
– ersh
Jan 14 at 19:19




1




1




$begingroup$
@ersh So if you're happy with this answer please consider accepting it so the question wouldn't be left in the unanswered queue.
$endgroup$
– BigbearZzz
Jan 14 at 19:23




$begingroup$
@ersh So if you're happy with this answer please consider accepting it so the question wouldn't be left in the unanswered queue.
$endgroup$
– BigbearZzz
Jan 14 at 19:23












$begingroup$
Sorry. I forgot. I am quite well satisfied with your example.
$endgroup$
– ersh
Jan 14 at 20:28




$begingroup$
Sorry. I forgot. I am quite well satisfied with your example.
$endgroup$
– ersh
Jan 14 at 20:28


















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Human spaceflight From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Space traveler" redirects here. For other uses, see Space traveler (disambiguation). @media all and (max-width:720px){.mw-parser-output .tmulti>.thumbinner{width:100%!important;max-width:none!important}.mw-parser-output .tmulti .tsingle{float:none!important;max-width:none!important;width:100%!important;text-align:center}} Apollo 11 crewmember Buzz Aldrin on the Moon, 1969 International Space Station crewmember Tracy Caldwell Dyson views the Earth, 2010 Space Shuttle Discovery heads into space with a crew aboard, STS-121 in 2006 Inside a space suit on the Canadarm, 1993 Human spaceflight (also referred to as crewed spaceflight or manned spaceflight ) is space travel with a crew or passengers aboard the spacecraft. Spacecraft carrying people may be operated directly, by human crew, or it may be e...
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Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

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0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB] ...
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張江高科駅

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張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
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